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

Answer

D

Explanation

When voltage is set to `V_0/3`,charge supplied by battry `(CV_0)/3`
When voltage raised to `(2V_0)/3`.additional charge supplied `CV_0-(2CV_0)/3-(CV_0)/3=(CV_0)/3`
When Voltage is reaised to `V_0` additional charge supplied `=CV_0-(2CV_0)/3=(CV_0)/3`
Total energy supplied by cell `=(V_0)/3((CV_0)/3)+(2V_0)/3+((CV_0)/3)+V_0((CV_0)/3)=2/3CV_0^(2)`
Final charge on capicitor`=CV_0` Energy stored in capacitor `=1/2CV_0^(2)` `therefore` Energy dissipated across resistor `E_D=2/3CV_0^( 2)-1/2=1/6CV_0^( 2)`