JEE Advanced2017PhysicsCapacitanceActual
Paragraph: Consider a simple R C 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 V₀ (i.e., charging continues for time T>>R C ). 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 s
Options
- A(E_C=E_D )
- B(E_C=E_D 2 )
- C(E_C= 1 2 E_D )
- D(E_C=2 E_D )
Correct answer
A. (E_C=E_D )
Step-by-step solution
When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be q = C V Energy stored in capacitor E C = 1 2 C V 2 .....(i) Work done by battery W = V q = V C V = C V 2 Dissipated across resistance E D = (work done by battery) – (energy store) E D = C V 2 - 1 2 C V 2 = 1 2 C V 2 .....(ii) From (i) and (ii) E D =   E C