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Capacitor question

2017 · Shift 2 · Q52

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JEE Advanced
Physics
Capacitor
2017 · Shift 2 · Q52

Capacitor question

2017 · Shift 2 · Q52

JEE AdvancedPhysicsCapacitorMCQ+3 / −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 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

{{2{V_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. JEE Advanced 2017 Paper 2 Offline Physics - Capacitor Question 8 English Comprehension

JEE Advanced 2017 Paper 2 Offline Physics - Capacitor Question 8 English Comprehension

In Process 2, total energy dissipated across the resistance ED is

A
${E_D} = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$
B
${E_D} = 3\left( {{1 \over 2}CV_0^2} \right)$
C
${E_D} = 3CV_0^2$
D
${E_D} = {1 \over 2}CV_0^2$

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