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 current in the circuit 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 10⁻⁶ = 12.0 s The current in the circuit during the charging of a capacitor is given by: I = I₀ e^ -t/ The initial current I₀ is: I₀ = E R = 6.00 1.00 = 6.00 A At t = 12.0 s , we have t = . Substituting this into the current equation: I = 6.00 e⁻¹ = 6.00 e Using e 2.718 : I = 6.00 2.718 2.21 A