JEE Advanced 2017 · Paper 2 · Physics Q15

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PhysicsSingle correct+3Question 15 of 54

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PARAGRAPH 1

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 $(E_D)$ 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 $T \gg RC$ . Then the voltage is raised to $\frac{2V_0}{3}$ 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.

Figure 1: a source V, switch S, resistor R and capacitor C in series. Figure 2: source voltage against time: Process 1 holds V₀ from t = 0; Process 2 steps from V₀/3 (0 to T) to 2V₀/3 (T to 2T) to V₀ (after 2T), with T ≫ RC.

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

E_C = E_D

E_C = E_D\ln 2

E_C = \frac{1}{2}E_D

E_C = 2E_D

Answer and worked solution

Official answer: A

The battery delivers charge $CV_0$ at constant voltage $V_0$ , so it does work $CV_0^2$ .

Stored: $E_C=\tfrac12CV_0^2$ .
The rest is dissipated in $R$ : $E_D=\tfrac12CV_0^2$ .

So $E_C=E_D$ , whatever the value of $R$ .

(A)

Topics:RC circuits Energy stored in a capacitor

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JEE Advanced 2017 · Paper 2, Physics question 15