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Difficulty: MediumCapacitors and Capacitance

Three capacitors, each of capacitance 12 μF12\text{ }\mu\text{F}, are arranged such that two of them are connected in parallel, and this combination is connected in series with the third capacitor. If the entire network is connected across a 20 V20\text{ V} d.c. power supply, what is the total energy stored in the network?

  1. A
    0.8×10−3 J0.8 \times 10^{-3}\text{ J}
  2. 1.6×10−3 J1.6 \times 10^{-3}\text{ J}Answer
  3. C
    3.6×10−3 J3.6 \times 10^{-3}\text{ J}
  4. D
    7.2×10−3 J7.2 \times 10^{-3}\text{ J}

Answer

The total energy stored in the network is 1.6×10−3 J1.6 \times 10^{-3}\text{ J}.
Combining two 12 μF12\text{ }\mu\text{F} capacitors in parallel gives a parallel capacitance of 24 μF24\text{ }\mu\text{F}. Connecting this combination in series with the third 12 μF12\text{ }\mu\text{F} capacitor results in an equivalent network capacitance of Ceq=24×1224+12=8 μFC_{\text{eq}} = \frac{24 \times 12}{24 + 12} = 8\text{ }\mu\text{F}. Substituting this into the energy formula E=12CeqV2E = \frac{1}{2} C_{\text{eq}} V^2 yields E=12×8×10−6×400=1.6×10−3 JE = \frac{1}{2} \times 8 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}.

Step-by-Step Solution

1
Calculate the equivalent capacitance of the parallel branch.
Cp=12 μF+12 μF=24 μFC_p = 12\text{ }\mu\text{F} + 12\text{ }\mu\text{F} = 24\text{ }\mu\text{F}
Capacitors in parallel add directly.
2
Calculate the total equivalent capacitance of the network.
Ceq=Cp×C3Cp+C3=24×1224+12=28836=8 μF=8×10−6 FC_{\text{eq}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{24 \times 12}{24 + 12} = \frac{288}{36} = 8\text{ }\mu\text{F} = 8 \times 10^{-6}\text{ F}
The parallel combination is connected in series with the third capacitor.
3
Calculate the total electrical energy stored in the combination.
E=12CeqV2=12×(8×10−6 F)×(20 V)2=4×10−6×400=1.6×10−3 JE = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times (8 \times 10^{-6}\text{ F}) \times (20\text{ V})^2 = 4 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}
The formula for energy stored in a capacitor network is E=12CV2E = \frac{1}{2} C V^2.

Key Concept

Mixed capacitor networks and energy storage
Estimated Time:1m 30s
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Topics

Capacitors and Capacitance

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