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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

  1. A(E_D= 1 2 C V₀^2 )
  2. B(E_D=3 ( 1 2 C V₀^2 ) )
  3. C(E_D= 1 3 ( 1 2 C V₀^2 ) )
  4. D(E_D=3 C V₀^2 )

Correct answer

C. (E_D= 1 3 ( 1 2 C V₀^2 ) )

Step-by-step solution

For process (i) Charge on capacitor = C V 0   3 Energy stored in capacitor = 1 2 C V 0 2 9 = C V 0 2 18 Work done by battery = C V 0 3 × V 3 = C V 0 2 9 Heat loss = C V 0 2 9 - C V 0 2 18 = C V 0 2 18 For process (ii) Charge on capacitor = 2 C V 0 3 Extra charge flow through battery = C V 0 3 Work done by battery: C V 0 3   . 2 V 0 3 = 2 C V 0 2 9 Final energy store in capacitor: 1 2 C 2 V 0 3 2 = 4 C V 0 2 18 Energy store in process 2: 4 C V 0 2 18 - C V 0 2 18 = 3 C V 0 2 18 Heat loss in process (i

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