Q52 / 54 · JEE Advanced 2017 (Paper 2) · Physics

Paragraph: 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 $\left(E_{D}\right)$ 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 $T>>R C$. 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 $2 .$ Figure
Figure

Question: In Process 2, total energy dissipated across the resistance $E_{D}$ is:

  • A
    \(E_D=\frac{1}{2} C V_0^2\)
  • B
    \(E_D=3\left(\frac{1}{2} C V_0^2\right)\)
  • Correct answer
    C
    \(E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right)\)
  • D
    \(E_D=3 C V_0^2\)

Solution

For process (i)
Charge on capacitor =CV0 3
Energy stored in capacitor =12CV029=CV0218
Work done by battery =CV03×V3=CV029
Heat loss  =CV029-CV0218=CV0218
For process (ii)
Charge on capacitor =2CV03
Extra charge flow through battery =CV03
Work done by battery:  CV03 .2V03=2CV029
Final energy store in capacitor: 12C2V032=4CV0218
Energy store in process 2: 4CV0218-CV0218=3CV0218
Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)
=2CV029-3CV0218=CV0218
For process (iii)
Charge on capacitor =CV0
Extra charge flow through battery: CV0-2CV03=CV03
Work done by battery in this process: CV03 V0=CV023
Find energy store in capacitor:  12CV02
Energy stored in this process:  12CV02-4CV0218=5CV0218
Heat loss in process (iii):  CV023-5CV0218=CV0218
Now total heat loss  ED: CV0218+CV0218+CV0218=CV026
Final energy store in capacitor: 12CV02
So we can say that ED=1312CV02

Hence, the correct answer is option: 'C'.

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