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 10 ms after the connections are made.

Step-by-step solution

Total resistance of the circuit, R = R_L + r = 10 + 2 = 12 Time constant of the circuit, = L R = 120 10⁻³ 12 = 10 ms Steady state current, I₀ = E R = 6 12 = 0.5 A The current at any time t is given by: i = I₀(1 - e^ -t/ ) The charge flown 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 = 10 ms = : q = I₀ ( - (1 - e⁻¹)) = I₀ e⁻¹ q = 0.5 10 10⁻³ 1 e Using e 2.718 : q = 5 10⁻³ 2.718 1.84 10⁻³ C = 1.84 mC The closest value is 1.