Question

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 $V_e$ (i.e. charging continues for time $T>>RC$ ). In the process some dissipation ($E_D$) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $E_c$.

Process 2 :

In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time T>>RC. Then the voltage is raised to $\frac{2V_0}{3}$ without discharging the capacitor and again maintained for a time $T>>RC$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final voltage $V_0$ as in Process $1.$

These two processes are depicted in Figure 2.

In Process 1, the energy stored in the capacitor $E_c$ and heat dissipated across resistance $E_D$ are related by

• A. $E_C=E_D\ln 2$
• B. $E_C=\frac{1}{2}{E}_D$
• C. $E_c=E_D$
• D. $E_c=2E_b$

Answer 1

The correct option is C $E_c=E_D$

$U=\frac{q_0^2}{2C}=$ energy stored

$E_C=\frac{1}{2}{Cv}_0^2$

Work done by battery $=q{V}_0$

Work done by battery $={CV}_0^2$

$\therefore$ heat loss $={CV}_0^2-\frac{q_0^2}{2C}$

$={CV}_0^2-\frac{1}{2}{CV}_0^2$

$E_D=\frac{1}{2}{CV}_0^2$

$\therefore \frac{E_D}{E_C}=\frac{\frac{1}{2}{CV}_0^2}{\frac{1}{2}{CV}_0^2}=\frac{1}{1}$