Easy Current Division Rule Formula and Example

Current division rule is used in the current divider circuit. If the series circuit acts as a voltage divider circuit and uses voltage division rule, the parallel circuit acts as a current divider circuit and uses current division rule. In a parallel circuit, the voltage across each branch has the same value.

Current Division Rule

The current flowing each branch depends on the resistance of the branch. You can observe the circuit below.

current division rule

Now let’s remember what Kirchhoff’s Current Law tells us when looking at the circuit above. The Kirchhoff’s Current Law states that:

The algebraic sum of all currents entering a junction is equal to the algebraic sum of all current leaving a junction.

Or

The algebraic sum of all currents entering and leaving a junction equals to zero.

The current entering junction a is equal to the current leaving junction b. Between junction a and b, the current is divided into two branches, I1 and I2 with resistances R1 and R2 respectively with V as the voltage drops across the both resistances (R1 and R2).

If the main characteristic of a series circuit is the currents flowing through each resistor have the same value (I1=I2=I3…), then the main characteristic of a parallel circuit is the voltage drops across each branch have the same value (V1=V2=V3…).

Current Division Rule Formula

We already know that:

    \begin{align*}V=I\times R\end{align*}

Or

(1)   \begin{align*}I=\frac{V}{R}\end{align*}

Then we can find the current for each branch:

    \begin{align*}I_{1}=\frac{V}{R_{1}}\\I_{2}=\frac{V}{R_{2}}\end{align*}

The equivalent resistance for parallel resistor can be calculated from:

(2)   \begin{align*}R=\frac{R_{1}R_{2}}{R_{1}+R_{2}}\end{align*}

Substituting equation (2) into (1) produces

(3)   \begin{align*}I=\frac{V(R_{1}+R_{2})}{R_{1}R_{2}}\end{align*}

And

(4)   \begin{align*}V=I_{1}R_{1}=I_{2}R_{2}\end{align*}

Substituting equation (4) into (3) we get the equation:

(5)   \begin{align*}I&=\frac{I_{1}R_{1}(R_{1}+R_{2})}{R_{1}R_{2}}\\&=\frac{I_{1}}{R_{2}}(R_{1}+R_{2})\end{align*}

If we use the I2 and R2, then:

(6)   \begin{align*}I&=\frac{I_{2}R_{2}(R_{1}+R_{2})}{R_{1}R_{2}}\\&=\frac{I_{1}}{R_{1}}(R_{1}+R_{2})\end{align*}

The equations we get in (5) and (6) can be used to find the value of I1 and I2, where:

    \begin{align*}&I_{1}=I\frac{R_{2}}{R_{1}+R_{2}}\\&\mbox{and}\\&I_{2}=I\frac{R_{1}}{R_{1}+R_{2}}\end{align*}

The current division rule states:

The current flowing in parallel branches is equal to the ratio of opposite resistance to the total resistance and multiplied by total current.

Current Division Rule Example

1. We have the simple circuit below with only two branches.

current division rule

The equivalent resistance will be:

    \begin{align*}R&=\frac{R_{1}R_{2}}{R_{1}+R_{2}}\\&=\frac{3\times6}{3+6}\\&=2 \Omega\end{align*}

The total current will be

    \begin{align*}I&=\frac{V}{R}\\&=\frac{10}{2}\\&=5A\end{align*}

Now we calculate I1:

    \begin{align*}I_{1}&=I\frac{R_{2}}{R_{1}+R_{2}}\\&=\frac{5\times6}{3+6}\\&=\frac{10}{3}A\end{align*}

Then we calculate I2:

    \begin{align*}I_{2}&=I\frac{R_{1}}{R_{1}+R_{2}}\\&=\frac{5\times3}{3+6}\\&=\frac{5}{3}A\end{align*}

This clarifies that

    \begin{align*}I&=I_{1}+I_{2}\\5A&=\frac{10}{3}+\frac{5}{3}\\5A&=5A\end{align*}

2. Now we will find the current values for a circuit which has three branches.

current division rule

We start by calculating the equivalent resistance:

    \begin{align*}\frac{1}{R}&=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\\&=\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\\&=\frac{12}{36}+\frac{6}{36}+\frac{4}{36}\\\frac{1}{R}&=\frac{22}{36}\\R&=\frac{36}{22}=\frac{18}{11}\Omega\end{align*}

The total current will be

    \begin{align*}I&=\frac{V_{S}}{R}\\&=\frac{36}{18/11}\\&=22A\end{align*}

The voltage drop across the is

    \begin{align*}V&=I\times R\\&=22\times\frac{18}{11}\\&=36V\end{align*}

current division rule

Since parallel circuit shares the same voltage drop across the parallel resistors then:

    \begin{align*}V=V_{1}=V_{2}=V_{3}=36\end{align*}

Now we will find the current for each branch:

    \begin{align*}I_{1}&=\frac{V}{R_{1}}=\frac{36}{3}=12A\\I_{2}&=\frac{V}{R_{2}}=\frac{36}{6}=6A\\I_{3}&=\frac{V}{R_{3}}=\frac{36}{9}=4A\end{align*}

There is another easy method to use if we have the total current already. From the calculation above we know that the I = 22 A.

We can easily find the current for each branch using:

    \begin{align*}I_{1}&=I\times\frac{R}{R_{1}}=22\times\frac{18/11}{3}=12A\\I_{2}&=I\times\frac{R}{R_{2}}=22\times\frac{18/11}{6}=6A\\I_{3}&=I\times\frac{R}{R_{3}}=22\times\frac{18/11}{9}=4A\end{align*}

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