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Difficulty: MediumParallel Circuits

An aircraft DC power distribution bus supplies 36 V36\text{ V} across two parallel branch circuits. The first branch contains a resistive load of 12 Ω12\ \Omega, and an ammeter at the main bus measures a total circuit current of 5 A5\text{ A}. What is the resistance of the load in the second branch?

  1. A
    7.2 Ω7.2\ \Omega
  2. 18 Ω18\ \OmegaAnswer
  3. C
    12 Ω12\ \Omega
  4. D
    72 Ω72\ \Omega

Answer

The resistance of the second branch load is 18 Ω18\ \Omega.
In a parallel circuit, each branch receives the full source voltage of 36 V36\text{ V}. Using Ohm's Law, the first branch draws I1=36 V/12 Ω=3 AI_1 = 36\text{ V} / 12\ \Omega = 3\text{ A}. Subtracting this from the total current of 5 A5\text{ A} gives a second branch current of I2=2 AI_2 = 2\text{ A}. Dividing the 36 V36\text{ V} bus voltage by 2 A2\text{ A} yields a resistance of 18 Ω18\ \Omega for the second branch.

Step-by-Step Solution

1
Calculate the current flowing through the first branch using Ohm's Law (I1=V/R1I_1 = V / R_1).
I1=36 V/12 Ω=3 AI_1 = 36\text{ V} / 12\ \Omega = 3\text{ A}
In a parallel circuit, the full supply voltage of 36 V36\text{ V} is present across every parallel branch.
2
Determine the current flowing through the second branch using Kirchhoff's Current Law (I2=Itotal−I1I_2 = I_{\text{total}} - I_1).
I2=5 A−3 A=2 AI_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}
Total circuit current equals the sum of all individual branch currents in a parallel circuit.
3
Calculate the resistance of the second branch using Ohm's Law (R2=V/I2R_2 = V / I_2).
R2=36 V/2 A=18 ΩR_2 = 36\text{ V} / 2\text{ A} = 18\ \Omega
Dividing the common branch voltage by the branch current yields the branch resistance.

Key Concept

Parallel branch current relationships and individual component resistance calculations using Ohm's Law and Kirchhoff's Current Law.

Alternative Method

Find the total circuit equivalent resistance first: Req=36 V/5 A=7.2 ΩR_{\text{eq}} = 36\text{ V} / 5\text{ A} = 7.2\ \Omega. Then apply the parallel resistance formula: 1/R2=1/Req−1/R1=1/7.2−1/12=5/36−3/36=2/36=1/181/R_2 = 1/R_{\text{eq}} - 1/R_1 = 1/7.2 - 1/12 = 5/36 - 3/36 = 2/36 = 1/18, which gives R2=18 ΩR_2 = 18\ \Omega.
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Topics

Parallel Circuits

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