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Difficulty: MediumParallel Circuits

A 72 V72\text{ V} direct-current (DC) power supply energizes an industrial sensor interface consisting of two resistors connected in parallel. The first resistor (R1R_1) has a resistance of 24 Ω24\ \Omega. If the power supply delivers a total current of 5 A5\text{ A} to the circuit, what is the resistance of the second resistor (R2R_2)?

  1. A
    9.6 Ω9.6\ \Omega
  2. 36 Ω36\ \OmegaAnswer
  3. C
    14.4 Ω14.4\ \Omega
  4. D
    28.8 Ω28.8\ \Omega

Answer

36 Ω36\ \Omega
In a parallel circuit, the voltage across every parallel branch is identical to the source voltage (72 V72\text{ V}). The current through the known resistor is I1=72 V24 Ω=3 AI_1 = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}. Since the total current entering the parallel junction is 5 A5\text{ A}, the remaining current flowing through the second branch must be I2=5 A−3 A=2 AI_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}. Using Ohm's law for the second branch yields R2=72 V2 A=36 ΩR_2 = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega.

Step-by-Step Solution

1
Calculate the current flowing through the first branch resistor (R1R_1)
I1=VR1=72 V24 Ω=3 AI_1 = \frac{V}{R_1} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}
In a parallel circuit, each branch experiences the full supply voltage of 72 V72\text{ V}.
2
Determine the current flowing through the second branch (I2I_2)
I2=Itotal−I1=5 A−3 A=2 AI_2 = I_{\text{total}} - I_1 = 5\text{ A} - 3\text{ A} = 2\text{ A}
By Kirchhoff's Current Law, total current in a parallel circuit equals the sum of the individual branch currents.
3
Calculate the resistance of the second resistor (R2R_2) using Ohm's law
R2=VI2=72 V2 A=36 ΩR_2 = \frac{V}{I_2} = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega
Applying Ohm's law (R=VIR = \frac{V}{I}) with the branch voltage and branch current gives the branch resistance.

Key Concept

Parallel branch voltage equality, Kirchhoff's Current Law, and Ohm's law branch calculations
Estimated Time:1m 0s
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Topics

Parallel Circuits

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