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Zorluk: OrtaCapacitors and Capacitance

A 4.0 μF4.0\text{ }\mu\text{F} capacitor and a 6.0 μF6.0\text{ }\mu\text{F} capacitor are connected in parallel. This parallel combination is connected in series with a 10.0 μF10.0\text{ }\mu\text{F} capacitor across a 60.0 V60.0\text{ V} d.c. power supply. What is the total energy stored in the combination?

  1. 9.0×10−3 J9.0 \times 10^{-3}\text{ J}Cevap
  2. B
    3.6×10−2 J3.6 \times 10^{-2}\text{ J}
  3. C
    3.48×10−3 J3.48 \times 10^{-3}\text{ J}
  4. D
    2.23×10−2 J2.23 \times 10^{-2}\text{ J}

Cevap

The total energy stored in the combination is 9.0×10−3 J9.0 \times 10^{-3}\text{ J}.
First, find the equivalent capacitance of the parallel branch by adding 4.0 μF4.0\text{ }\mu\text{F} and 6.0 μF6.0\text{ }\mu\text{F} to get 10.0 μF10.0\text{ }\mu\text{F}. Next, combine this result in series with the 10.0 μF10.0\text{ }\mu\text{F} capacitor to obtain a total circuit capacitance of 5.0 μF5.0\text{ }\mu\text{F} (or 5.0×10−6 F5.0 \times 10^{-6}\text{ F}). Substituting Ceq=5.0×10−6 FC_{eq} = 5.0 \times 10^{-6}\text{ F} and V=60.0 VV = 60.0\text{ V} into E=12CeqV2E = \frac{1}{2} C_{eq} V^2 gives 9.0×10−3 J9.0 \times 10^{-3}\text{ J}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the parallel branch.
Cp=C1+C2=4.0 μF+6.0 μF=10.0 μFC_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}
Capacitors in parallel add directly.
2
Calculate the total equivalent capacitance of the entire circuit.
Ceq=Cp×C3Cp+C3=10.0 μF×10.0 μF10.0 μF+10.0 μF=5.0 μF=5.0×10−6 FC_{eq} = \frac{C_p \times C_3}{C_p + C_3} = \frac{10.0\text{ }\mu\text{F} \times 10.0\text{ }\mu\text{F}}{10.0\text{ }\mu\text{F} + 10.0\text{ }\mu\text{F}} = 5.0\text{ }\mu\text{F} = 5.0 \times 10^{-6}\text{ F}
The parallel combination CpC_p is connected in series with C3C_3.
3
Calculate the total energy stored using E=12CeqV2E = \frac{1}{2} C_{eq} V^2.
E=12(5.0×10−6 F)(60.0 V)2=9.0×10−3 JE = \frac{1}{2} (5.0 \times 10^{-6}\text{ F}) (60.0\text{ V})^2 = 9.0 \times 10^{-3}\text{ J}
The energy stored in a capacitive circuit depends on the net capacitance and applied potential difference.

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