Question

Difficulty: HardCapacitors and Capacitance

In an electric circuit, two capacitors C1=12 μFC_1 = 12\text{ }\mu\text{F} and C2=6 μFC_2 = 6\text{ }\mu\text{F} are connected in series. This series combination is then connected in parallel with a third capacitor C3C_3 of unknown value. When a direct-current potential difference of 100 V100\text{ V} is applied across the entire network, the total electrostatic energy stored in the circuit is 100 mJ100\text{ mJ}. What is the capacitance of C3C_3 in microfarads (μF\mu\text{F})?

Answer: 16 μF

Answer

The capacitance of C3C_3 is 16 μF16\text{ }\mu\text{F}.
First, the series combination of 12 μF12\text{ }\mu\text{F} and 6 μF6\text{ }\mu\text{F} yields an equivalent branch capacitance of 4 μF4\text{ }\mu\text{F}. Second, using E=12CeqV2E = \frac{1}{2} C_{eq} V^2 with E=0.100 JE = 0.100\text{ J} and V=100 VV = 100\text{ V} gives a total circuit equivalent capacitance of 20 μF20\text{ }\mu\text{F}. Finally, subtracting the branch capacitance from the total parallel equivalent capacitance gives C3=20 μF4 μF=16 μFC_3 = 20\text{ }\mu\text{F} - 4\text{ }\mu\text{F} = 16\text{ }\mu\text{F}.

Step-by-Step Solution

1
Calculate the effective capacitance of the series branch containing C1C_1 and C2C_2
C12=4 μFC_{12} = 4\text{ }\mu\text{F}
Capacitors in series combine reciprocally: 1C12=1C1+1C2=112+16=312    C12=4 μF\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} \implies C_{12} = 4\text{ }\mu\text{F}.
2
Determine the total equivalent capacitance CeqC_{eq} of the circuit using the given stored energy and voltage
Ceq=20 μFC_{eq} = 20\text{ }\mu\text{F}
Energy stored in a capacitor network is E=12CeqV2E = \frac{1}{2} C_{eq} V^2. Rearranging gives Ceq=2EV2=2×0.100 J(100 V)2=20×106 F=20 μFC_{eq} = \frac{2E}{V^2} = \frac{2 \times 0.100\text{ J}}{(100\text{ V})^2} = 20 \times 10^{-6}\text{ F} = 20\text{ }\mu\text{F}.
3
Calculate the unknown capacitance C3C_3 from the parallel combination formula
C3=16 μFC_3 = 16\text{ }\mu\text{F}
Because the branch C12C_{12} and C3C_3 are in parallel, Ceq=C12+C3    20 μF=4 μF+C3    C3=16 μFC_{eq} = C_{12} + C_3 \implies 20\text{ }\mu\text{F} = 4\text{ }\mu\text{F} + C_3 \implies C_3 = 16\text{ }\mu\text{F}.

Key Concept

Series and parallel combinations of capacitors combined with electrostatic energy storage
Estimated Time:2m 0s
Rate this question