Current Divider Guided Learning Q6 – Worked example

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This article is part of a guided learning series exploring the theoretical and practical aspects of resistors in parallel.

:pushpin: Canonical Article: Current Divider Formula and Derivation
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You are reading: Question 6

Worked example for the current divider

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

Solution

This is a counterexample where it is best to avoid the current divider equations. Instead, we will 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.

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.

Conceptually, this allows us to disregard the circuit to the left of the red line. With a 5 A source and a 5A load as R2, there is no current flowing across the red vertical line. 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 1.33 Ω.

  3. The voltage drop is the same across all parallel elements: 10 A x 1.33 Ω = 13.3 VDC

  4. The new R2 value is 13.3 V / 5 A = 2.67 Ω

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 convince as you continue your circuit analysis.

A fun counterexample for the current divider

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 we have all 15 A flowing through R2 Consequently R2 is 0 Ω.

Tech Tip: My old instructor loved to assign type of problem. If you are clever and recognize the pattern, it takes second to find the solution with zero calculations. If you don’t, you are in for a long struggle.

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