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Difficulty: HardParallel Circuits

An aircraft auxiliary power bus supplies a constant 24 V24\text{ V} DC across three parallel branch circuits. Branch 1 draws a current of 3 A3\text{ A}, Branch 2 contains an unknown resistor R2R_2, and Branch 3 contains a 12 Ω12\ \Omega resistor. If the total electrical power consumed by all three branches combined is 192 W192\text{ W}, what is the resistance of resistor R2R_2?

  1. A
    4 Ω4\ \Omega
  2. 8 Ω8\ \OmegaAnswer
  3. C
    6 Ω6\ \Omega
  4. D
    3 Ω3\ \Omega

Answer

The resistance of resistor R2R_2 is 8 Ω8\ \Omega.
The correct resistance is 8 Ω8\ \Omega. First, finding total current from the total power gives Itotal=192 W24 V=8 AI_{\text{total}} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}. Since the voltage across each parallel branch is 24 V24\text{ V}, Branch 3 draws I3=24 V12 Ω=2 AI_3 = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}. Subtracting the known branch currents (3 A3\text{ A} and 2 A2\text{ A}) from the total current (8 A8\text{ A}) leaves 3 A3\text{ A} flowing through Branch 2. Applying Ohm's Law yields R2=24 V3 A=8 ΩR_2 = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega.

Step-by-Step Solution

1
Calculate the total circuit current using total power and supply voltage.
Itotal=PtotalV=192 W24 V=8 AI_{\text{total}} = \frac{P_{\text{total}}}{V} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}
In any DC circuit, total power is the product of total voltage and total current (P=V×IP = V \times I).
2
Calculate the current drawn by Branch 3 using Ohm's Law.
I3=VR3=24 V12 Ω=2 AI_3 = \frac{V}{R_3} = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}
In a parallel circuit, full supply voltage (24 V24\text{ V}) appears across every parallel branch.
3
Determine the current flowing through Branch 2 using Kirchhoff's Current Law for parallel circuits.
I2=Itotal−I1−I3=8 A−3 A−2 A=3 AI_2 = I_{\text{total}} - I_1 - I_3 = 8\text{ A} - 3\text{ A} - 2\text{ A} = 3\text{ A}
Total current in a parallel circuit equals the sum of all individual branch currents (Itotal=I1+I2+I3I_{\text{total}} = I_1 + I_2 + I_3).
4
Calculate the resistance of R2R_2 using Ohm's Law.
R2=VI2=24 V3 A=8 ΩR_2 = \frac{V}{I_2} = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega
Branch resistance is found by dividing the branch voltage by the branch current (R=VIR = \frac{V}{I}).

Key Concept

Parallel circuit rules: constant branch voltage (Vtotal=V1=V2=V3V_{\text{total}} = V_1 = V_2 = V_3), additive branch currents (Itotal=∑InI_{\text{total}} = \sum I_n), and power relationships (P=V×IP = V \times I).

Alternative Method

Calculate individual branch powers directly: Branch 1 consumes P1=V×I1=24 V×3 A=72 WP_1 = V \times I_1 = 24\text{ V} \times 3\text{ A} = 72\text{ W}. Branch 3 consumes P3=V2R3=57612=48 WP_3 = \frac{V^2}{R_3} = \frac{576}{12} = 48\text{ W}. The power dissipated in Branch 2 is P2=Ptotal−P1−P3=192 W−72 W−48 W=72 WP_2 = P_{\text{total}} - P_1 - P_3 = 192\text{ W} - 72\text{ W} - 48\text{ W} = 72\text{ W}. Using the power-resistance formula, R2=V2P2=57672=8 ΩR_2 = \frac{V^2}{P_2} = \frac{576}{72} = 8\ \Omega.
Estimated Time:1m 30s
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Parallel Circuits

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