At t = 0 , an LR circuit with L = 1.0 ext H and R = 20 ext is connected across an emf of 2.0 ext V . What is the value of the self-induced emf in the circuit at t = 200 ext ms ?

Step-by-step solution

The current in an LR circuit during growth is given by I = I₀(1 - e^ -Rt/L ) . The magnitude of the self-induced emf is given by e = L dI dt . Differentiating the current expression with respect to time, we get e = L ( I₀ R L e^ -Rt/L ) = E e^ -Rt/L , where E is the applied emf. Given E = 2.0 V , R = 20 , L = 1.0 H , and t = 200 ms = 0.2 s . Substituting the values, we get e = 2.0 e^ -20 0.2 / 1.0 = 2.0 e⁻⁴ . Using the value e⁻⁴ 0.0183 , the induced emf is e = 2.0 0.0183 = 0.0366 V 0.037 V .