# Current Divider

## Current Dividers

Current dividers are the inverse of voltage dividers.  Voltage dividers work with series circuits where current remains constant; any parallel components must be combined before the voltage divider equation works.  In contrast, current dividers work with parallel circuits where the voltage is the same across all components and any series components must be combined before applying the current divider equation.

The general equation for a current divider is given below and illustrated in Figure 1.  Io is the measured current, Is is the source current (or the current entering the node, Ro is the resistance of the path Io flows through, and RT is the equivalent resistance of the circuit.

### General Current Divider Equation:

$I_{0} = I_{s} ( \frac{R_{t}}{R_{0}})$

Notice that this equation is nearly identical to the voltage divider equation, except that RT and Ro are inverted.  RT is in the numerator because current dividers work exclusively with parallel resistors.  Therefore, RT is always smaller than any of the individual resistors because of the nature of the parallel resistance formula.  The result is that Io is always less than or equal to Is.

The current divider equation is derived in the same fashion as the voltage divider equation.  Consider Figure 2.  All three components are connected in parallel, meaning all three have the same voltage across them.

$V_{s} = V_{1} = V_{2}$

Use Ohm’s law to place the equation above in terms of resistance and current.

Ohm’s Law: V = IR

$\rightarrow I_{s} = (R_{1}||R_{2}) = I_{1}R_{1} = I_{2}R_{2}$

Now use these equations to find I1 and I2.

$I_{1} = I_{s} ( \frac{ R_{1} || R_{2} } {R_{1} } )$                  $I_{2} = I_{s} ( \frac{R_{1} || R_{2} }{ R_{2} } )$

### Current Divider Example

Example: Find the current Io for the circuit in Figure 3.

Solution: This problem requires two current dividers to solve.  The current first splits between R1 and the rest of the circuit and then again between R3 and R4.  Because Io is part of the second split, the current through R2 must be found first.  Remember, though, that the current through R2 also passes through R3 and R4.  The current divider equation cannot be applied without taking the resistance of the entire path into consideration.

Combine R2, R3, and R4 before applying the first current divider equation.

$R_{234} = R_{2} + R_{3} || R_{4} = 10 + \frac{6 \cdot 3}{6 + 3} = 12 \Omega$

$I_{2} = I_{s} ( \frac{ R_{1} || R_{234} } {R_{234}} ) = 4( \frac{ \frac{36 \cdot 12}{36 + 12}}{12} ) = 4( \frac{9}{12}) = 3A$

The equation above shows that the 36 Ω resistor takes 1 A of the 4 A available to the circuit.  Therefore, the second current divider must use 3 A as the source current because that is the amount of current flowing into the node.  Also, do not try to incorporate R2 into the second equation.  The current splits between R3 and R4; R2 has nothing to do with where the current flows once it reaches the node.

$I_{0} = I_{s} ( \frac{ R_{3} || R_{4} } {R_{4}} ) = 3( \frac{ \frac{6 \cdot 3}{6 + 3}}{3} ) = 3( \frac{2}{3}) = 2A$

Thanks to Ryan Eatinger (reatinge@ksu.edu) for the contribution of this post.