Through resistanceless leads, a capacitor of capacitance 12.0 F is connected to a battery of emf 6.00 V and internal resistance 1.00 . Determine the power delivered by the battery 12.0 s after the connections are made.

Step-by-step solution

The time constant of the RC circuit is given by: = RC = 1.00 12.0 10⁻⁶ = 12.0 s The current in the circuit as a function of time is: I(t) = E R e^ -t/ At t = 12.0 s , we have t = . Substituting the values: I = 6.00 1.00 e⁻¹ = 6.00 e 2.207 A The total power delivered by the battery is the power supplied by its emf, which is given by: P = EI P = 6.00 2.207 13.24 W Rounding to three significant figures, the power delivered is 13.2 W .