Process 1 : In the circuit the switch S is closed at and the capacitor is fully charged to voltage `v_(0)` (i.e., charging continues for time T >> 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 `(v_(0))/(3)` maintained for a charging time T >> RC. Then the voltage is raised to `(2V_(0))/(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 `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.
In Process 2, total energy dissipated across the resistance `E_(D)` is :

Options

A. `E_(C)=1/2E_(D)`
B. `E_(C)=2E_(D)`
C. `E_(C)=E_(D)`
D. `E_(C)=E_(D) ln 2`

Answer

C

Explanation

Energy supplied by cell `=V_0(CV_0)=CV_0^(2)` , Energy stored in capacitor `(E_c)=1/2CV_0^(2)`
`therefore` Energy dissipited across resistor `(E_D)=CV_0^(2)-1/2CV_0^(2)=1/2CV_0^(2) therefore E_c=E_D`