An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E / 2 , the ratio of the work done by the battery to the heat dissipated by the resistor is given by:-

Step-by-step solution

i = E R e - t / R C , Q = C E 1 - e - t / R C Capacitor is charged to E 2 So Q = C E 2 ∴ C E 2 = C E 1 - e - t / R C 1 2 = e - t / R C t = R C ln 2 Work done by battery = Q flown  ( Δ V ) = C E 2 ( E ) = C E 2 2 Heat dissipated = ∫ 0 R C ℓ n 2 i 2 R d t = E 2 R ∫ 0 R C l n 2 e - 2 t / R C · d t = 3 4 C E 2 2  Work done   Heat dissipated  = C E 2 / 2 3 4 C E 2 2 = 4 3