An inductor-coil with a resistance of 10 and an inductance of 120 mH is connected to a battery having an emf of 6 V and an internal resistance of 2 . Determine the charge that flows through the inductor-coil in 100 ms after the connections are made.

Step-by-step solution

Total resistance of the circuit, R = 10 + 2 = 12 Steady state current, I₀ = E R = 6 12 = 0.5 A Time constant, = L R = 120 10⁻³ 12 = 0.01 s The current in the circuit at any time t is given by i = I₀(1 - e^ -t/ ) The charge flowing through the inductor in time t is: q = ₀^ t i dt = ₀^ t I₀(1 - e^ -t/ ) dt q = I₀ [ t + e^ -t/ ]₀^ t = I₀ ( t - (1 - e^ -t/ ) ) Substituting t = 100 ms = 0.1 s , = 0.01 s , and I₀ = 0.5 A : q = 0.5 ( 0.1 - 0.01(1 - e⁻¹⁰) ) Since e⁻¹⁰ 0 , we get: q 0.5 (0.1 - 0.01) = 0.5 0.09 = 0.045 C = 4