Examkin

Question

Difficulty: MediumCapacitors and Capacitance

Two capacitors with capacitances of 3 μF3\text{ }\mu\text{F} and 6 μF6\text{ }\mu\text{F} are connected in parallel. This combination is then connected in series with a 9 μF9\text{ }\mu\text{F} capacitor across a 12 V12\text{ }\text{V} direct current source. What is the total electrical energy stored in the network?

  1. 3.24×10−4 J3.24 \times 10^{-4}\text{ }\text{J}Answer
  2. B
    6.48×10−4 J6.48 \times 10^{-4}\text{ }\text{J}
  3. C
    7.92×10−4 J7.92 \times 10^{-4}\text{ }\text{J}
  4. D
    1.30×10−3 J1.30 \times 10^{-3}\text{ }\text{J}

Answer

The total electrical energy stored in the network is 3.24×10−4 J3.24 \times 10^{-4}\text{ }\text{J}.
First, find the equivalent capacitance of the parallel branch (3 μF+6 μF=9 μF3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}). Next, combine this with the 9 μF9\text{ }\mu\text{F} capacitor in series, yielding Ceq=9×99+9=4.5 μF=4.5×10−6 FC_{eq} = \frac{9 \times 9}{9 + 9} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}. Finally, substituting into the stored energy formula E=12CeqV2E = \frac{1}{2} C_{eq} V^2 gives E=0.5×4.5×10−6×144=3.24×10−4 JE = 0.5 \times 4.5 \times 10^{-6} \times 144 = 3.24 \times 10^{-4}\text{ }\text{J}.

Step-by-Step Solution

1
Calculate equivalent capacitance of the parallel section
Cp=C1+C2=3 μF+6 μF=9 μFC_p = C_1 + C_2 = 3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}
Capacitors in parallel add directly.
2
Calculate total equivalent capacitance of the network
1Ceq=1Cp+1C3=19 μF+19 μF=29 μF  ⟹  Ceq=4.5 μF=4.5×10−6 F\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{9\text{ }\mu\text{F}} + \frac{1}{9\text{ }\mu\text{F}} = \frac{2}{9\text{ }\mu\text{F}} \implies C_{eq} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}
Capacitors in series combine via reciprocals.
3
Calculate total stored electrical energy
E=12CeqV2=12×(4.5×10−6 F)×(12 V)2=3.24×10−4 JE = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times (4.5 \times 10^{-6}\text{ }\text{F}) \times (12\text{ }\text{V})^2 = 3.24 \times 10^{-4}\text{ }\text{J}
Energy stored in a capacitor network depends on equivalent capacitance and potential difference across it.

Key Concept

Equivalent capacitance of series-parallel combinations and energy stored in a capacitor
Rate this question

Topics

Capacitors and Capacitance

More questions from this topic