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Difficulty: MediumParallel Circuits

A direct-current (DC) power supply delivers a constant 90 V90\text{ V} to two resistors connected in parallel. An ammeter measures the total circuit current leaving the source as 5 A5\text{ A}. If the resistance of the first resistor (R1R_1) is 30 Ω30\ \Omega, what is the resistance of the second resistor (R2R_2)?

  1. A
    12 Ω12\ \Omega
  2. B
    18 Ω18\ \Omega
  3. C
    36 Ω36\ \Omega
  4. 45 Ω45\ \OmegaAnswer

Answer

The resistance of the second resistor (R2R_2) is 45 Ω45\ \Omega.
In any parallel circuit, the voltage across each branch is identical to the source voltage (90 V90\text{ V}). The current in the first branch is I1=90 V/30 Ω=3 AI_1 = 90\text{ V} / 30\ \Omega = 3\text{ A}. According to Kirchhoff's Current Law, the total current equals the sum of branch currents (Itotal=I1+I2I_{\text{total}} = I_1 + I_2). Therefore, the current in the second branch is I2=5 A−3 A=2 AI_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}. Applying Ohm's Law to the second branch gives R2=90 V/2 A=45 ΩR_2 = 90\text{ V} / 2\text{ A} = 45\ \Omega.

Step-by-Step Solution

1
Determine the current flowing through the first parallel branch (I1I_1)
I1=VR1=90 V30 Ω=3 AI_1 = \frac{V}{R_1} = \frac{90\text{ V}}{30\ \Omega} = 3\text{ A}
In a parallel circuit, each branch experiences the full source voltage (90 V90\text{ V}).
2
Calculate the current flowing through the second branch (I2I_2) using Kirchhoff's Current Law
I2=Itotal−I1=5 A−3 A=2 AI_2 = I_{\text{total}} - I_1 = 5\text{ A} - 3\text{ A} = 2\text{ A}
The sum of individual branch currents equals the total current supplied by the source.
3
Calculate the resistance of the second resistor (R2R_2) using Ohm's Law
R2=VI2=90 V2 A=45 ΩR_2 = \frac{V}{I_2} = \frac{90\text{ V}}{2\text{ A}} = 45\ \Omega
Ohm's Law defines resistance as the voltage drop across the component divided by the current through it.

Key Concept

Parallel branch voltage equality and Kirchhoff's Current Law
Estimated Time:1m 15s
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Topics

Parallel Circuits

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