This article explains the current divider rule with diagrams, covering the following topics.

• What the current divider rule is
• Current divider formulas, their derivation, and worked examples
• The difference between the current divider rule and the voltage divider rule

What Is the Current Divider Rule?]

The current divider rule is used to calculate the current through each resistor in a parallel circuit.

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\) connected in parallel.

Let \(I{\mathrm{[A]}}\) be the current entering the parallel circuit, and let \(R{\mathrm{[{\Omega}]}}\) be its equivalent resistance. The current \(I_1{\mathrm{[A]}}\) through resistor \(R_1{\mathrm{[{\Omega}]}}\) and the current \(I_2{\mathrm{[A]}}\) through resistor \(R_2{\mathrm{[{\Omega}]}}\) are given by the following equations.

\begin{eqnarray}
I_1&=&\frac{R}{R_1}×I=\frac{R_2}{R_1+R_2}×I{\mathrm{[A]}}\tag{1}\\
\\
I_2&=&\frac{R}{R_2}×I=\frac{R_1}{R_1+R_2}×I{\mathrm{[A]}}\tag{2}
\end{eqnarray}

Equations (1) and (2) are the current divider formulas. In the forms \(I_1=(R/R_1)I\) and \(I_2=(R/R_2)I\), the denominator is the resistance of the resistor whose current you want to calculate.

Resistors \(R_1\) and \(R_2\) divide the current \(I\) into \(I_1\) and \(I_2\). This is called current division.

The general relationship used to derive equations (1) and (2) can be expressed in words as follows.

\begin{eqnarray}
\mbox{Target current}=\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\tag{3}
\end{eqnarray}

Equivalent resistance is the resistance of a single resistor that can replace multiple resistors. In equations (1) and (2), \(R_1\) and \(R_2\) are the individual resistance values, and \(R\) is the equivalent resistance of the entire parallel circuit.

Derivation of the Current Divider Formulas

Let us derive equations (1) and (2) from equation (3).

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\) is given by the following equation.

\begin{eqnarray}
R=\frac{1}{\displaystyle\frac{1}{R_1}+\displaystyle\frac{1}{R_2}}=\frac{R_1×R_2}{R_1+R_2}{\mathrm{[{\Omega}]}}\tag{4}
\end{eqnarray}

Substituting equation (4) into equation (3) gives the following expression for the current \(I_1{\mathrm{[A]}}\) through resistor \(R_1{\mathrm{[{\Omega}]}}\), yielding equation (1).

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_1&=&\frac{\displaystyle\frac{R_1×R_2}{R_1+R_2}}{R_1}×I\\
\\
&=&\frac{R_2}{R_1+R_2}×I{\mathrm{[A]}}\tag{5}
\end{eqnarray}

Similarly, substituting equation (4) into equation (3) gives the following expression for the current \(I_2{\mathrm{[A]}}\) through resistor \(R_2{\mathrm{[{\Omega}]}}\), yielding equation (2).

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_2&=&\frac{\displaystyle\frac{R_1×R_2}{R_1+R_2}}{R_2}×I\\
\\
&=&\frac{R_1}{R_1+R_2}×I{\mathrm{[A]}}\tag{6}
\end{eqnarray}

Thus, when resistors are connected in parallel, the current through each resistor can be calculated using the following equation.

\begin{eqnarray}
\mbox{Target current}=\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\tag{7}
\end{eqnarray}

With this equation, you can calculate the current through each resistor even when three or more resistors are connected in parallel, by first finding the equivalent resistance.

Current Divider Example (2 Resistors in Parallel)

Example

In the circuit shown above, the current \(I\) entering the parallel circuit is \(10{\mathrm{[A]}}\), resistor \(R_1\) is \(4{\mathrm{[{\Omega}]}}\), and resistor \(R_2\) is \(6{\mathrm{[{\Omega}]}}\). Find the following:

• What is the equivalent resistance \(R\) of resistors \(R_1\) and \(R_2\), in \({\mathrm{{\Omega}}}\)?
• What is the current \(I_1\) through resistor \(R_1\), in \({\mathrm{A}}\)?
• What is the current \(I_2\) through resistor \(R_2\), in \({\mathrm{A}}\)?

Solution

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1\) and \(R_2\) is given by the following equation.

\begin{eqnarray}
R=\frac{1}{\displaystyle\frac{1}{R_1}+\displaystyle\frac{1}{R_2}}=\frac{R_1×R_2}{R_1+R_2}=\frac{4×6}{4+6}=2.4{\mathrm{[{\Omega}]}}\tag{8}
\end{eqnarray}

Therefore, the current \(I_1{\mathrm{[A]}}\) through resistor \(R_1\) is calculated as follows.

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_1&=&\frac{2.4}{4}×10\\
\\
&=&6{\mathrm{[A]}}\tag{9}
\end{eqnarray}

Similarly, the current \(I_2{\mathrm{[A]}}\) through resistor \(R_2\) is calculated as follows.

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_2&=&\frac{2.4}{6}×10\\
\\
&=&4{\mathrm{[A]}}\tag{10}
\end{eqnarray}

The answers are an equivalent resistance of \(2.4{\mathrm{[\Omega]}}\), \(I_1\) of \(6{\mathrm{[A]}}\), and \(I_2\) of \(4{\mathrm{[A]}}\). Resistor \(R_1\), which has the lower resistance, carries the larger current. We can also confirm that the two currents add up to the current of \(10{\mathrm{[A]}}\) entering the parallel circuit.

Current Divider Rule: 3 Resistors in Parallel

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\) connected in parallel.

Let \(I{\mathrm{[A]}}\) be the current entering the parallel circuit, and let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of the 3 resistors. The currents \(I_1{\mathrm{[A]}}\), \(I_2{\mathrm{[A]}}\), and \(I_3{\mathrm{[A]}}\) through resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\), respectively, are given by the following equations.

\begin{eqnarray}
I_1&=&\frac{R}{R_1}×I=\frac{R_2×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}×I{\mathrm{[A]}}\tag{11}\\
\\
I_2&=&\frac{R}{R_2}×I=\frac{R_1×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}×I{\mathrm{[A]}}\tag{12}\\
\\
I_3&=&\frac{R}{R_3}×I=\frac{R_1×R_2}{R_2×R_3+R_1×R_3+R_1×R_2}×I{\mathrm{[A]}}\tag{13}
\end{eqnarray}

Equations (11) through (13) can also be derived by substituting the equivalent resistance of the entire parallel circuit into the following equation.

\begin{eqnarray}
\mbox{Target current}=\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\tag{14}
\end{eqnarray}

Let us derive equation (11) from equation (14).

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\) is given by the following equation.

\begin{eqnarray}
R&=&\frac{1}{\displaystyle\frac{1}{R_1}+\displaystyle\frac{1}{R_2}+\displaystyle\frac{1}{R_3}}\\
\\
&=&\frac{R_1×R_2×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}\tag{15}
\end{eqnarray}

Substituting equation (15) into equation (14) gives the following expression for the current \(I_1{\mathrm{[A]}}\) through resistor \(R_1{\mathrm{[{\Omega}]}}\), yielding equation (11).

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_1&=&\frac{\displaystyle\frac{R_1×R_2×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}}{R_1}×I\\
\\
&=&\frac{R_2×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}×I{\mathrm{[A]}}\tag{16}
\end{eqnarray}

Equations (12) and (13) can be derived in the same way.

Current Divider Example (3 Resistors in Parallel)

Example

In the circuit shown above, the current \(I\) entering the parallel circuit is \(10{\mathrm{[A]}}\), resistor \(R_1\) is \(10{\mathrm{[{\Omega}]}}\), resistor \(R_2\) is \(50{\mathrm{[{\Omega}]}}\), and resistor \(R_3\) is \(200{\mathrm{[{\Omega}]}}\). Find the following:

• What is the equivalent resistance \(R\) of resistors \(R_1\), \(R_2\), and \(R_3\), in \({\mathrm{{\Omega}}}\)?
• What is the current \(I_1\) through resistor \(R_1\), in \({\mathrm{A}}\)?
• What is the current \(I_2\) through resistor \(R_2\), in \({\mathrm{A}}\)?
• What is the current \(I_3\) through resistor \(R_3\), in \({\mathrm{A}}\)?

Solution

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1\), \(R_2\), and \(R_3\) is given by the following equation.

\begin{eqnarray}
R&=&\frac{R_1×R_2×R_3}{R_2×R_3+R_1×R_3+R_1×R_2}\\
\\
&=&\frac{10×50×200}{50×200+10×200+10×50}\\
\\
&=&8{\mathrm{[{\Omega}]}}\tag{17}
\end{eqnarray}

Therefore, the current \(I_1{\mathrm{[A]}}\) through resistor \(R_1\) is calculated as follows.

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_1&=&\frac{8}{10}×10\\
\\
&=&8{\mathrm{[A]}}\tag{18}
\end{eqnarray}

Similarly, the current \(I_2{\mathrm{[A]}}\) through resistor \(R_2\) is calculated as follows.

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_2&=&\frac{8}{50}×10\\
\\
&=&1.6{\mathrm{[A]}}\tag{19}
\end{eqnarray}

Similarly, the current \(I_3{\mathrm{[A]}}\) through resistor \(R_3\) is calculated as follows.

\begin{eqnarray}
\mbox{Target current}&=&\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\\
\\
{\Leftrightarrow}I_3&=&\frac{8}{200}×10\\
\\
&=&0.4{\mathrm{[A]}}\tag{20}
\end{eqnarray}

The answers are an equivalent resistance of \(8{\mathrm{[\Omega]}}\) and currents of \(8{\mathrm{[A]}}\), \(1.6{\mathrm{[A]}}\), and \(0.4{\mathrm{[A]}}\), respectively. With three resistors, we can use the same method by first finding the equivalent resistance of the entire parallel circuit. The three currents add up to the current of \(10{\mathrm{[A]}}\) entering the parallel circuit.

Current Divider Rule: N Resistors in Parallel

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), …, \(R_N{\mathrm{[{\Omega}]}}\) connected in parallel.

Let \(I{\mathrm{[A]}}\) be the current entering the parallel circuit, and let \(R{\mathrm{[{\Omega}]}}\) be its equivalent resistance. The current through each resistor can be calculated using the following equation.

\begin{eqnarray}
\mbox{Target current}=\frac{\mbox{Equivalent resistance}}{\mbox{Resistance of the target resistor}}×\mbox{Current entering the parallel circuit}\tag{21}
\end{eqnarray}

Therefore, the currents \(I_1{\mathrm{[A]}}\), \(I_2{\mathrm{[A]}}\), …, \(I_N{\mathrm{[A]}}\) through resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), …, \(R_N{\mathrm{[{\Omega}]}}\), respectively, are given by the following equations.

\begin{eqnarray}
I_1&=&\frac{R}{R_1}×I{\mathrm{[A]}}\tag{22}\\
\\
I_2&=&\frac{R}{R_2}×I{\mathrm{[A]}}\tag{23}\\
\\
I_N&=&\frac{R}{R_N}×I{\mathrm{[A]}}\tag{24}
\end{eqnarray}

Here, \(N\) is the number of resistors connected in parallel. Regardless of the number of resistors, the numerator contains the equivalent resistance of the entire parallel circuit, and the denominator contains the resistance of the resistor whose current you want to calculate. The currents through all the resistors add up to the current \(I\) entering the parallel circuit.

Difference Between the Current Divider Rule and the Voltage Divider Rule

The formulas for the current divider rule and the voltage divider rule are very similar.

The current divider rule calculates currents in a parallel circuit, while the voltage divider rule calculates voltages in a series circuit. Although the formulas are similar, they apply to different circuit connections and calculate different quantities.

The symbol \(R\) in these two equations does not represent the same equivalent resistance. In equation (25), \(R\) is the equivalent resistance of the parallel circuit; in equation (26), it is the equivalent resistance of the series circuit.

Equations (25) and (26) look similar. Note that resistance \(R_N{\mathrm{[{\Omega}]}}\) is in the denominator in the current divider rule, whereas resistance \(R_N{\mathrm{[{\Omega}]}}\) is in the numerator in the voltage divider rule.

Summary

This article explained the following key points about the current divider rule.

• The current divider rule calculates the current through each resistor in a parallel circuit.
• The current through each resistor is found by multiplying the current entering the parallel circuit by the equivalent resistance of the entire parallel circuit divided by the resistance of that resistor.
• Because the same voltage appears across all resistors in a parallel circuit, a resistor with a lower resistance carries a larger current.
• Current division is used to calculate currents in a parallel circuit, while voltage division is used to calculate voltages in a series circuit.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

This article explains the voltage divider rule with diagrams, covering the following topics.

• What the voltage divider rule is
• Voltage divider formulas, their derivation, and worked examples
• The difference between the voltage divider rule and the current divider formula

What Is the Voltage Divider Rule?

The voltage divider rule is a rule for calculating the voltage across each resistor when resistors are connected in series.

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\) connected in series, with a supply voltage \(V{\mathrm{[V]}}\) applied across the circuit.

Let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\). The voltage \(V_1{\mathrm{[V]}}\) across resistor \(R_1{\mathrm{[{\Omega}]}}\) and the voltage \(V_2{\mathrm{[V]}}\) across resistor \(R_2{\mathrm{[{\Omega}]}}\) are given by the following equations.

\begin{eqnarray}
V_1&=&\frac{R_1}{R}×V=\frac{R_1}{R_1+R_2}×V{\mathrm{[V]}}\tag{1}\\
\\
V_2&=&\frac{R_2}{R}×V=\frac{R_2}{R_1+R_2}×V{\mathrm{[V]}}\tag{2}
\end{eqnarray}

Equations (1) and (2) are the voltage divider formulas. The numerator contains the resistance of the resistor whose voltage you want to calculate.

Resistors \(R_1\) and \(R_2\) divide the voltage \(V\) into \(V_1\) and \(V_2\). This is called voltage division.

In words, equations (1) and (2) can be expressed as follows.

\begin{eqnarray}
\mbox{Target voltage}=\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\tag{3}
\end{eqnarray}

Equivalent resistance is the resistance of a single resistor that can replace multiple resistors. In equations (1) and (2), \(R_1\) and \(R_2\) are the individual resistance values, and \(R\) is the equivalent resistance of the entire series circuit.

Derivation of the Voltage Divider Formulas

Let us derive equations (1) and (2).

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\) connected in series, with a supply voltage \(V{\mathrm{[V]}}\) applied across the circuit.

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1{\mathrm{[{\Omega}]}}\) and \(R_2{\mathrm{[{\Omega}]}}\) in this circuit is given by the following equation.

\begin{eqnarray}
R=R_1+R_2{\mathrm{[{\Omega}]}}\tag{4}
\end{eqnarray}

Therefore, using Ohm’s law, the current \(I{\mathrm{[A]}}\) flowing through this circuit is given by the following equation.

\begin{eqnarray}
I=\frac{V}{R}=\frac{V}{R_1+R_2}{\mathrm{[A]}}\tag{5}
\end{eqnarray}

Also, using Ohm’s law, the voltage \(V_1{\mathrm{[V]}}\) across resistor \(R_1{\mathrm{[{\Omega}]}}\) and the voltage \(V_2{\mathrm{[V]}}\) across resistor \(R_2{\mathrm{[{\Omega}]}}\) are given by the following equations.

\begin{eqnarray}
V_1&=&R_1×I{\mathrm{[V]}}\tag{6}\\
\\
V_2&=&R_2×I{\mathrm{[V]}}\tag{7}
\end{eqnarray}

Substituting equation (5) into equations (6) and (7) gives the following expressions, which yield equations (1) and (2).

\begin{eqnarray}
V_1&=&R_1×\frac{V}{R_1+R_2}\\
\\
&=&\frac{R_1}{R_1+R_2}×V{\mathrm{[V]}}\tag{8}\\
\\
V_2&=&R_2×\frac{V}{R_1+R_2}\\
\\
&=&\frac{R_2}{R_1+R_2}×V{\mathrm{[V]}}\tag{9}
\end{eqnarray}

Thus, we can find the voltage across each resistor by first calculating the current through the entire circuit and then multiplying that current by each resistance value.

Voltage Divider Example (Two Resistors in Series)

Example

In the circuit shown above, the supply voltage \(V\) is \(10{\mathrm{[V]}}\), resistor \(R_1\) is \(2{\mathrm{[{\Omega}]}}\), and resistor \(R_2\) is \(3{\mathrm{[{\Omega}]}}\). Find the following:

• What is the equivalent resistance \(R\) of resistors \(R_1\) and \(R_2\), in \({\mathrm{{\Omega}}}\)?
• What is the voltage \(V_1\) across resistor \(R_1\), in \({\mathrm{V}}\)?
• What is the voltage \(V_2\) across resistor \(R_2\), in \({\mathrm{V}}\)?

Solution

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1\) and \(R_2\) is given by the following equation.

\begin{eqnarray}
R=R_1+R_2=2+3=5{\mathrm{[{\Omega}]}}\tag{10}
\end{eqnarray}

Therefore, the voltage \(V_1{\mathrm{[V]}}\) across resistor \(R_1\) is calculated as follows.

\begin{eqnarray}
\mbox{Target voltage}&=&\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\\
\\
{\Leftrightarrow}V_1&=&\frac{R_1}{R}×V\\
\\
&=&\frac{2}{5}×10\\
\\
&=&4{\mathrm{[V]}}\tag{11}
\end{eqnarray}

Similarly, the voltage \(V_2{\mathrm{[V]}}\) across resistor \(R_2\) is calculated as follows.

\begin{eqnarray}
\mbox{Target voltage}&=&\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\\
\\
{\Leftrightarrow}V_2&=&\frac{R_2}{R}×V\\
\\
&=&\frac{3}{5}×10\\
\\
&=&6{\mathrm{[V]}}\tag{12}
\end{eqnarray}

The answers are an equivalent resistance of \(5{\mathrm{[\Omega]}}\), \(V_1\) of \(4{\mathrm{[V]}}\), and \(V_2\) of \(6{\mathrm{[V]}}\). Resistor \(R_2\), which has the higher resistance, has the larger voltage across it. We can also confirm that the two voltages add up to the supply voltage of \(10{\mathrm{[V]}}\).

Voltage Divider Rule: Three Resistors in Series

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\) connected in series, with a supply voltage \(V{\mathrm{[V]}}\) applied across the circuit.

Let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\). The voltages \(V_1{\mathrm{[V]}}\), \(V_2{\mathrm{[V]}}\), and \(V_3{\mathrm{[V]}}\) across resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), and \(R_3{\mathrm{[{\Omega}]}}\), respectively, are given by the following equations.

\begin{eqnarray}
V_1&=&\frac{R_1}{R}×V=\frac{R_1}{R_1+R_2+R_3}×V{\mathrm{[V]}}\tag{13}\\
\\
V_2&=&\frac{R_2}{R}×V=\frac{R_2}{R_1+R_2+R_3}×V{\mathrm{[V]}}\tag{14}\\
\\
V_3&=&\frac{R_3}{R}×V=\frac{R_3}{R_1+R_2+R_3}×V{\mathrm{[V]}}\tag{15}
\end{eqnarray}

Voltage Divider Example (Three Resistors in Series)

Example

In the circuit shown above, the supply voltage \(V\) is \(10{\mathrm{[V]}}\), resistor \(R_1\) is \(2{\mathrm{[{\Omega}]}}\), resistor \(R_2\) is \(3{\mathrm{[{\Omega}]}}\), and resistor \(R_3\) is \(5{\mathrm{[{\Omega}]}}\). Find the following:

• What is the equivalent resistance \(R\) of resistors \(R_1\), \(R_2\), and \(R_3\), in \({\mathrm{{\Omega}}}\)?
• What is the voltage \(V_1\) across resistor \(R_1\), in \({\mathrm{V}}\)?
• What is the voltage \(V_2\) across resistor \(R_2\), in \({\mathrm{V}}\)?
• What is the voltage \(V_3\) across resistor \(R_3\), in \({\mathrm{V}}\)?

Solution

The equivalent resistance \(R{\mathrm{[{\Omega}]}}\) of resistors \(R_1\), \(R_2\), and \(R_3\) is given by the following equation.

\begin{eqnarray}
R=R_1+R_2+R_3=2+3+5=10{\mathrm{[{\Omega}]}}\tag{16}
\end{eqnarray}

Therefore, the voltage \(V_1{\mathrm{[V]}}\) across resistor \(R_1\) is calculated as follows.

\begin{eqnarray}
\mbox{Target voltage}&=&\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\\
\\
{\Leftrightarrow}V_1&=&\frac{R_1}{R}×V\\
\\
&=&\frac{2}{10}×10\\
\\
&=&2{\mathrm{[V]}}\tag{17}
\end{eqnarray}

Similarly, the voltage \(V_2{\mathrm{[V]}}\) across resistor \(R_2\) is calculated as follows.

\begin{eqnarray}
\mbox{Target voltage}&=&\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\\
\\
{\Leftrightarrow}V_2&=&\frac{R_2}{R}×V\\
\\
&=&\frac{3}{10}×10\\
\\
&=&3{\mathrm{[V]}}\tag{18}
\end{eqnarray}

Similarly, the voltage \(V_3{\mathrm{[V]}}\) across resistor \(R_3\) is calculated as follows.

\begin{eqnarray}
\mbox{Target voltage}&=&\frac{\mbox{Resistance of the target resistor}}{\mbox{Equivalent resistance}}×\mbox{Voltage across the series circuit}\\
\\
{\Leftrightarrow}V_3&=&\frac{R_3}{R}×V\\
\\
&=&\frac{5}{10}×10\\
\\
&=&5{\mathrm{[V]}}\tag{19}
\end{eqnarray}

The answers are an equivalent resistance of \(10{\mathrm{[\Omega]}}\) and voltages of \(2{\mathrm{[V]}}\), \(3{\mathrm{[V]}}\), and \(5{\mathrm{[V]}}\), respectively. With three resistors, we can use the same method by putting the equivalent resistance of the entire circuit in the denominator. The three voltages add up to the supply voltage of \(10{\mathrm{[V]}}\).

Voltage Divider Rule: N Resistors in Series

The diagram above shows resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), …, \(R_N{\mathrm{[{\Omega}]}}\) connected in series, with a supply voltage \(V{\mathrm{[V]}}\) applied across the circuit.

Let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of resistors \(R_1{\mathrm{[{\Omega}]}}\) through \(R_N{\mathrm{[{\Omega}]}}\). The voltages \(V_1{\mathrm{[V]}}\), \(V_2{\mathrm{[V]}}\), …, \(V_N{\mathrm{[V]}}\) across resistors \(R_1{\mathrm{[{\Omega}]}}\), \(R_2{\mathrm{[{\Omega}]}}\), …, \(R_N{\mathrm{[{\Omega}]}}\), respectively, are given by the following equations.

\begin{eqnarray}
V_1&=&\frac{R_1}{R}×V=\frac{R_1}{R_1+R_2+{\;}{\cdots}{\;}+R_N}×V{\mathrm{[V]}}\tag{20}\\
\\
V_2&=&\frac{R_2}{R}×V=\frac{R_2}{R_1+R_2+{\;}{\cdots}{\;}+R_N}×V{\mathrm{[V]}}\tag{21}\\
\\
V_N&=&\frac{R_N}{R}×V=\frac{R_N}{R_1+R_2+{\;}{\cdots}{\;}+R_N}×V{\mathrm{[V]}}\tag{22}
\end{eqnarray}

Here, \(N\) is the number of resistors connected in series. Regardless of the number of resistors, the denominator contains the sum of all the resistance values, and the numerator contains the resistance of the resistor whose voltage you want to calculate. The voltages across all the resistors add up to the voltage \(V\) across the entire series circuit.

Difference Between the Voltage Divider Rule and the Current Divider Rule

The formulas for the voltage divider rule and the current divider rule are very similar.

The voltage divider rule calculates voltages in a series circuit, while the current divider rule calculates currents in a parallel circuit. Although the formulas are similar, they apply to different circuit connections and calculate different quantities.

The symbol \(R\) in these two equations does not represent the same equivalent resistance. In equation (23), \(R\) is the equivalent resistance of the series circuit; in equation (24), it is the equivalent resistance of the parallel circuit.

Equations (23) and (24) look similar. Note that resistance \(R_N{\mathrm{[{\Omega}]}}\) is in the numerator in the voltage divider rule, whereas resistance \(R_N{\mathrm{[{\Omega}]}}\) is in the denominator in the current divider rule.

Summary

This article explained the following key points about the voltage divider rule.

• The voltage divider rule calculates the voltage across each resistor in a series circuit.
• The voltage across each resistor is found by multiplying the supply voltage by the resistance of that resistor divided by the equivalent resistance of the entire series circuit.
• Because the same current flows through all resistors in a series circuit, a resistor with a higher resistance has a larger voltage across it.
• Voltage division is used to calculate voltages in a series circuit, while current division is used to calculate currents in a parallel circuit.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

This article explains the following aspects of battery internal resistance using diagrams:

• The meaning of battery internal resistance \(r\) and its equivalent circuit
• The difference between electromotive force and terminal voltage, and the equation relating them
• How to read the terminal voltage–current graph
• Current and terminal voltage under open-circuit and short-circuit conditions

What Is Battery Internal Resistance?

A battery has a very small resistance \(r\) inside it. This resistance \(r\) is called internal resistance.

As shown in the diagram above, a battery can therefore be represented by an ideal source with an electromotive force \(E\), connected in series with an internal resistance \(r\).

The electromotive force \(E\) is the voltage produced by the battery’s chemical reactions. Its unit is \({\mathrm{[V]}}\) (volts). The terminal voltage \(V\), on the other hand, is the voltage between the battery’s positive and negative terminals. This is the voltage applied to a device or resistor connected to the battery. Like electromotive force, terminal voltage is measured in \({\mathrm{[V]}}\).

Relationship Between Battery Electromotive Force \(E\) and Terminal Voltage \(V\)

The left side of the diagram above shows a battery with an electromotive force of \(E{\mathrm{[V]}}\) and an internal resistance of \(r{\mathrm{[{\Omega}]}}\), connected to a resistor with a resistance of \(R{\mathrm{[{\Omega}]}}\).

On the right side, the graph shows the battery terminal voltage \(V\) on the vertical axis and the current \(I\) flowing from the battery on the horizontal axis.

For an ideal battery with an internal resistance \(r\) of \(0{\mathrm{[{\Omega}]}}\), no voltage drop \(rI\) occurs across the internal resistance, even when current \(I\) flows. The electromotive force \(E\) and terminal voltage \(V\) are therefore equal (the blue line in the graph above).

However, a real battery has an internal resistance \(r\). When current \(I\) flows, a voltage drop \(rI\) occurs across this resistance, making the battery terminal voltage \(V\) lower than its electromotive force \(E\) (\(V {\leq} E\)).

In this circuit, the electromotive force \(E\) equals the sum of the voltage \(rI\) across the internal resistance and the terminal voltage \(V\) across the external resistor. The relationship between electromotive force and terminal voltage is therefore expressed by the following equation:

\begin{eqnarray}
E&=&rI+V\\
\\
{\Leftrightarrow}V&=&E-rI\tag{1}
\end{eqnarray}

Equation (1) shows that the terminal voltage \(V\) equals the electromotive force \(E\) minus the voltage drop \(rI\) across the internal resistance \(r\). The unit of \(rI\) is \({\mathrm{[V]}}\). It represents the reduction in voltage caused by current \(I\) flowing through the internal resistance \(r\).

If the electromotive force \(E\) and internal resistance \(r\) remain constant, increasing the current \(I\) increases the voltage drop \(rI\), reducing the battery terminal voltage \(V\). Plotting Equation (1) gives the red line in the graph above.

When the current \(I\) is \(0{\mathrm{[A]}}\), the terminal voltage \(V\) equals the electromotive force \(E\) (\(V=E\)). When the current \(I\) is \(\displaystyle\frac{E}{r}{\mathrm{[A]}}\), the terminal voltage \(V\) is \(0{\mathrm{[V]}}\). Equation (1) also shows that the slope of the line is \(-r\). The larger the internal resistance \(r\), the greater the decrease in terminal voltage \(V\) for a given increase in current \(I\).

Next, let us look more closely at how to draw this graph.

When the Current \(I\) Flowing from the Battery Is \(0{\mathrm{[A]}}\)

Substituting \(I=0{\mathrm{[A]}}\) into Equation (1) gives the point where the red line intersects the graph’s vertical axis (terminal voltage \(V\)).

Let us calculate this point.

Substituting \(I=0{\mathrm{[A]}}\) into Equation (1) gives the following terminal voltage \(V\):

\begin{eqnarray}
V&=&E-rI\\
\\
&=&E-r×0\\
\\
&=&E{\mathrm{[V]}}\tag{2}
\end{eqnarray}

The red line therefore intersects the vertical axis (terminal voltage \(V\)) at \(E{\mathrm{[V]}}\).

When the current \(I\) flowing from the battery is \(0{\mathrm{[A]}}\), the voltage drop \(rI\) across the internal resistance \(r\) is \(0{\mathrm{[V]}}\). The terminal voltage \(V\) and electromotive force \(E\) are therefore equal (\(V=E\)).

When the Terminal Voltage \(V\) Is \(0{\mathrm{[V]}}\)

Substituting \(V=0{\mathrm{[V]}}\) into Equation (1) gives the point where the red line intersects the graph’s horizontal axis (current \(I\)).

Let us calculate this point.

Substituting \(V=0{\mathrm{[V]}}\) into Equation (1) gives the following current \(I\) flowing from the battery:

\begin{eqnarray}
V&=&E-rI\\
\\
{\Leftrightarrow}I&=&\frac{E-V}{r}\\
\\
&=&\frac{E-0}{r}\\
\\
&=&\frac{E}{r}\tag{3}
\end{eqnarray}

The red line therefore intersects the horizontal axis (current \(I\)) at \(\displaystyle\frac{E}{r}{\mathrm{[A]}}\).

Connecting the point where the current \(I\) flowing from the battery is \(0{\mathrm{[A]}}\) (where \(V=E{\mathrm{[V]}}\)) to the point where the terminal voltage \(V\) is \(0{\mathrm{[V]}}\) (where \(I=\displaystyle\frac{E}{r}{\mathrm{[A]}}\)) gives the red line representing the relationship between the battery terminal voltage \(V\) and the current \(I\) flowing from the battery.

Summary

This article explained the following points about battery internal resistance and the relationship between electromotive force and terminal voltage:

• A battery can be represented by an equivalent circuit consisting of an ideal source with an electromotive force \(E\), connected in series with an internal resistance \(r\).
• During discharge, the terminal voltage is \(V=E-rI\). If the electromotive force \(E\) and internal resistance \(r\) remain constant, the terminal voltage decreases as the current \(I\) increases.
• No current flows in an open circuit, so the terminal voltage \(V\) equals the electromotive force \(E\) in this equivalent circuit.
• Under ideal short-circuit conditions with an internal resistance \(r>0\), the terminal voltage \(V\) is zero and the current is \(\displaystyle\frac{E}{r}\). Do not short-circuit a real battery.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

Electric potential is the electrical “height” relative to a reference point, while voltage is the difference in electric potential between two points. This article uses diagrams to explain electric potential and voltage, covering the following topics:

• The meanings of electric potential and voltage, and how they differ
• Calculations for a circuit with dry cells connected in series
• Understanding electric potential and voltage through elevation and elevation difference
• How changing the reference point affects electric potential and voltage

The Difference Between Electric Potential and Voltage

Electric potential is the electrical “height” relative to a reference point (usually earth ground) defined as \(0{\mathrm{V}}\).

In contrast, voltage is the difference in electric potential between any two points. Voltage is also called potential difference.

Both electric potential and voltage (potential difference) are measured in \({\mathrm{V}}\) (volts).

Consider a circuit with two \(1.5{\mathrm{V}}\) dry cells connected in series, as shown above. First, taking point C as the \(0{\mathrm{V}}\) reference point (earth ground), the electric potentials at points A, B, and C are as follows.

Electric potential at point A, \(V_A\):
From the reference point, there are two \(1.5{\mathrm{V}}\) dry cells connected in series, so the electric potential is higher by the combined voltage of the two cells.

\begin{eqnarray}
V_A=3.0{\mathrm{V}}({\Leftarrow}{\;}1.5{\mathrm{V}}+1.5{\mathrm{V}})
\end{eqnarray}

Electric potential at point B, \(V_B\):
From the reference point, there is one \(1.5{\mathrm{V}}\) dry cell, so the electric potential is higher by the voltage of one cell.

\begin{eqnarray}
V_B=1.5{\mathrm{V}}
\end{eqnarray}

Electric potential at point C, \(V_C\):
Point C is the reference point, so its electric potential is as follows.

\begin{eqnarray}
V_C=0{\mathrm{V}}
\end{eqnarray}

Next, we calculate the voltages (potential differences) between A and B, A and C, and B and C. In this article, \(V_{AB}\) means “the voltage at point A relative to point B,” calculated by subtracting the electric potential at B from that at A. For \(V_{AC}\) and \(V_{BC}\), we likewise subtract the potential at the second point in the subscript from the potential at the first.

Voltage between A and B, \(V_{AB}\):
Subtract the electric potential at B (\(V_B=1.5{\mathrm{V}}\)) from the electric potential at A (\(V_A=3.0{\mathrm{V}}\)).

\begin{eqnarray}
V_{AB}=V_A-V_B=3.0{\mathrm{V}}-1.5{\mathrm{V}}=1.5{\mathrm{V}}
\end{eqnarray}

Voltage between A and C, \(V_{AC}\):
Subtract the electric potential at C (\(V_C=0{\mathrm{V}}\)) from the electric potential at A (\(V_A=3.0{\mathrm{V}}\)).

\begin{eqnarray}
V_{AC}=V_A-V_C=3.0{\mathrm{V}}-0{\mathrm{V}}=3.0{\mathrm{V}}
\end{eqnarray}

Voltage between B and C, \(V_{BC}\):
Subtract the electric potential at C (\(V_C=0{\mathrm{V}}\)) from the electric potential at B (\(V_B=1.5{\mathrm{V}}\)).

\begin{eqnarray}
V_{BC}=V_B-V_C=1.5{\mathrm{V}}-0{\mathrm{V}}=1.5{\mathrm{V}}
\end{eqnarray}

For example, \(V_{AB}=1.5{\mathrm{V}}\) means that the electric potential at A is \(1.5{\mathrm{V}}\) higher than at B. Reversing the order of the two points reverses the sign of the voltage. In this example, the voltage at B relative to A, \(V_{BA}\), is \(-1.5{\mathrm{V}}\).

Understanding Electric Potential and Voltage Through Elevation and Elevation Difference

The difference between electric potential and voltage is easier to visualize using elevation and elevation difference, as shown below.

Electric potential (electrical “height” relative to a reference point) is like elevation (height above sea level), while voltage (potential difference between any two points) is like elevation difference (the difference in height between any two points).

In the diagram, the elevations relative to sea level are \(300{\mathrm{m}}\) and \(100{\mathrm{m}}\), so the elevation difference is \(200{\mathrm{m}}\).

Electric Potential and Voltage with Point B as the Reference Point (Earth Ground)

Previously, we used point C as the reference point (earth ground). Now, let us calculate the electric potentials and voltages with point B as the \(0{\mathrm{V}}\) reference point, as shown below.

Taking point B as the reference point (earth ground), the electric potentials at points A, B, and C are as follows.

Electric potential at point A, \(V_A\):
From the reference point, there is one \(1.5{\mathrm{V}}\) dry cell, so the electric potential is higher by the voltage of one cell.

\begin{eqnarray}
V_A=1.5{\mathrm{V}}
\end{eqnarray}

Electric potential at point B, \(V_B\):
Point B is the reference point, so its electric potential is as follows.

\begin{eqnarray}
V_B=0{\mathrm{V}}
\end{eqnarray}

Electric potential at point C, \(V_C\):
From the reference point, there is one \(1.5{\mathrm{V}}\) dry cell, so the electric potential is lower by the voltage of one cell.

\begin{eqnarray}
V_C=-1.5{\mathrm{V}}
\end{eqnarray}

Since the positive terminal of the lower dry cell (point B) is defined as \(0{\mathrm{V}}\), its negative terminal (point C) has an electric potential of \(-1.5{\mathrm{V}}\). A negative electric potential means that the potential is lower than at the chosen reference point.

We also calculate the voltages (potential differences) between A and B, A and C, and B and C by subtracting the electric potentials in the same order as before.

Voltage between A and B, \(V_{AB}\):
Subtract the electric potential at B (\(V_B=0{\mathrm{V}}\)) from the electric potential at A (\(V_A=1.5{\mathrm{V}}\)).

\begin{eqnarray}
V_{AB}=V_A-V_B=1.5{\mathrm{V}}-0{\mathrm{V}}=1.5{\mathrm{V}}
\end{eqnarray}

Voltage between A and C, \(V_{AC}\):
Subtract the electric potential at C (\(V_C=-1.5{\mathrm{V}}\)) from the electric potential at A (\(V_A=1.5{\mathrm{V}}\)).

\begin{eqnarray}
V_{AC}=V_A-V_C=1.5{\mathrm{V}}-(-1.5{\mathrm{V}})=3.0{\mathrm{V}}
\end{eqnarray}

Voltage between B and C, \(V_{BC}\):
Subtract the electric potential at C (\(V_C=-1.5{\mathrm{V}}\)) from the electric potential at B (\(V_B=0{\mathrm{V}}\)).

\begin{eqnarray}
V_{BC}=V_B-V_C=0{\mathrm{V}}-(-1.5{\mathrm{V}})=1.5{\mathrm{V}}
\end{eqnarray}

Thus, when only the reference point is changed in the same circuit, the electric potential at each point changes, but the voltage between the same two points does not.

When we change the reference point from C to B, the electric potentials at A, B, and C all decrease by \(1.5{\mathrm{V}}\). Since both points’ potentials change by the same amount, their difference—the voltage—does not change. This is the same idea as changing the reference level for elevation without changing the difference in height between two locations.

Summary

This article explained the difference between electric potential and voltage. The key points are as follows:

• Electric potential is the electrical “height” relative to a chosen reference point, which does not have to be the earth.
• Voltage is the potential difference between any two points. Reversing their order reverses its sign.
• Both electric potential and voltage are measured in \({\mathrm{V}}\) (volts).
• Electric potential is easier to understand as elevation, and voltage as the elevation difference between two locations.
• Changing only the reference point in the same circuit does not change the voltage between two points compared in the same order.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

These terms are commonly used, but if you are just starting to learn about electrical circuits, their meanings may not be clear.

This article uses diagrams and equations to explain the following aspects of open and short circuits:

• What open and short circuits mean and how they differ
• How current and voltage change when a circuit is opened or shorted
• The relationship between an OFF switch and an open circuit
• A short-circuit current calculation example and precautions

The Difference Between Open and Short Circuits

First, let us compare disconnecting a resistor from a circuit with connecting a wire across its terminals. The following diagram shows the difference in connections when resistor \(R_2\) is open-circuited or short-circuited.

An open circuit is a condition in which part of the circuit is disconnected.

In the diagram above, “leaving resistor \(R_2\) open” means “disconnecting resistor \(R_2\) from the circuit.”

A short circuit is a condition in which two points in a circuit are connected by a conductor with very low resistance, such as a wire.

In the diagram above, “shorting resistor \(R_2\)” means “connecting the two terminals of resistor \(R_2\) (points A and B) with a conductor that has very low resistance, such as a wire.”

The differences between an “open circuit” and a “short circuit” are summarized below. We will explain them in more detail later.

Let us now take a closer look at each term in turn.

What Is an Open Circuit?

An open circuit is a condition in which part of the circuit is disconnected. It is also called an “open.”

The following diagram shows a simple circuit consisting of a voltage source \(V\), resistor \(R_1\), and resistor \(R_2\). In this circuit, “leaving resistor \(R_2\) open” means “disconnecting resistor \(R_2\) from the circuit.”

Disconnecting resistor \(R_2\) from the circuit creates a gap between points A and B, opening the circuit. There is no longer a complete path for current to flow around the circuit, so no current flows.

Impedance is a quantity that represents the opposition to current flow. A higher impedance makes it harder for current to flow, and no current flows through a disconnected part of a circuit. For this reason, an open circuit is sometimes described as a condition in which “the impedance becomes infinite.”

How Current and Voltage Change Before and After Opening the Circuit

Let us compare the current \(I\) flowing through the circuit and the voltage \(V_{AB}\) between points A and B before and after disconnecting resistor \(R_2\).

Here, \(V\) is the voltage of the voltage source in \({\mathrm{[V]}}\), \(I\) is the current flowing through the circuit in \({\mathrm{[A]}}\), and \(R_1\) and \(R_2\) are the resistance values of the two resistors in \({\mathrm{[\Omega]}}\). \(V_{AB}\) is the voltage at point A relative to point B, expressed in \({\mathrm{[V]}}\).

Before Disconnecting Resistor \(R_2\)

Before the circuit is opened, resistors \(R_1\) and \(R_2\) are connected in series. Their equivalent resistance is \(R_1+R_2\), so Ohm’s law gives the current \(I\) flowing through the circuit as follows:

\begin{eqnarray}
I=\frac{V}{R_1+R_2}
\end{eqnarray}

The voltage \(V_{AB}\) between points A and B is the voltage across resistor \(R_2\). The current through resistor \(R_2\) is also \(I\), so the voltage is given by the following equation:

\begin{eqnarray}
V_{AB}=R_2I=\frac{R_2}{R_1+R_2}V
\end{eqnarray}

The voltage \(V\) of the voltage source is divided between the two resistors. Therefore, before the circuit is opened, \(V_{AB}\) is lower than \(V\).

After Disconnecting Resistor \(R_2\)

When resistor \(R_2\) is disconnected, the resistance between points A and B can be considered infinite. If we denote this resistance by \(R_{AB}\), Ohm’s law gives the current \(I\) flowing through the circuit as follows:

\begin{eqnarray}
I=\frac{V}{R_1+R_{AB}}=\frac{V}{R_1+{\infty}}=0
\end{eqnarray}

Disconnecting resistor \(R_2\) makes the circuit current \(I\) zero, so the voltage drop across resistor \(R_1\) also becomes zero. As a result, the voltage \(V\) of the voltage source equals the voltage \(V_{AB}\) between points A and B. This can be expressed as follows:

\begin{eqnarray}
V_{AB}=V
\end{eqnarray}

Thus, in this circuit, “opening the circuit stops current from flowing” and “the voltage across the open section increases.”

An OFF Switch Also Creates an Open Circuit

The term “open circuit” does not apply only to removing a component. The following diagram shows a circuit consisting of a voltage source \(V\), a switch \(SW\), and a resistor \(R_1\).

When switch \(SW\) is turned OFF, its contacts separate, disconnecting part of the circuit and stopping current flow.

For this reason, turning a switch OFF is also described as “opening” the switch.

What Is a Short Circuit?

A short circuit is a condition in which two points in a circuit are connected by a conductor with very low resistance, such as a wire. It is also called a “short.”

The following diagram shows a simple circuit consisting of a voltage source \(V\), resistor \(R_1\), and resistor \(R_2\). In this circuit, “shorting resistor \(R_2\)” means “connecting the two terminals of resistor \(R_2\) (points A and B) with a conductor that has very low resistance, such as a wire.”

There is nothing to impede current flow through the shorted section. For this reason, a short circuit is sometimes described as a condition in which “the impedance becomes zero.”

How Current and Voltage Change Before and After Shorting the Circuit

Let us compare the current \(I\) flowing through the circuit and the voltage \(V_{AB}\) between points A and B before and after shorting resistor \(R_2\).

Here, \(V\) is the voltage of the voltage source in \({\mathrm{[V]}}\), \(I\) is the current flowing through the circuit in \({\mathrm{[A]}}\), and \(R_1\) and \(R_2\) are the resistance values of the two resistors in \({\mathrm{[\Omega]}}\). \(V_{AB}\) is the voltage at point A relative to point B, expressed in \({\mathrm{[V]}}\).

Before Shorting Resistor \(R_2\)

Before the circuit is shorted, resistors \(R_1\) and \(R_2\) are connected in series. Their equivalent resistance is \(R_1+R_2\), so Ohm’s law gives the current \(I\) flowing through the circuit as follows:

\begin{eqnarray}
I=\frac{V}{R_1+R_2}
\end{eqnarray}

The voltage \(V_{AB}\) between points A and B is the voltage across resistor \(R_2\). The current through resistor \(R_2\) is also \(I\), so the voltage is given by the following equation:

\begin{eqnarray}
V_{AB}=R_2I=\frac{R_2}{R_1+R_2}V
\end{eqnarray}

After Shorting Resistor \(R_2\)

When the two terminals of resistor \(R_2\) are shorted with an ideal conductor, the resistance between points A and B can be considered zero. If we denote this resistance by \(R_{AB}\), Ohm’s law gives the current \(I\) flowing through the circuit as follows:

\begin{eqnarray}
I=\frac{V}{R_1+R_{AB}}=\frac{V}{R_1+0}=\frac{V}{R_1}
\end{eqnarray}

Also, because the equivalent resistance \(R_{AB}\) between points A and B is zero, the voltage \(V_{AB}\) between them is also zero. This can be expressed as follows:

\begin{eqnarray}
V_{AB}=R_{AB}I=0×I=0
\end{eqnarray}

Thus, in this circuit, “shorting the circuit makes it easier for current to flow” and “there is no voltage across the shorted section.”

Caution: A Short Circuit Can Cause a Very Large Current

The following diagram shows a simple circuit consisting of a voltage source \(V=10{\mathrm{[V]}}\) and a resistor \(R_1=100{\mathrm{[\Omega]}}\). Let us consider a circuit in which resistor \(R_1\) alone determines the current.

We will calculate the current \(I\) flowing through the circuit before and after shorting resistor \(R_1\) with a low-resistance conductor whose resistance is \(R_2=0.1{\mathrm{[\Omega]}}\).

Before Shorting with a Low-Resistance Conductor (\(R_2=0.1{\mathrm{[\Omega]}}\))

Before the resistor is shorted, current can flow only through resistor \(R_1\). The circuit current \(I\) is therefore equal to the current \(I_1\) through resistor \(R_1\). Using Ohm’s law, we obtain:

\begin{eqnarray}
I=I_1=\frac{V}{R_1}=\frac{10}{100}=0.1{\mathrm{[A]}}
\end{eqnarray}

After Shorting with a Low-Resistance Conductor (\(R_2=0.1{\mathrm{[\Omega]}}\))

After the resistor is shorted, resistor \(R_1\) and the conductor resistance \(R_2\) are connected in parallel. Both have \(10{\mathrm{[V]}}\) across them, so current \(I_1\) flows through resistor \(R_1\), and current \(I_2\) flows through the conductor. The current \(I\) supplied by the voltage source is the sum of these currents, giving the following value:

\begin{eqnarray}
I=I_1+I_2=\frac{V}{R_1}+\frac{V}{R_2}=\frac{10}{100}+\frac{10}{0.1}=0.1+100=100.1{\mathrm{[A]}}
\end{eqnarray}

Before shorting, the circuit current \(I\) is \(0.1{\mathrm{[A]}}\). After shorting, the circuit current \(I\) is \(100.1{\mathrm{[A]}}\). This shows that shorting a circuit can cause a very large current to flow.

The current that flows during a short circuit is called “short-circuit current.” This current can melt the wire insulation (the insulating polymer covering the wire). Sparks produced during a short circuit can also cause a fire or other serious problems.

Summary

This article explained the following points about “the difference between open and short circuits“:

• An open circuit (open) is a condition in which part of the circuit is disconnected.
• A short circuit (short) is a condition in which two points are connected by a conductor with very low resistance, such as a wire.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

This article explains the following topics using diagrams and equations.

• Differences between voltage sources and current sources
• Conditions for voltage sources and current sources to be equivalent
• Source transformation methods and calculation examples

Differences Between Voltage Sources and Current Sources

The diagram below compares voltage source and current source models that include internal resistance.

A voltage source supplies voltage to a load. Its basic equivalent circuit consists of an ideal voltage source \(V\) (a source that provides a constant voltage) in series with an internal resistance \(r_v\).

A current source supplies current to a load. Its basic equivalent circuit consists of an ideal current source \(I\) (a source that provides a constant current) in parallel with an internal resistance \(r_i\).

These voltage sources and current sources can be transformed into equivalent circuits when certain conditions are met. Let us now look at how to perform this transformation.

How to Transform Voltage Sources and Current Sources

To transform a voltage source into a current source, or vice versa, we first need to determine the conditions under which a voltage source and a current source are equivalent.

Conditions for Voltage Sources and Current Sources to Be Equivalent

The diagram below shows both models connected to the same load resistance, along with the conditions for equivalence.

A voltage source and a current source are equivalent when connecting the same load resistance \(R\) produces the same current \(I_R\) through the load resistance \(R\) before and after the transformation. If the current \(I_R\) through the load resistance \(R\) is the same, the voltage \(V_R\) across the load resistance \(R\) is also the same.

When a load resistance \(R\) is connected to the voltage source, the current \(I_R\) is given by the following equation.

In the voltage source model, the internal resistance \(r_v\) and the load resistance \(R\) are in series. The current is therefore found by dividing the voltage \(V\) by the sum of these resistances.

\begin{eqnarray}
I_R=\frac{V}{r_v+R}\tag{1}
\end{eqnarray}

When a load resistance \(R\) is connected to the current source, the current \(I_R\) is given by the following equation.

In the current source model, the ideal source current \(I\) divides between the internal resistance \(r_i\) and the load resistance \(R\). Equation (2) gives the portion of this current that flows through the load.

\begin{eqnarray}
I_R=\frac{r_i}{r_i+R}I\tag{2}
\end{eqnarray}

Here, \(V\) is the ideal voltage source voltage \({\mathrm{[V]}}\), \(I\) is the ideal current source current \({\mathrm{[A]}}\), \(r_v\) and \(r_i\) are the respective internal resistances \({\mathrm{[Ω]}}\), \(R\) is the load resistance \({\mathrm{[Ω]}}\), and \(I_R\) is the load current \({\mathrm{[A]}}\). For the voltage source and current source to be equivalent, equations (1) and (2) must be equal. The conditions are as follows.

Equation (3) describes the relationship between voltage and current, while equation (4) shows that the internal resistance retains the same value after transformation. These conditions allow us to transform a voltage source into an equivalent current source, or vice versa. Let us try both transformations.

How to Transform a Voltage Source into a Current Source

The diagram below shows a voltage source transformed into a current source. The resistor changes from a series connection to a parallel connection.

We transform a voltage source consisting of an ideal voltage source \(V\) (a source that provides a constant voltage) in series with an internal resistance \(r_v\) into a current source.

For reference, the conditions for voltage sources and current sources to be equivalent are repeated below.

Using these conditions, the ideal current source \(I\) and internal resistance \(r_i\) after transforming the voltage source into a current source are given by the following equations.

\begin{eqnarray}
I&=&\frac{V}{r_v}\tag{5}\\
\\
r_i&=&r_v\tag{6}
\end{eqnarray}

The conditions for voltage sources and current sources to be equivalent therefore allow us to transform a voltage source into a current source.

For example, suppose the ideal voltage source \(V\) has a voltage of \(V=5{\mathrm{[V]}}\), and the voltage source internal resistance \(r_v\) is \(r_v=0.5{\mathrm{[Ω]}}\). After transformation into a current source, the ideal current source \(I\) and internal resistance \(r_i\) have the following values.

\begin{eqnarray}
I&=&\frac{V}{r_v}=\frac{5}{0.5}=10{\mathrm{[A]}}\tag{7}\\
\\
r_i&=&r_v=0.5{\mathrm{[Ω]}}\tag{8}
\end{eqnarray}

How to Transform a Current Source into a Voltage Source

The diagram below shows a current source transformed into a voltage source. The resistor changes from a parallel connection to a series connection.

We transform a current source consisting of an ideal current source \(I\) (a source that provides a constant current) in parallel with an internal resistance \(r_i\) into a voltage source.

For reference, the conditions for voltage sources and current sources to be equivalent are repeated below.

Equations (3) and (4) give the ideal voltage source \(V\) and internal resistance \(r_v\) after transforming the current source into a voltage source.

The conditions for voltage sources and current sources to be equivalent therefore allow us to transform a current source into a voltage source.

For example, suppose the ideal current source \(I\) has a current of \(I=5{\mathrm{[A]}}\), and the current source internal resistance \(r_i\) is \(r_i=0.5{\mathrm{[Ω]}}\). After transformation into a voltage source, the ideal voltage source \(V\) and internal resistance \(r_v\) have the following values.

\begin{eqnarray}
V&=&r_iI=0.5×5=2.5{\mathrm{[V]}}\tag{9}\\
\\
r_v&=&r_i=0.5{\mathrm{[Ω]}}\tag{10}
\end{eqnarray}

Worked Example: Voltage Sources and Current Sources

Worked Example

Consider the circuit below, in which two voltage sources, each with an internal resistance, are connected in parallel.

In the circuit shown above, the voltage sources \(V_1=5{\mathrm{[V]}}(r_{v1}=0.2{\mathrm{[Ω]}})\) and \(V_2=6{\mathrm{[V]}}(r_{v2}=0.25{\mathrm{[Ω]}})\) are connected in parallel, with a load resistance of \(R=1{\mathrm{[Ω]}}\).

For this circuit, consider the following currents:

• The current \(I_{R1}\) flowing from voltage source \(V_1\) toward the load
• The current \(I_{R2}\) flowing from voltage source \(V_2\) toward the load
• The current \(I_R\) through the load resistance \(R\)

Let us calculate these currents.

Solution

In the diagram, the positive directions for \(I_{R1}\) and \(I_{R2}\) are from the voltage sources toward the upper node. The positive direction for \(I_R\) is downward through the load resistor. We calculate the currents using the following steps.

Transform the Voltage Sources into Current Sources

The diagram below shows the circuit after each of the two voltage sources in the example has been transformed into a current source.

Applying equation (5) to each voltage source gives the following currents for the ideal current sources \(I_1\) and \(I_2\) after transformation.

\begin{eqnarray}
I_1&=&\frac{V_1}{r_{v1}}=\frac{5}{0.2}=25{\mathrm{[A]}}\tag{11}\\
\\
I_2&=&\frac{V_2}{r_{v2}}=\frac{6}{0.25}=24{\mathrm{[A]}}\tag{12}
\end{eqnarray}

The two ideal current sources supply current in the same direction. The combined current \(I\) from the ideal current sources \(I_1\) and \(I_2\) is therefore found by adding their currents.

\begin{eqnarray}
I&=&I_1+I_2=25+24=49{\mathrm{[A]}}\tag{13}
\end{eqnarray}

The current source internal resistances \(r_{i1}\) and \(r_{i2}\) have the following values.

\begin{eqnarray}
r_{i1}&=&r_{v1}=0.2{\mathrm{[Ω]}}\tag{14}\\
\\
r_{i2}&=&r_{v2}=0.25{\mathrm{[Ω]}}\tag{15}
\end{eqnarray}

Calculate the Current Through Each Resistor

Resistors connected in parallel have the same voltage across them, so a smaller resistance carries a larger current. The quantity that describes how easily current flows is conductance (the reciprocal of resistance), measured in \({\mathrm{[S]}}\) (siemens). For resistors connected in parallel, current divides in proportion to the conductance of each resistor. The currents through the resistors are therefore as follows.

\begin{eqnarray}
I_{r1}&=&\frac{\displaystyle\frac{1}{r_{i1}}}{\displaystyle\frac{1}{r_{i1}}+\displaystyle\frac{1}{r_{i2}}+\displaystyle\frac{1}{R}}×I\\
\\
&=&\frac{\displaystyle\frac{1}{0.2}}{\displaystyle\frac{1}{0.2}+\displaystyle\frac{1}{0.25}+\displaystyle\frac{1}{1}}×49\\
\\
&=&\frac{5}{5+4+1}×49\\
\\
&=&24.5{\mathrm{[A]}}\tag{16}\\
\\
I_{r2}&=&\frac{\displaystyle\frac{1}{r_{i2}}}{\displaystyle\frac{1}{r_{i1}}+\displaystyle\frac{1}{r_{i2}}+\displaystyle\frac{1}{R}}×I\\
\\
&=&\frac{\displaystyle\frac{1}{0.25}}{\displaystyle\frac{1}{0.2}+\displaystyle\frac{1}{0.25}+\displaystyle\frac{1}{1}}×49\\
\\
&=&\frac{4}{5+4+1}×49\\
\\
&=&19.6{\mathrm{[A]}}\tag{17}\\
\\
I_{R}&=&\frac{\displaystyle\frac{1}{R}}{\displaystyle\frac{1}{r_{i1}}+\displaystyle\frac{1}{r_{i2}}+\displaystyle\frac{1}{R}}×I\\
\\
&=&\frac{\displaystyle\frac{1}{1}}{\displaystyle\frac{1}{0.2}+\displaystyle\frac{1}{0.25}+\displaystyle\frac{1}{1}}×49\\
\\
&=&\frac{1}{5+4+1}×49\\
\\
&=&4.9{\mathrm{[A]}}\tag{18}
\end{eqnarray}

Use Kirchhoff’s Law to Calculate the Source Currents

The diagram below shows the relationship between the current flowing from each source toward the load and the current diverted through its internal resistance.

Kirchhoff’s current law states that the total current entering a node equals the total current leaving it. Subtracting the current through each parallel internal resistance from the current supplied by its ideal current source gives the currents \(I_{R1}\) and \(I_{R2}\) flowing from the original voltage sources toward the load.

\begin{eqnarray}
I_{R1}&=I_1-I_{r1}=25-24.5=0.5{\mathrm{[A]}}\tag{19}\\
\\
I_{R2}&=I_2-I_{r2}=24-19.6=4.4{\mathrm{[A]}}\tag{20}
\end{eqnarray}

The resulting current \(I_{R1}\) through voltage source \(V_1\), current \(I_{R2}\) through voltage source \(V_2\), and current \(I_R\) through the load resistance \(R\) are shown in the diagram below.

Summary

This article covered the following points about voltage sources and current sources.

• An ideal voltage source maintains a constant voltage, while an ideal current source maintains a constant current, regardless of the load
• In models that include internal resistance, the resistance is connected in series with a voltage source and in parallel with a current source
• Voltage sources and current sources can be transformed into equivalent circuits when \(V=r_iI\) and \(r_v=r_i\) are satisfied

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

A current source supplies current to a circuit. However, ideal and practical current sources differ in how their output current behaves when a load is connected.

For example, an ideal current source maintains a constant output current even when the load resistance \(R\) changes. In contrast, when a practical current source is modeled using an equivalent circuit with internal resistance, its output current decreases as the load resistance increases.

This article explains the following topics about current sources:

• What Is a Current Source?
• Load Current When the Current Source Internal Resistance r Is Infinite
• Load Current with Current Source Internal Resistance r
• Current Source Symbols
• Voltage Across a Current Source

We use diagrams to make these topics easy to understand. We hope you find this article helpful.

What Is a Current Source?

A current source is a power source that maintains a constant output current \(I_R\) regardless of the load resistance \(R\). In particular, an ideal power source that maintains a constant output current regardless of the load is called an “ideal current source.”

A practical current source can be represented by a basic equivalent circuit consisting of an “ideal current source \(I\) (a power source that supplies a constant current)” connected in parallel with an “internal resistance \(r\).”

A current source with an infinite internal resistance \(r\) (\(\infty{\mathrm{[\Omega]}}\)) is an ideal current source. In this case, the current \(I_r\) through the internal resistance \(r\) is \(0{\mathrm{[A]}}\), so the current \(I_R\) through the load resistance \(R\) is expressed by the following equation.

\begin{eqnarray}
I&=&I_r+I_R\\
\\
{\Leftrightarrow}I_R&=&I-I_r\\
\\
&=&I-0\\
\\
&=&I{\mathrm{[A]}}\tag{1}
\end{eqnarray}

Here, \(I\) is the ideal current source current \({\mathrm{[A]}}\), \(I_r\) is the current through the internal resistance \({\mathrm{[A]}}\), and \(I_R\) is the current through the load resistance \({\mathrm{[A]}}\).

Equation (1) does not contain the load resistance \(R\). Therefore, for an ideal current source, the current \(I_R\) through the load resistance \(R\) always remains constant (\(I{\mathrm{[A]}}\)), regardless of the value of \(R\).

Next, let us consider how the current \(I_R\) through the load resistance \(R\) behaves when the internal resistance \(r\) is taken into account.

Load Current with Current Source Internal Resistance r

The following diagram shows the equivalent circuit of a current source with internal resistance \(r\), along with how the output current changes with the load resistance \(R\).

For a practical current source, the internal resistance \(r\) cannot always be treated as infinite (\(\infty{\mathrm{[\Omega]}}\)). In this case, the current supplied by the “ideal current source \(I\) (a power source that supplies a constant current)” divides between the internal resistance \(r\) and the load resistance \(R\). The current \(I_R\) through the load resistance \(R\) is therefore given by the following equation.

\begin{eqnarray}
I_R=\frac{r}{r+R}I{\mathrm{[A]}}\tag{2}
\end{eqnarray}

Here, \(I\) is the ideal current source current \({\mathrm{[A]}}\), \(r\) is the current source internal resistance \({\mathrm{[\Omega]}}\), \(R\) is the load resistance \({\mathrm{[\Omega]}}\), and \(I_R\) is the current through the load resistance \({\mathrm{[A]}}\).

Equation (2) contains the load resistance \(R\). Therefore, when the value of the load resistance \(R\) changes, the current \(I_R\) through it also changes. As the load resistance \(R\) increases, the terminal voltage rises, and the current \(I_r\) through the internal resistance \(r\) also increases.

For an ideal current source, the internal resistance \(r\) is infinite (\(\infty{\mathrm{[\Omega]}}\)), so the current \(I_R\) through the load resistance \(R\) can be approximated as follows.

\begin{eqnarray}
I_R&=&\frac{r}{r+R}I\\
\\
&=&\frac{1}{1+\displaystyle\frac{R}{r}}I\\
\\
&=&\frac{1}{1+\displaystyle\frac{R}{∞}}I\\
\\
&{\approx}&\frac{1}{1+0}I\\
\\
&{\approx}&I{\mathrm{[A]}}\tag{3}
\end{eqnarray}

When the internal resistance \(r\) is infinite (∞)[Ω], the load resistance \(R\) disappears from the equation. Therefore, for an ideal current source, the current \(I_R\) through the load resistance \(R\) always remains constant (\(I{\mathrm{[A]}}\)), regardless of the value of \(R\).

Current Source Symbols

The following diagram shows common symbols used for current sources.

Current sources are represented by symbols such as those shown above. The “circle containing an arrow” shown on the far left is commonly used. Each symbol is explained below.

• A circle containing an arrow
  • For DC, the arrow indicates the direction of positive current flow. For AC, it indicates the positive reference direction for the current. If the current value is negative, the current flows in the direction opposite to the arrow.
• A circle containing a horizontal line
  • This is the graphical symbol specified in JIS C 0617 / IEC 60617 for an ideal current source. It is used for both DC and AC. Parameters such as the current value and its reference direction are indicated near the symbol as needed.
• Two offset circles
  • This symbol is sometimes used to represent a current source. The current value and its reference direction are indicated near the symbol as needed.

Voltage Across a Current Source

The following diagram shows how the voltage \(V_R\) across a current source changes when the load resistance \(R\) is varied.

For an ideal current source, the current \(I_R\) through the load resistance \(R\) remains constant, regardless of the value of \(R\).

For example, if a load resistance \(R\) of \(5{\mathrm{[\Omega]}}\) is connected to an ideal current source that supplies \(10{\mathrm{[A]}}\), the voltage \(V_R\) across the load resistance \(R\) (and across the current source) is given by Ohm’s law as follows.

\begin{eqnarray}
V_R=R\times I_R=5\times10=50{\mathrm{[V]}}\tag{4}
\end{eqnarray}

If the load resistance \(R\) doubles to \(10{\mathrm{[\Omega]}}\), the current remains at \(10{\mathrm{[A]}}\), so the voltage \(V_R\) across the current source also doubles.

\begin{eqnarray}
V_R=R\times I_R=10\times10=100{\mathrm{[V]}}\tag{5}
\end{eqnarray}

Thus, for an ideal current source, the voltage \(V_R\) across the current source increases as the load resistance \(R\) increases.

Voltage Across a Current Source When the Load Resistance \(R\) Is Open-Circuited

What happens if the load resistance \(R\) connected to an ideal current source supplying \(10{\mathrm{[A]}}\) is open-circuited? The following diagram illustrates the concept of increasing the load resistance without limit.

Open-circuiting the load resistance \(R\) is ideally equivalent to making its resistance infinite (\(\infty{\mathrm{[\Omega]}}\)).

In this case, as the load resistance \(R\) increases without limit while the current is kept constant, the required terminal voltage \(V_R\) of the current source increases without bound, as shown below.

\begin{eqnarray}
V_R=R×I_R=∞×10=∞{\mathrm{[V]}}\tag{6}
\end{eqnarray}

Because the internal resistance of an ideal current source is also assumed to be infinite, this means that, in the ideal circuit, there is no upper limit to the voltage required to maintain the current. A practical power source, however, has a maximum output voltage, and some power sources also provide overvoltage protection. Its actual terminal voltage therefore does not become infinite. Once the output voltage limit is reached, the source can no longer maintain the set current.

Current Source vs. Voltage Source

A “current source” is often compared with a “voltage source”.

A current source supplies current to a load, whereas a voltage source supplies voltage to a load.

An ideal current source is modeled as maintaining a constant output current even when the load changes. In contrast, an ideal voltage source is modeled as maintaining a constant output voltage even when the load changes.

Summary

This article covered the following key points about current sources:

• An ideal DC current source supplies a constant current regardless of the load.
• The internal resistance \(r\) of an ideal current source is infinite. A practical current source can be represented by an ideal current source connected in parallel with an internal resistance.
• When the internal resistance \(r\) is taken into account, the load current is \(I_R=\frac{r}{r+R}I\).
• For an ideal current source, the terminal voltage rises as the load resistance \(R\) increases. For a practical power source, the maximum output voltage must be taken into account.
• An ideal current source maintains a constant output current, whereas an ideal voltage source maintains a constant output voltage.

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

When studying electric circuits, you may come across the term voltage source.

A voltage source supplies voltage to a circuit. However, ideal and real voltage sources behave differently when a load is connected.

For example, an ideal voltage source maintains a constant output voltage even when the load resistance \(R\) changes. A real voltage source has internal resistance, so its output voltage decreases as the load current increases.

This article uses diagrams to explain the following topics:

• What a voltage source is
• Voltage across the load when the source’s internal resistance \(r\) is zero
• Voltage across the load when the source’s internal resistance \(r\) is taken into account
• Voltage source symbols
• Current flowing through a voltage source

We hope these explanations help you understand voltage sources.

What Is a Voltage Source?

A voltage source is a power source whose output voltage \(V_R\) remains constant regardless of the load resistance \(R\). In particular, an ideal power source that maintains a constant terminal voltage regardless of the load is called an ideal voltage source.

A real voltage source can be represented by a basic equivalent circuit consisting of an ideal voltage source \(V\), which provides a constant voltage, connected in series with an internal resistance \(r\).

An ideal voltage source has an internal resistance \(r\) of \(0{\mathrm{[\Omega]}}\). In this case, the voltage \(V_r\) across the internal resistance is \(0{\mathrm{[V]}}\), so the voltage \(V_R\) across the load resistance \(R\) is given by:

\begin{eqnarray}
V&=&V_r+V_R\\
\\
{\Leftrightarrow}V_R&=&V-V_r\\
\\
&=&V-0\\
\\
&=&V{\mathrm{[V]}}\tag{1}
\end{eqnarray}

Equation (1) does not contain the load resistance \(R\). Therefore, for an ideal voltage source, the voltage \(V_R\) across the load resistance \(R\) remains constant at \(V{\mathrm{[V]}}\), regardless of the value of \(R\).

Next, let’s consider how the voltage \(V_R\) across the load res

Voltage Across the Load When Internal Resistance \(r\) Is Taken into Account

For a real voltage source, the internal resistance \(r\) cannot always be treated as \(0{\mathrm{[\Omega]}}\). In this case, the voltage of the ideal voltage source \(V\), which provides a constant voltage, is divided between the internal resistance \(r\) and the load resistance \(R\). The voltage \(V_R\) across the load resistance \(R\) is therefore given by:

\begin{eqnarray}
V_R=\frac{R}{r+R}V{\mathrm{[V]}}\tag{2}
\end{eqnarray}

Here, \(V\) is the ideal source voltage \({\mathrm{[V]}}\), \(r\) is the source’s internal resistance \({\mathrm{[\Omega]}}\), \(R\) is the load resistance \({\mathrm{[\Omega]}}\), and \(V_R\) is the voltage across the load resistance \({\mathrm{[V]}}\).

Equation (2) contains the load resistance \(R\). Therefore, when the value of \(R\) changes, the voltage \(V_R\) across the load resistance also changes. As \(R\) decreases, the current increases, causing a larger voltage drop across the internal resistance \(r\).

For an ideal voltage source, the internal resistance \(r\) is \(0{\mathrm{[\Omega]}}\), so the voltage \(V_R\) across the load resistance \(R\) becomes:

\begin{eqnarray}
V_R&=&\frac{R}{r+R}V\\
\\
&=&\frac{R}{0+R}V\\
\\
&=&V{\mathrm{[V]}}\tag{3}
\end{eqnarray}

When the internal resistance \(r\) is \(0{\mathrm{[\Omega]}}\), the load resistance \(R\) cancels out of the equation. This shows that, for an ideal voltage source, the voltage \(V_R\) across the load resistance \(R\) remains constant at \(V{\mathrm{[V]}}\), regardless of the value of \(R\).

Voltage Source Symbols

Voltage sources are represented by symbols such as those shown above. Two parallel lines of different lengths are commonly used for a DC source, while a circle containing a sine wave is commonly used for an AC voltage source. Each symbol is explained below.

• Two parallel lines of different lengths
  • The longer line represents the positive (+) terminal, and the shorter line represents the negative (−) terminal.
  • The shorter line is sometimes drawn slightly thicker. However, this is not necessary as long as the difference in length clearly indicates the polarity.
• A circle containing a vertical line
  • This is the graphical symbol specified in JIS C 0617 / IEC 60617 for an ideal voltage source. Parameters such as polarity and voltage are indicated near the symbol as needed.
• A circle containing a sine wave
  • This is a common symbol used to represent a sinusoidal AC voltage source.

Current Flowing Through a Voltage Source

or an ideal voltage source, the voltage \(V_R\) across the load resistance \(R\) remains constant regardless of the value of \(R\).

For example, suppose a load resistance \(R\) of \(10{\mathrm{[\Omega]}}\) is connected to an ideal voltage source that provides \(10{\mathrm{[V]}}\). According to Ohm’s law, the current \(I_R\) through the load resistance is:

\begin{eqnarray}
I_R=\frac{V}{R}=\frac{10}{10}=1{\mathrm{[A]}}\tag{4}
\end{eqnarray}

If the load resistance \(R\) is halved to \(5{\mathrm{[\Omega]}}\), the voltage remains at \(10{\mathrm{[V]}}\), so the current \(I_R\) doubles:

\begin{eqnarray}
I_R=\frac{V}{R}=\frac{10}{5}=2{\mathrm{[A]}}\tag{5}
\end{eqnarray}

Thus, for an ideal voltage source, the current \(I_R\) flowing through the source increases as the load resistance \(R\) decreases.

Voltage Across the Voltage Source When the Load Resistance \(R\) Is Short-Circuited

What happens if the load resistance \(R\) connected to an ideal voltage source providing \(10{\mathrm{[V]}}\) is short-circuited?

Short-circuiting the load resistance \(R\) is ideally equivalent to reducing \(R\) to \(0{\mathrm{[\Omega]}}\).

In this case, the calculated current \(I_R\) through the voltage source is expressed as follows, indicating an extremely large current:

\begin{eqnarray}
I_R=\frac{V}{R}=\frac{10}{0}=∞{\mathrm{[A]}}\tag{6}
\end{eqnarray}

Because an ideal voltage source is also assumed to have an internal resistance of \(0{\mathrm{[\Omega]}}\), this means that the short-circuit current has no upper limit in the ideal circuit model. Real power sources, however, have internal resistance, wiring resistance, and other resistances. Depending on the source, overcurrent protection or current limiting may also operate. Therefore, the actual short-circuit current is not infinite.

Voltage Source vs. Current Source

A “voltage source” is often compared with a “current source”.

A voltage source supplies voltage to a load, whereas a current source supplies current to a load.

An ideal voltage source is modeled as maintaining a constant output voltage even when the load changes. In contrast, an ideal current source is modeled as maintaining a constant output current even when the load changes.

Summary

This article explained the following topics about voltage sources:

• What a voltage source is
• Voltage across the load when the source’s internal resistance \(r\) is zero
• Voltage across the load when the source’s internal resistance \(r\) is taken into account
• Voltage source symbols
• Current flowing through a voltage source

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

• How to obtain RMS Value and Average Value

RMS Value and Average Value of Pulse Wave

The first illustration shows the root mean square (RMS) value and average value of a pulse wave (with a maximum value of \(V_M\) and a period of \(T\)).

Now, let’s explain how each of these values are calculated (we strive to include as many intermediate steps as possible).

Waveform of a Pulse Wave

To find the root mean square (RMS) value and average value of a pulse wave, we first need to express the pulse wave as a formula.

The pulse wave has different formulas for the red region \(\left(0 \leq t \lt T_1\right)\) and the blue region \(\left(T_1 \leq t \lt T\right)\).

Let’s calculate the formulas for the red region and blue region.

Red Region

The region “\(0 \leq t \lt T_1\)” corresponds to the red region.

The red line is always \(V_M\) during the time \(T_1\).

Therefore, the red region can be represented by the following formula.

\begin{eqnarray}
v(t)=V_M
\end{eqnarray}

Blue Region

The region “\(T_2 \leq t \lt T\)” corresponds to the blue region.

The blue rgsion is zero.

Therefore, the blue region can be represented by the following formula.

\begin{eqnarray}
v(t)=0
\end{eqnarray}

By combining the red region and the blue region, the pulse wave can be represented by the following formula.

\begin{eqnarray}
v(t) = \begin{cases}
V_M & \left(0 \leq t \lt T_1\right) \\
\\
0 & \left(T_1 \leq t \lt T\right)
\end{cases}
\end{eqnarray}

RMS Value of a Pulse Wave

The root mean square (RMS) value \(V_{RMS}\) of a waveform \(v(t)\) is the square root of the mean of the square of \(v(t)\). It can be represented by the following formula:

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{T} \displaystyle \int_{0}^{T}v(t)^2dt}
\end{eqnarray}

The red region is \(0 \leq t \lt T_1\)“, and the blue region is “\(T_1 \leq t \lt T\)“.

In the formula for finding the absolute value, if we separate the red region and the blue region, we get the following formula.

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{T} \left(\displaystyle \int_{0}^{T_1}v(t)^2dt+\displaystyle \int_{T_1}^{T}v(t)^2dt\right )}\\
\\
&=& \sqrt{\displaystyle\frac{1}{T} \left(\displaystyle \int_{0}^{T_1}V_M^2dt+\displaystyle \int_{T_1}^{T}0\right )}\\
\\
&=& \sqrt{\displaystyle\frac{1}{T} \displaystyle \int_{0}^{T_1}V_M^2dt}
\end{eqnarray}

\(V_M^2\) is not a variable that changes with time \(t\), but a constant. Therefore, it can be taken out of the integration. As a result, the following equation is obtained.

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{T_1} V_M^2\displaystyle \int_{0}^{T}1dt}\\
\\
&=& V_M\sqrt{\displaystyle\frac{1}{T_1} \displaystyle \int_{0}^{T}1dt}
\end{eqnarray}

When calculating the above equation, the following formula is obtained.

\begin{eqnarray}
V_{RMS} &=& V_M\sqrt{\displaystyle\frac{1}{T} \left[t\right]_{0}^{T_1}}\\
\\
&=& V_M\sqrt{\displaystyle\frac{1}{T} \left(T_1-0\right)}\\
\\
&=& \sqrt{\displaystyle\frac{T_1}{T}}V_M
\end{eqnarray}

Therefore, the root mean square (RMS) value \(V_{RMS}\) of the pulse wave is as follows.

\begin{eqnarray}
V_{RMS} =\sqrt{\displaystyle\frac{T_1}{T}}V_M
\end{eqnarray}

Average Value of a Pulse Wave

The average value \(V_{AVE}\) of a waveform \(v(t)\) is the average value of the absolute value \(|v(t)|\) of \(v(t)\) and can be represented by the following formula:

This formula for calculating the average value uses the absolute value \(|v(t)|\) of \(v(t)\).

Therefore, if there is a negative region in the waveform, it needs to be converted so that the negative region becomes positive.

In the case of a pulse wave, there is no negative region, so the formula for \(v(t)\) and the formula for the absolute value \(|v(t)|\) of \(v(t)\) become the same.

Therefore, the absolute value \(|v(t)|\) of \(v(t)\) can be represented by the following formula.

\begin{eqnarray}
|v(t)| = \begin{cases}
V_M & \left(0 \leq t \lt T_1\right) \\
\\
0 & \left(T_1 \leq t \lt T\right)
\end{cases}
\end{eqnarray}

In the formula to find the average value, if we separate each term, we get the following formula.

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{T} \displaystyle \int_{0}^{T}|v(t)|dt\\
\\
&=& \displaystyle\frac{1}{T} \left(\displaystyle \int_{0}^{T_1}|v(t)|dt+\displaystyle \int_{T_1}^{T}|v(t)|dt\right )\\
\end{eqnarray}

Since the pulse wave is zero in the region \(T_1 \leq t \lt T\), the above equation becomes:

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{T} \left(\displaystyle \int_{0}^{T_1}|v(t)|dt+\displaystyle \int_{T_1}^{T}0dt\right )\\
\\
&=&\displaystyle\frac{1}{T} \displaystyle \int_{0}^{T_1}|v(t)|dt\\
\end{eqnarray}

In the above equation, when substituting \(v(t)=V_M\), the following formula is obtained.

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{T} \displaystyle \int_{0}^{T_1}V_Mdt\\
\\
&=& \displaystyle\frac{1}{T} V_M\displaystyle \int_{0}^{T_1}1dt\\
\\
&=& \displaystyle\frac{1}{T}V_M \left[t\right]_{0}^{T_1}\\
\\
&=& \displaystyle\frac{1}{T}V_M \left(T_1-0\right)\\
\\
&=&\displaystyle\frac{T_1}{T}V_M
\end{eqnarray}

Therefore, the average value \(V_{AVE}\) of the pulse wave is as follows.

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{T_1}{T}V_M
\end{eqnarray}

Summary

In this article, the following information on the “Pulse Wave“ was explained.

• How to obtain RMS Value and Average Value

Thank you for reading.

Copyright © 2026 Electrical Information All Rights Reserved.

• How to obtain RMS Value, Average Value, Form Factor, and Crest Factor

RMS Value, Average Value, Form Factor, and Crest Factor of Half-Wave Rectified Sine-Wave

The first illustration shows the root mean square (RMS) value, average value, form factor, and crest factor of a half-wave rectified sine-wave (with a maximum value of \(V_M\) and a period of \(T\)).

Now, let’s explain how each of these values are calculated (we strive to include as many intermediate steps as possible).

Waveform of a Half-Wave Rectified Sine-Wave

To find the root mean square (RMS) value and average value of a half-wave rectified sine-wave, we first need to express the half-wave rectified sine-wave as a formula.

The half-wave rectified sine-wave has different formulas for the red region \(\left(0 \leq t \lt \displaystyle \frac{T}{2}\right)\) and the blue region \(\left(\displaystyle \frac{T}{2} \leq t \lt T\right)\).

Let’s calculate the formulas for the red region and blue region.

Red Region

The region “\(0 \leq t \lt \displaystyle \frac{T}{2}\)” corresponds to the red region.

The red region is part of the waveform of \(V_M\sin{{\omega}t}\).

Therefore, the red region can be represented by the following formula.

\begin{eqnarray}
v(t)=V_M\sin{{\omega}t}
\end{eqnarray}

Blue Region

The region “\(\displaystyle \frac{T}{2} \leq t \lt T\)” corresponds to the blue region.

The blue region is zero.

Therefore, the blue region can be represented by the following formula.

\begin{eqnarray}
v(t)=0
\end{eqnarray}

By combining the red region and the blue region, the half-wave rectified sine-wave can be represented by the following formula.

\begin{eqnarray}
v(t) = \begin{cases}
V_M\sin{{\omega}t} & \left(0 \leq t \lt \displaystyle \frac{T}{2}\right) \\
\\
0 & \left(\displaystyle \frac{T}{2} \leq t \lt T\right)
\end{cases}
\end{eqnarray}

Next, we will calculate the RMS value, average value, form factor, and crest factor of the half-wave rectified sine-wave. However, using the above formula can make the calculation slightly complex.

Therefore, to simplify the calculation this time, we will convert the time axis (horizontal axis \(t\)) to the phase axis (horizontal axis \({{\omega}t}\)).

When the time axis (horizontal axis \(t\)) is converted to the phase axis (horizontal axis \({{\omega}t}\)), the half-wave rectified sine-wave becomes the following formula.

\begin{eqnarray}
v({\omega}t) = \begin{cases}
V_M\sin{{\omega}t} & \left(0 \leq {\omega}t \lt \pi\right) \\
\\
0 & \left(\pi \leq {\omega}t \lt 2\pi\right)
\end{cases}
\end{eqnarray}

RMS Value of a Half-Wave Rectified Sine-Wave

The root mean square (RMS) value \(V_{RMS}\) of a waveform \(v({{\omega}t})\) is the square root of the mean of the square of \(v({{\omega}t})\). It can be represented by the following formula:

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{2\pi} \displaystyle \int_{0}^{2\pi}v({{\omega}t})^2d({{\omega}t}})
\end{eqnarray}

The red region is \(0 \leq {\omega}t \lt \pi\)“, and the blue region is “\(\pi \leq {\omega}t \lt 2\pi\)“.

In the formula for finding the absolute value, if we separate the red region and the blue region, we get the following formula.

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{2\pi} \left(\displaystyle \int_{0}^{\pi}v({{\omega}t})^2d({\omega}t)+\displaystyle \int_{\pi}^{2\pi}v({{\omega}t})^2d({\omega}t)\right )}\\
\\
&=& \sqrt{\displaystyle\frac{1}{2\pi} \left(\displaystyle \int_{0}^{\pi}(V_M\sin{{\omega}t})^2d({\omega}t)+\displaystyle \int_{\pi}^{2\pi}0^2d({\omega}t)\right )}\\
\\
&=& \sqrt{\displaystyle\frac{1}{2\pi}\displaystyle \int_{0}^{\pi}(V_M\sin{{\omega}t})^2d({\omega}t)}
\end{eqnarray}

The formula for the red region is \(V_M\sin{{\omega}t}\), and the formula for the blue region is \(-V_M\sin{{\omega}t}\), but in the calculation of the effective value, we square it, so we can combine the above formulas. If we combine them, we get the following formula.

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{\pi} \displaystyle \int_{0}^{2\pi}(V_M\sin{{\omega}t})^2d({\omega}t)}
\end{eqnarray}

In the above formula, if you let

\begin{eqnarray}
X &=& \displaystyle \int_{0}^{\pi}(V_M\sin{{\omega}t})^2 d({\omega}t)
\end{eqnarray}

then the root mean square (RMS) value \(V_{RMS}\) of a waveform \(v({{\omega}t})\) can be represented by the following formula:

\begin{eqnarray}
V_{RMS} &=& \sqrt{\displaystyle\frac{1}{2\pi} X}
\end{eqnarray}

Next, we calculate the value of \(X\).

\begin{eqnarray}
X &=& \displaystyle \int_{0}^{\pi}v({{\omega}t})^2 d({\omega}t)\\
\\
&=& \displaystyle \int_{0}^{\pi}{V_M}^2{\sin}^2{\omega}t d({\omega}t)\\
\\
&=& {V_M}^2\displaystyle \int_{0}^{\pi}{\sin}^2{\omega}t d({\omega}t)\\
\\
&=& {V_M}^2\displaystyle \int_{0}^{\pi}\displaystyle \frac{1-{\cos2{\omega}t}}{2} d({\omega}t)\\
\\
&=& {V_M}^2 \left[\displaystyle\frac{1}{2}{{\omega}t}-\frac{1}{4}{\sin2{\omega}t} \right]_{0}^{\pi}\\
\\
&=& {V_M}^2 \left( \displaystyle\frac{1}{2}×{\pi}-\frac{1}{4}{\sin{2\pi}}\right)\\
\\
&=& \displaystyle\frac{{V_M}^2 {\pi}}{2}
\end{eqnarray}

Therefore, the root mean square (RMS) value \(V_{RMS}\) of a waveform \(v({{\omega}t})\) becomes:

\begin{eqnarray}
V_{RMS} &=&\sqrt{\displaystyle\frac{1}{2\pi} X}\\
\\
&=&\sqrt{\displaystyle\frac{1}{2\pi} \displaystyle\frac{{V_M}^2 {\pi}}{2}}\\
\\
&=&\displaystyle\frac{1}{2}V_M
\end{eqnarray}

Average Value of a Half-Wave Rectified Sine-Wave

The average value \(V_{AVE}\) of a waveform \(v({{\omega}t})\) is the average value of the absolute value \(|v({{\omega}t})|\) of \(v({{\omega}t})\) and can be represented by the following formula:

This formula for calculating the average value uses the absolute value \(|v({{\omega}t})|\) of \(v({{\omega}t})\).

Therefore, if there is a negative region in the waveform, it needs to be converted so that the negative region becomes positive.

In the case of a half-wave rectified sine-wave, there is no negative region, so the formula for \(v({{\omega}t})\) and the formula for the absolute value \(|v({{\omega}t})|\) of \(v({{\omega}t})\) become the same.

Therefore, the absolute value \(|v({{\omega}t})|\) of \(v({{\omega}t})\) can be represented by the following formula.

\begin{eqnarray}
|v({{\omega}t})| = \begin{cases}
V_M\sin{{\omega}t} & \left(0 \leq {\omega}t \lt \pi\right) \\
\\
0 & \left(\pi \leq {\omega}t \lt 2\pi\right)
\end{cases}
\end{eqnarray}

In the formula to find the average value, if we separate each term, we get the following formula.

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{2\pi} \displaystyle \int_{0}^{2\pi}|v({\omega}t)|d({{\omega}t})\\
\\
&=& \displaystyle\frac{1}{2\pi}\left(\displaystyle \int_{0}^{\pi}|v({\omega}t)|d({{\omega}t})+ \int_{\pi}^{2\pi}|v({\omega}t)|d({{\omega}t}) \right)\\
\end{eqnarray}

Since the half-wave rectified sine-wave is zero in the region \(\pi \leq {\omega}t \lt 2\pi\), the above equation becomes:

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{2\pi}\displaystyle \int_{0}^{\pi}|v({\omega}t)|d({{\omega}t})
\end{eqnarray}

If we calculate the above formula, the average value \(V_{AVE}\) of the half-wave rectified sine-wave becomes:

\begin{eqnarray}
V_{AVE} &=& \displaystyle\frac{1}{2\pi}\displaystyle \int_{0}^{\pi}V_M\sin{{\omega}t}d({{\omega}t})\\
\\
&=& \displaystyle\frac{1}{2\pi}V_M\left[-\cos{\omega}t \right]_{0}^{\pi}\\
\\
&=& \displaystyle\frac{V_M}{2\pi}\left(-\cos\pi + \cos0 \right)\\
\\
&=& \displaystyle\frac{1}{\pi}V_M
\end{eqnarray}

Form Factor of a Half-Wave Rectified Sine-Wave

The form factor can be represented by the following formula:

\begin{eqnarray}
\mbox{Form Factor} &=& \displaystyle\frac{\mbox{Root Mean Square Value}~~V_{RMS}}{\mbox{Average Value}~~V_{AVE}}
\end{eqnarray}

Since we’ve already calculated the root mean square (RMS) value \(V_{RMS}\) and average value \(V_{AVE}\) of the sine wave, we can substitute these values into the formula to calculate the form factor of the half-wave rectified sine-wave.

\begin{eqnarray}
\mbox{Form Factor} = \displaystyle\frac{\displaystyle\frac{1}{2}V_M}{\displaystyle\frac{1}{\pi}V_M} = \displaystyle\frac{\pi}{2}
\end{eqnarray}

Maximum Value of a Half-Wave Rectified Sine-Wave

As can be seen from the waveform, the maximum value \(V_{PEAK}\)​ of a half-wave rectified sine-wave is represented by the following value:

\begin{eqnarray}
\mbox{Maximum Value}~~V_{PEAK}=V_M
\end{eqnarray}

Crest Factor of a Half-Wave Rectified Sine-Wave

The Crest factor can be represented by the following formula:

\begin{eqnarray}
\mbox{Crest Factor} &=& \displaystyle\frac{\mbox{Peak Value}~~V_{PEAK}}{\mbox{Root Mean Square Value}~~V_{RMS}}
\end{eqnarray}

Since we’ve already calculated the root mean square (RMS) value \(V_{RMS}\) and peak value \(V_{PEAK}\) of the sine wave, we can substitute these values into the formula to calculate the peak factor (crest factor) of the half-wave rectified sine-wave.

The Crest factor of the half-wave rectified sine-wave is represented by the following value:

\begin{eqnarray}
\mbox{Crest Factor} = \displaystyle\frac{V_M}{\displaystyle\frac{1}{2}V_M} = 2
\end{eqnarray}

Summary

In this article, the following information on the “Half-Wave Rectified Sine-Wave“ was explained.

• How to obtain RMS Value, Average Value, Form Factor, and Crest Factor

Thank you for reading.

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