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Difficulty: HardCapacitors and Capacitance

Two capacitors with capacitances C1=4.0 μFC_1 = 4.0\text{ }\mu\text{F} and C2=12.0 μFC_2 = 12.0\text{ }\mu\text{F} are connected in series across a 120.0 V120.0\text{ V} d.c. power supply. After becoming fully charged, the capacitors are disconnected from the supply and reconnected in parallel with plates of like polarity connected together. What is the total electrostatic energy lost in the reconnection process?

  1. 5.4×10−3 J5.4 \times 10^{-3}\text{ J}Answer
  2. B
    1.62×10−2 J1.62 \times 10^{-2}\text{ J}
  3. C
    1.76×10−2 J1.76 \times 10^{-2}\text{ J}
  4. D
    2.16×10−2 J2.16 \times 10^{-2}\text{ J}

Answer

The total electrostatic energy lost in the reconnection process is 5.4×10−3 J5.4 \times 10^{-3}\text{ J}.
The correct answer is derived by finding the total initial energy stored in series (2.16×10−2 J2.16 \times 10^{-2}\text{ J}), determining the total combined charge (720 μC720\text{ }\mu\text{C}) and parallel capacitance (16.0 μF16.0\text{ }\mu\text{F}) upon reconnection to find the final stored energy (1.62×10−2 J1.62 \times 10^{-2}\text{ J}), and calculating the difference of 5.4×10−3 J5.4 \times 10^{-3}\text{ J}.

Step-by-Step Solution

1
Calculate the equivalent capacitance and charge of the initial series arrangement.
Cs=C1C2C1+C2=4.0×12.04.0+12.0=3.0 μFC_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{4.0 \times 12.0}{4.0 + 12.0} = 3.0\text{ }\mu\text{F}, so charge on each capacitor Q=CsV=(3.0×10−6 F)(120 V)=360 μCQ = C_s V = (3.0 \times 10^{-6}\text{ F})(120\text{ V}) = 360\text{ }\mu\text{C}.
Capacitors in series store identical charge equal to the product of equivalent series capacitance and total applied voltage.
2
Calculate the initial total electrostatic energy stored.
Ei=12CsV2=12(3.0×10−6 F)(120 V)2=2.16×10−2 JE_i = \frac{1}{2} C_s V^2 = \frac{1}{2} (3.0 \times 10^{-6}\text{ F})(120\text{ V})^2 = 2.16 \times 10^{-2}\text{ J}.
Initial stored energy is determined by the series combination connected across the supply voltage.
3
Determine total charge and equivalent capacitance after parallel reconnection.
Qp=Q1+Q2=360 μC+360 μC=720 μCQ_p = Q_1 + Q_2 = 360\text{ }\mu\text{C} + 360\text{ }\mu\text{C} = 720\text{ }\mu\text{C} and Cp=C1+C2=4.0 μF+12.0 μF=16.0 μFC_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 12.0\text{ }\mu\text{F} = 16.0\text{ }\mu\text{F}.
Connecting like-polarity plates aggregates the individual charges and sums the capacitances in parallel.
4
Calculate the final potential difference and final stored energy.
Vp=QpCp=720 μC16.0 μF=45 VV_p = \frac{Q_p}{C_p} = \frac{720\text{ }\mu\text{C}}{16.0\text{ }\mu\text{F}} = 45\text{ V}, so Ef=12CpVp2=12(16.0×10−6 F)(45 V)2=1.62×10−2 JE_f = \frac{1}{2} C_p V_p^2 = \frac{1}{2} (16.0 \times 10^{-6}\text{ F})(45\text{ V})^2 = 1.62 \times 10^{-2}\text{ J}.
Charge redistributes until both capacitors reach a common potential difference VpV_p.
5
Calculate the energy lost during reconnection.
ΔE=Ei−Ef=2.16×10−2 J−1.62×10−2 J=5.4×10−3 J\Delta E = E_i - E_f = 2.16 \times 10^{-2}\text{ J} - 1.62 \times 10^{-2}\text{ J} = 5.4 \times 10^{-3}\text{ J}.
The energy dissipated as heat and spark during charge redistribution is the difference between initial and final total energies.

Key Concept

Energy dissipation during charge sharing between reconnected capacitors
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Topics

Capacitors and Capacitance

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