Question

When a battery of emf $E$ volts is connected across a capacitor of capacitance $C$, then after some time the potential difference between the plates of the capacitor becomes equal to the battery voltage. The ratio of the work done by the battery and the energy stored in the capacitor when it is fully charged is :

• A. 1:1
• B. 1:2
• C. 2:1
• D. 4:1

Answer 1

Answer: (C)

2:1

Answer 2

When a capacitor of capacitance $C$ is charged by a battery of emf $E$ volts, the potential difference across the capacitor plates eventually becomes equal to the battery voltage, $V = E$.

1. Energy stored in the capacitor ($U_C$):

   The energy stored in a fully charged capacitor is given by the formula: $U_C = \frac{1}{2}CV^2$

   Since the potential difference across the fully charged capacitor is $E$, we have: $U_C = \frac{1}{2}CE^2$

2. Work done by the battery ($W_B$):

   The total charge transferred by the battery to charge the capacitor is $Q = CV$. Since $V=E$, the charge transferred is $Q = CE$. The work done by the battery is the product of the total charge transferred and the battery's emf: $W_B = Q \times E$ Substituting $Q = CE$: $W_B = (CE) \times E = CE^2$

3. Ratio of the work done by the battery to the energy stored in the capacitor:

   We need to find the ratio $\frac{W_B}{U_C}$: $\frac{W_B}{U_C} = \frac{CE^2}{\frac{1}{2}CE^2}$ $\frac{W_B}{U_C} = \frac{1}{\frac{1}{2}}$ $\frac{W_B}{U_C} = 2$

   Therefore, the ratio of the work done by the battery to the energy stored in the capacitor is 2:1.