At t = 0 , a capacitor having capacitance C is connected through a resistance R to a battery of emf E . Determine the maximum rate of energy storage in the capacitor, and find the time at which this maximum rate is reached.

Step-by-step solution

The charge on the capacitor at time t is given by q = C E (1 - e^ -t/RC ) . The current in the circuit is i = dq dt = E R e^ -t/RC . The rate of energy storage in the capacitor is P = dU dt = d dt ( q^2 2C ) = q C i . Substituting the expressions for q and i gives: P = C E (1 - e^ -t/RC ) C E R e^ -t/RC = E ^2 R (e^ -t/RC - e^ -2t/RC ) . To find the maximum rate of energy storage, let x = e^ -t/RC . Then P = E ^2 R (x - x^2) . For maximum P , dP dx = 0 1 - 2x = 0 x = 1 2 . Thus, e^ -t/RC = 1 2 t = CR 2 . The maximu