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 = 0 .

Step-by-step solution

The current in an L-R circuit at time t is given by: i = E R (1 - e^ -Rt/L ) The induced emf e across the inductor is: e = -L di dt Differentiating i with respect to t : di dt = E L e^ -Rt/L Thus, the induced emf is: e = -L ( E L e^ -Rt/L ) = -E e^ -Rt/L The rate of change of the induced emf is: de dt = d dt (-E e^ -Rt/L ) = -E ( - R L ) e^ -Rt/L = ER L e^ -Rt/L At t = 0 , the magnitude of the rate of change of the induced emf is: | de dt | = ER L Substituting the given values E = 5.0 V , R = 10 , and L = 20 mH = 2