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 20 ms after the connections are made.

Step-by-step solution

The total resistance of the circuit is R = 10 + 2 = 12 . The time constant of the circuit is = L R = 120 10⁻³ 12 = 0.01 s . The steady-state current is i₀ = E R = 6 12 = 0.5 A . The current at any time t is given by: i(t) = i₀(1 - e^ -t/ ) The charge q that flows in time t is the integral of current with respect to time: q = ₀^ t i(t) dt = ₀^ t i₀(1 - e^ -t/ ) dt q = i₀ [ t + e^ -t/ ]₀^ t = i₀ ( t - + e^ -t/ ) Substituting t = 20 ms = 0.02 s , = 0.01 s , and i₀ = 0.5 A : q = 0.5 ( 0.02 - 0.01 + 0.01 e^ -0.02/0.01 )