An inductor-coil with an inductance of 20 mH and a resistance of 10 is connected to an ideal battery having an emf of 5.0 V . Determine the rate of change of the induced emf at t = 1.0 s .

Step-by-step solution

The current in an LR circuit as a function of time is given by: i = E₀ R (1 - e^ -Rt/L ) The induced emf in the inductor is: e = L di dt = E₀ e^ -Rt/L The rate of change of the induced emf is: de dt = -E₀ R L e^ -Rt/L Given values are E₀ = 5.0 V , R = 10 , and L = 20 mH = 20 10⁻³ H . The time constant is = L R = 20 10⁻³ 10 = 2 10⁻³ s . At t = 1.0 s , the exponent is - Rt L = - 1.0 2 10⁻³ = -500 . Substituting these values into the rate of change equation: de dt = -5.0 500 e⁻⁵⁰⁰ Since e⁻⁵⁰⁰ 0 , the rate of change of