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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


  1. ED=13(12CV20)
  2. ED=3(12cv2o)
  3. ED=12CV2o
  4. ED=3CV20


Solution

The correct option is A ED=13(12CV20)

Wbattery=v03(CV03)=CV209

Energy of capacitor =CV2018
Heat loss  E1=CV209CV2018E1=118CV20



Uf=q22C=418CV20Uf=29CV20ΔU=CV206

Wbattery=2V03(CV03)              =29CV20 heat loss =29CV20CV206

                     =354CV20

E2=CV2018




 Work done by battery =V0(13CV0)=CV203ΔU=518CV20

Total heat loss E3=CV203518CV20
E3=118CV20

Total heat dissipated

E=E1+E2+E3=118CV20+118CV20+118CV20=16CV20=13(12CV20)
 

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