Current Divider Examples Explained

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.

current-flow

[adrotate banner=”1″]
The general equation for a current divider is given below and illustrated in Figure 1. I_o is the measured current, I_s is the source current (or the current entering the node, R_o is the resistance of the path I_o flows through, and R_T 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 R_T and R_o are inverted. R_T is in the numerator because current dividers work exclusively with parallel resistors. Therefore, R_T is always smaller than any of the individual resistors because of the nature of the parallel resistance formula. The result is that I_o is always less than or equal to I_s.

parallel-current-flow

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 I_o for the circuit in Figure 3.

example-current-divider-circuit

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 I_o 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.

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

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

example-current-divider-circuit-current-flow

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.

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

Current Dividers

General Current Divider Equation:

Current Divider Example

Leave a Reply Cancel reply