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Difficulty: HardElectricity and Magnetism

A direct-current circuit is powered by a 12 V12\text{ V} battery connected in series with a 3 Ω3\,\Omega resistor, followed by two 6 Ω6\,\Omega resistors connected in parallel with each other. If one of the 6 Ω6\,\Omega parallel resistors burns out and creates an open circuit, what is the total electrical power consumed by the remaining active circuit?

  1. A
    24 W24\text{ W}
  2. B
    8 W8\text{ W}
  3. 16 W16\text{ W}Answer
  4. D
    48 W48\text{ W}

Answer

16 W16\text{ W}
When one of the parallel 6 Ω6\,\Omega resistors opens, the parallel network is reduced to the single remaining 6 Ω6\,\Omega resistor. This resistor is in series with the 3 Ω3\,\Omega resistor, giving a total equivalent resistance of R=3 Ω+6 Ω=9 ΩR = 3\,\Omega + 6\,\Omega = 9\,\Omega. Applying the power formula P=V2RP = \frac{V^2}{R} with a 12 V12\text{ V} battery gives P=1449=16 WP = \frac{144}{9} = 16\text{ W}.

Step-by-Step Solution

1
Determine the new circuit configuration after component failure
A single 3 Ω3\,\Omega resistor in series with the remaining 6 Ω6\,\Omega resistor
When one parallel branch burns out into an open circuit, no current flows through that branch, leaving the remaining parallel resistor in direct series with the first resistor.
2
Calculate the total equivalent resistance (RtotalR_{\text{total}})
Rtotal=3 Ω+6 Ω=9 ΩR_{\text{total}} = 3\,\Omega + 6\,\Omega = 9\,\Omega
Resistances in series add directly together.
3
Calculate total electrical power using the formula P=V2RP = \frac{V^2}{R}
P=(12 V)29 Ω=1449=16 WP = \frac{(12\text{ V})^2}{9\,\Omega} = \frac{144}{9} = 16\text{ W}
Electrical power is directly proportional to the square of the source voltage and inversely proportional to the total equivalent resistance.

Key Concept

Series-Parallel Circuit Analysis and Electrical Power Calculation
Estimated Time:1m 30s
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Topics

Electricity and Magnetism

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