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_(D)=3((1)/(2)CV_(0)^(2))`
B. `E_(D)=(1)/(3)((1)/(2)CV_(0)^(2))`
C. `E_(D)=3CV_(0)^(2)`
D. `E_(D)=(1)/(2)CV_(0)^(2)`

Answer

B