Question

A capacitor of capacitance 10mF is charged by connecting it to a battery of emf 50 V. The capacitor is then disconnected and reconnected to the same battery, with the polarity reversed.

Which of the following statements are correct?

• A. The initial energy stored in the capacitor is 12.5 mj
• B. The heat developed in the wire after reconnecting is 25 mj
• C. The heat developed in the wire after reconnecting is 50 mJ
• D. The final energy stored in the capacitor is 12.5 mj

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

The problem involves a capacitor being charged, then disconnected and reconnected with reversed polarity. We need to analyze the energy stored and heat dissipated during these processes.

Assumption: The capacitance given as "10mF" is likely a typo and should be "10$\mu$F". This is inferred because the options are in milliJoules (mJ), and using 10mF would result in energies in Joules (J), not milliJoules. The similar question also uses microfarads. Therefore, we will proceed with $C = 10 \mu F = 10 \times 10^{-6} F$.

Given values:

• Capacitance $C = 10 \mu F = 10 \times 10^{-6} F$
• Battery EMF $V = 50 V$

1. Initial Energy Stored in the Capacitor (Statement A): When the capacitor is charged by the battery of emf $V$, the initial energy stored is given by: $U_{initial} = \frac{1}{2}CV^2$ $U_{initial} = \frac{1}{2}(10 \times 10^{-6} F)(50 V)^2$ $U_{initial} = \frac{1}{2}(10 \times 10^{-6})(2500)$ $U_{initial} = (5 \times 10^{-6})(2500)$ $U_{initial} = 12500 \times 10^{-6} J$ $U_{initial} = 12.5 \times 10^{-3} J = 12.5 \text{ mJ}$ So, statement (A) is correct.

2. Final Energy Stored in the Capacitor (Statement D): When the capacitor is reconnected to the same battery with reversed polarity, it will again be charged to the same voltage $V = 50 V$, but with opposite polarity. The energy stored depends only on the magnitude of the capacitance and voltage, not on the polarity. $U_{final} = \frac{1}{2}CV^2$ Since $C$ and $V$ are the same as in the initial state, the final energy stored will be the same as the initial energy stored. $U_{final} = 12.5 \text{ mJ}$ So, statement (D) is correct.

3. Heat Developed in the Wire after Reconnecting (Statements B and C): To find the heat developed, we use the principle of energy conservation: Work done by battery ($W_{battery}$) = Change in energy stored in capacitor ($\Delta U$) + Heat dissipated ($H_{dissipated}$) $W_{battery} = \Delta U + H_{dissipated}$ Here, $\Delta U = U_{final} - U_{initial} = 12.5 \text{ mJ} - 12.5 \text{ mJ} = 0$. So, $H_{dissipated} = W_{battery}$.

Now, let's calculate the work done by the battery. The work done by the battery is $W_{battery} = V \times \Delta Q_{battery}$, where $\Delta Q_{battery}$ is the total charge supplied by the battery. Initially, the charge on the capacitor is $Q_{initial} = CV = (10 \times 10^{-6} F)(50 V) = 500 \times 10^{-6} C = 500 \mu C$. Let's say one plate has $+Q_{initial}$ and the other has $-Q_{initial}$. When the polarity is reversed, the plate that initially had $+Q_{initial}$ now has $-Q_{final}$ (where $Q_{final} = CV$). So, its charge changes from $+CV$ to $-CV$. The total charge that flows through the circuit (and thus through the battery) to achieve this change is $2CV$. $\Delta Q_{battery} = 2CV$ $\Delta Q_{battery} = 2 \times (10 \times 10^{-6} F) \times (50 V)$ $\Delta Q_{battery} = 2 \times 500 \times 10^{-6} C = 1000 \times 10^{-6} C = 1 \times 10^{-3} C = 1 \text{ mC}$.

Now, calculate the work done by the battery: $W_{battery} = V \times \Delta Q_{battery}$ $W_{battery} = (50 V) \times (1 \times 10^{-3} C)$ $W_{battery} = 50 \times 10^{-3} J = 50 \text{ mJ}$.

Since $H_{dissipated} = W_{battery}$, $H_{dissipated} = 50 \text{ mJ}$. So, statement (C) is correct. Statement (B) is incorrect.

Conclusion: Statements (A), (C), and (D) are correct.