Consider a simple RC circuit as shown in Figure 1. Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0 (i.e. charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC. Process 2 : In a different process the voltage is first set to V_0 \over 3 and maintained for a charging time T >> RC. Then, the voltage is raised to 2V_0 \over 3 without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2. In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are related by

Capacitor PYQ — JEE Advanced 2017 · Shift 2 · Q51 with Solution

View written solutionFree

Correct answer: B

For a capacitor $C$ charged to voltage $V_0$ ,

$E_C = \frac{1}{2}CV_0^2$

This is the final electrostatic energy stored in the capacitor.

When an initially uncharged capacitor is connected to a battery of emf $V_0$ through a resistor $R$ , the total energy supplied by the battery is

$E_{\text{battery}} = V_0 Q$

Since the final charge on the capacitor is

$Q = CV_0,$

we get

$E_{\text{battery}} = V_0(CV_0)=CV_0^2$

Out of this, the energy stored in the capacitor is

$E_C = \frac{1}{2}CV_0^2$

So the heat dissipated in the resistor is

$E_D = E_{\text{battery}} - E_C = CV_0^2 - \frac{1}{2}CV_0^2 = \frac{1}{2}CV_0^2$

Thus,

$E_D = E_C$

So,

$\boxed{E_C = E_D}$

The correct option is:

$\boxed{\text{B}}$

with relation

$\boxed{E_C = E_D}$