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# A capacitor of capacitance 12.0 μF is connected to a battery of emf 6.00 V and internal resistance 1.00 Ω through resistance less leads. 12.0 μs after the connections are made, what will be (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heat and (d) the rate at which the energy stored in the capacitor is increasing.

Jitender Pal
7 years ago
Sol. C = 12.0 μF = 12 * 10^-6 emf = 6.00 /v, R = 1 Ω T = 12 μc, i = I base 0 e^-t/RC = CV/T = e^-t/RC = 12 * 10^-6 * 6/12 * 10^-6 * e^-1 = 2.207 = 2.1 A b) Power delivered by battery We known, V = V base 0 e^-t/RC (where V and V base 0 are potential VI) VI = V base 0l e^-t/RC ⇒ VI = V base 0l * e^-1 = 6 * 6 e^-1 = 13.24 W c) U = CV^2/T (e^-t/RC)^2 [CV^2/T = energy drawing per unit time] = 12 * 10^-6 * 36/12 * 10^-6 *(e^-1)^2 = 4.872.