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JEE Advance 2017
Paper-2
Question
PARGRAPH - I
Consider a simple RC circuit as shown in figure 1.
Process-1: In the circuit the switch S is closed at $\mathrm{t}=0$ and the capcitor is fully charged to voltage $\mathrm{V}_0$ (i.e., charging continues for time $\mathrm{T} \gg \mathrm{RC}$ ). In the process some dissipation ( $\mathrm{E}_{\mathrm{D}}$ ) 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 set to $\frac{\mathrm{V}_0}{3}$ and maintained for a charging time $T \gg R C$. Then the voltage is raised to $\frac{2 v_0}{3}$ without discharging the capacitor and again maintained for a time $\mathrm{T} \gg \mathrm{RC}$. The process is repeated one more time by raising the voltage to $\mathrm{V}_0$ and the capacitor is charged to the same final voltage $\mathrm{V}_0$ 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 -
Select the correct option:
A
$E_C=E_D$
B
$\mathrm{E}_{\mathrm{C}}=\mathrm{E}_{\mathrm{D}} \ln 2$
C
$E_C=2 E_D$
D
$E_C=\frac{1}{2} E_D$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Advance
Physics
Easy
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