  Question

# Consider a simple RC 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 Ve (i.e. charging continues for time T>>RC ). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is Ec. Process 2 :  In a different process the voltage is first set to V03 and maintained for a charging time T>>RC. Then the voltage is raised to 2V03 without discharging the capacitor and again maintained for a time T>>RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2. In Process 2, total energy dissipated across the resistance ED is ED=13(12CV20)ED=3(12cv2o)ED=12CV2oED=3CV20

Solution

## The correct option is A ED=13(12CV20) Wbattery=v03(CV03)=CV209 Energy of capacitor =CV2018 Heat loss  E1=CV209−CV2018E1=118CV20 Uf=q22C=418CV20Uf=29CV20ΔU=CV206 Wbattery=2V03(CV03)              =29CV20∴ heat loss =29CV20−CV206                      =354CV20 E2=CV2018 Work done by battery =V0(13CV0)=CV203ΔU=518CV20 Total heat loss E3=CV203−518CV20 E3=118CV20 ∴ Total heat dissipated E=E1+E2+E3=118CV20+118CV20+118CV20=16CV20=13(12CV20)  Suggest corrections   