When resistors are connected in parallel, how much current flows through each resistor? The rule that lets us calculate these currents from the ratio of the resistance values is the current divider rule.
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 \({\mathrm{A}}\)?
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 \({\mathrm{A}}\)?
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.
Current Divider Rule
The current divider rule is used to calculate the current through each resistor in a parallel circuit.
Let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of the resistors connected in parallel, and let \(I{\mathrm{[A]}}\) be the current entering the parallel circuit. The current \(I_N{\mathrm{[A]}}\) through resistor \(R_N{\mathrm{[{\Omega}]}}\) is given by the following equation.
\begin{eqnarray}
I_N=\frac{R}{R_N}×I{\mathrm{[A]}}\tag{25}
\end{eqnarray}
Voltage Divider Rule
The voltage divider rule is used to calculate the voltage across each resistor in a series circuit.
Let \(R{\mathrm{[{\Omega}]}}\) be the equivalent resistance of the resistors connected in series, and let \(V{\mathrm{[V]}}\) be the voltage across the series circuit. The voltage \(V_N{\mathrm{[V]}}\) across resistor \(R_N{\mathrm{[{\Omega}]}}\) is given by the following equation.
\begin{eqnarray}
V_N=\frac{R_N}{R}×V{\mathrm{[V]}}\tag{26}
\end{eqnarray}
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.
Related Article
For more details on the voltage divider rule, see the following article. 続きを見る
Voltage Divider Rule: Formula, Derivation, and Examples
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.