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Question: A capacitor $C = 10\mu F$ is connected across a battery of emf 10V. After it is charged to 10V, the ...

A capacitor C=10μFC = 10\mu F is connected across a battery of emf 10V. After it is charged to 10V, the battery is disconnected and reconnected across the capacitor with reversed polarity. The heat (in mJ) dissipated in the circuit on reconnection is

Answer

2

Explanation

Solution

The problem involves a capacitor initially charged and then reconnected to a battery with reversed polarity. We need to calculate the heat dissipated during this process.

1. Initial State of the Capacitor: The capacitor has capacitance C=10μF=10×10−6FC = 10\mu F = 10 \times 10^{-6} F and is charged to a voltage V=10VV = 10V. The initial charge stored on the capacitor is Qinitial=CV=(10×10−6F)(10V)=100μCQ_{initial} = CV = (10 \times 10^{-6} F)(10 V) = 100 \mu C. The initial energy stored in the capacitor is Uinitial=12CV2U_{initial} = \frac{1}{2}CV^2. Uinitial=12(10×10−6F)(10V)2=12(10×10−6)(100)=500×10−6J=0.5mJU_{initial} = \frac{1}{2}(10 \times 10^{-6} F)(10 V)^2 = \frac{1}{2}(10 \times 10^{-6})(100) = 500 \times 10^{-6} J = 0.5 mJ.

2. Final State of the Capacitor: When the battery is reconnected with reversed polarity, the capacitor will eventually be charged to the same voltage V=10VV = 10V, but with the opposite polarity. The final energy stored in the capacitor is Ufinal=12CV2U_{final} = \frac{1}{2}CV^2. Ufinal=12(10×10−6F)(10V)2=0.5mJU_{final} = \frac{1}{2}(10 \times 10^{-6} F)(10 V)^2 = 0.5 mJ.

3. Charge Flow through the Battery: Let's consider one plate of the capacitor, say plate A. Initially, if plate A was connected to the positive terminal of the battery, its charge was +Qinitial=+CV+Q_{initial} = +CV. When reconnected with reversed polarity, plate A is now connected to the negative terminal of the battery. Therefore, its final charge will be −Qfinal=−CV-Q_{final} = -CV. The change in charge on plate A is ΔQA=Qfinal−Qinitial=(−CV)−(CV)=−2CV\Delta Q_A = Q_{final} - Q_{initial} = (-CV) - (CV) = -2CV. This means that a charge of 2CV2CV flows out of plate A. This charge flows through the external circuit and the battery. Alternatively, consider the plate connected to the positive terminal of the battery in the final configuration. Let's say this is plate B. Initially, plate B had charge −CV-CV. Finally, plate B has charge +CV+CV. The charge supplied by the battery to plate B is ΔQbattery=(+CV)−(−CV)=2CV\Delta Q_{battery} = (+CV) - (-CV) = 2CV. ΔQbattery=2(10×10−6F)(10V)=200μC\Delta Q_{battery} = 2(10 \times 10^{-6} F)(10 V) = 200 \mu C.

4. Work Done by the Battery: The work done by the battery is Wbattery=V×ΔQbatteryW_{battery} = V \times \Delta Q_{battery}. Wbattery=(10V)(2CV)=2CV2W_{battery} = (10 V)(2CV) = 2CV^2. Wbattery=2(10×10−6F)(10V)2=2(10×10−6)(100)=2000×10−6J=2mJW_{battery} = 2(10 \times 10^{-6} F)(10 V)^2 = 2(10 \times 10^{-6})(100) = 2000 \times 10^{-6} J = 2 mJ.

5. Heat Dissipated: According to the principle of energy conservation, the work done by the battery is converted into the change in energy stored in the capacitor and the heat dissipated in the circuit. Wbattery=(Ufinal−Uinitial)+HdissipatedW_{battery} = (U_{final} - U_{initial}) + H_{dissipated} Hdissipated=Wbattery−(Ufinal−Uinitial)H_{dissipated} = W_{battery} - (U_{final} - U_{initial}) Since Ufinal=UinitialU_{final} = U_{initial} (both are 12CV2\frac{1}{2}CV^2), the change in stored energy is zero: Ufinal−Uinitial=0U_{final} - U_{initial} = 0. Therefore, the heat dissipated is equal to the work done by the battery: Hdissipated=Wbattery=2CV2H_{dissipated} = W_{battery} = 2CV^2. Hdissipated=2mJH_{dissipated} = 2 mJ.

The heat dissipated in the circuit on reconnection is 2 mJ.