A 10 F capacitor is connected to a battery with an emf of 2 V . It takes 50 ms for the charge on the capacitor to reach 12.6 C . Determine the resistance of the circuit.

Step-by-step solution

Given C = 10 F , V = 2 V , t = 50 ms , and q = 12.6 C . The maximum charge on the capacitor is q₀ = CV = 10 10⁻⁶ 2 = 20 C . The charge on a capacitor during charging is given by q = q₀(1 - e^ -t/RC ) . Substituting the values: 12.6 = 20(1 - e^ -t/RC ) 1 - e^ -t/RC = 12.6 20 = 0.63 e^ -t/RC = 0.37 Since e⁻¹ 0.37 , we have t RC = 1 . Therefore, RC = t = 50 ms . R = 50 10⁻³ 10 10⁻⁶ = 5 10^3 = 5 k .