JEE Advanced 2017 · Paper 2 · Physics Q16

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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 2, total energy dissipated across the resistance $E_D$ is:

E_D = \frac{1}{2}CV_0^2

E_D = 3\left(\frac{1}{2}CV_0^2\right)

E_D = \frac{1}{3}\left(\frac{1}{2}CV_0^2\right)

E_D = 3CV_0^2

Answer and worked solution

Official answer: C

Each step raises the capacitor's voltage by $\dfrac{V_0}3$ from a source held at the new value. As in Process 1, each step dissipates
$\tfrac12C\left(\frac{V_0}3\right)^2.$

Three steps.
$E_D=3\cdot\frac12C\frac{V_0^2}9=\frac13\left(\frac12CV_0^2\right).$

(C)

Topics:RC circuits Energy stored in a capacitor

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JEE Advanced 2017 · Paper 2, Physics question 16