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 dissipated as heat 12.0 s after the connections are made.

Step-by-step solution

The time constant of the RC circuit is given by = RC . Substituting the given values, = (1.00 ) (12.0 F ) = 12.0 s . The current in the circuit during charging is given by I(t) = V R e^ -t/ . At t = 12.0 s , we have t = . I = 6.00 1.00 e⁻¹ = 6.00 e A . The power dissipated as heat in the internal resistance is P = I^2 R . P = ( 6.00 e )^2 1.00 = 36.0 e^2 W . Using e 2.718 , we get P 36.0 7.389 4.87 W .