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 $V_{0}$ (i.e., charging continues for time $T \gg RC$). In the process some dissipation $\left(E_{D}\right)$ 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 $T \gg RC$. Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time $T \gg RC$. 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:

(C) $E_{C}=\frac{1}{2} E_{D}$

Current Electricity — JEE Advanced · 2017 · 21 May 2017 · Paper 2 Question with Solution

Passage

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_{0}

(i.e., charging continues for time

T ≫ R C

). 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 ≫ R C

. Then the voltage is raised to

\frac{2 V _{0}}{3}

without discharging the capacitor and again maintained for a time

T ≫ R C

. 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.

Consider a simple

R C

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_{0}

(i.e., charging continues for time

T ≫ R C

). 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 ≫ R C

. Then the voltage is raised to

\frac{2 V _{0}}{3}

without discharging the capacitor and again maintained for a time

T ≫ R C

. 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:

Answer: (A)

Work done by battery:

W_{b} = C V_{0} \times V_{0} = C V_{0}^{2}

. Energy stored in capacitor:

E_{C} = \frac{1}{2} C V_{0}^{2}

. Heat dissipated:

E_{D} = W_{b} - E_{C} = C V_{0}^{2} - \frac{1}{2} C V_{0}^{2} = \frac{1}{2} C V_{0}^{2}

. Therefore

E_{C} = E_{D}

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option