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 = 10 ms .

Step-by-step solution

The current in an L-R circuit during growth is given by: i = E R (1 - e^ -Rt/L ) The induced emf in the inductor is: = L di dt = L ( E R R L e^ -Rt/L ) = E e^ -Rt/L The rate of change of the induced emf is: d dt = -E ( R L ) e^ -Rt/L The magnitude of the rate of change of the induced emf is: | d dt | = ER L e^ -Rt/L Substituting the given values E = 5.0 V , R = 10 , L = 20 10⁻³ H , and t = 10 10⁻³ s : R L = 10 20 10⁻³ = 500 s ⁻¹ Rt L = 500 10 10⁻³ = 5 | d dt | = 5.0 500 e⁻⁵ = 2500 e⁻⁵ Using the approximate value e⁻