Current Divider Worked Example: Solving for an Unknown Resistor

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This worked example shows how to solve for an unknown resistor in a parallel current divider circuit. It also demonstrates that circuit intuition is faster than a direct application of algebra.

Given a parallel circuit consisting of 4 components, solve for R2 given the R2 branch current of 5 A:

  • 15 A constant current source
  • R1 = 2 Ω
  • R2 = unknown
  • R3 = 4 Ω

:pushpin: Canonical Article: Current Divider Formula and Derivation
:blue_book: Learning Companion (Q&A): Explore All Questions

Worked solution

This is a reminder to stop and study the circuit. The solution is easier if we split the constant current source into two pieces as shown in Figure 1:

  • To the left is a 5A source along with R2
  • To the right is the remaining 10 A along with R1 and R3

Figure 1: Hand sketch of the circuit with the constant current source split into two.
Figure 1: Hand sketch of the circuit with the constant current source split into two.1536×864 32.6 KB

Figure 1: Hand sketch of the circuit with the constant current source split into two.

Conceptually, we have constructed a self-contained circuit on the left side of the red line with a 5 A source feeding 5A to the R2 load. However, the two sides of the circuit are still connected and so share the same node voltage.

We then solve for R2 using these steps:

  1. With 5 A flowing through R2, we observe that R1 and R3 carry the remaining 10 A.

  2. R1 in parallel with R3 is 1.333 Ω.

  3. The voltage drop is the same across all parallel elements: 10 A x 1.33 Ω = 13.33 V

  4. The new R2 value is 13.33 V / 5 A = 2.67 Ω

Tech Tip: Recall that constant current sources may be operated in parallel. This fact allows us to conceptually split the current between two or more constant current sources. This little trick can help us visualize the circuit’s operation and help us solve the equation faster.

Techniques such as this are central to the engineering mindset. Don’t automatically jump to the equations if there is an easier way to solve the problem. It also helps us verify the problem as we have a deeper understanding of the interaction between elements.

Verify

We will use the current divider rule to verify the current through R2.

  1. Total parallel resistance with R1 || R2 || R3 = 0.89 Ω

  2. Using the generalized equation:

I_{R_X} = I_{Total} \left( \frac{R_{Total}}{R_X} \right)

I_{R_2} = 15 \left( \frac{0.89}{2.67} \right) = 5 A :white_check_mark: (verified correct)

Tech Tip: In the previous example we used shorthand notation to show resistors in parallel. The double bars symbol (||) in statements such as R1 || R2 || R3 is a compact way to describe circuit connections. You will appreciate this convenience as you continue your circuit analysis.

Another pattern recognition problem

Given a parallel circuit consisting of 4 components, solve for R2 given the R2 branch current of 15A:

  • 15 A constant current source:
  • R1 = 2 Ω
  • R2 = unknown
  • R3 = 4 Ω

In this case, all 15 A flows through R2. Consequently, R2 is 0 Ω.

Tech Tip: My old instructor loved to assign type of problem. If you recognize the pattern, it takes seconds to find the solution with zero calculations. If not the algebra isn’t fun.

:writing_hand: Article by Aaron Dahlen, LCDR USCG (Ret.), Application Engineer at DigiKey. Author bio.