Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectromagnetic Induction
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 = 1.0 ext s ?
Options
- A4.1 10⁻⁹ ext V
- B2.1 10⁻⁹ ext V
- C8.2 10⁻⁹ ext V
- D4.1 10⁻⁸ ext V
Correct answer
A. 4.1 10⁻⁹ ext V
Step-by-step solution
The current in an LR circuit at time t is given by I = E R (1 - e^ -Rt/L ) . The magnitude of the self-induced emf is e = L dI dt . Differentiating the expression for current with respect to time, we get: dI dt = E R R L e^ -Rt/L = E L e^ -Rt/L Thus, the self-induced emf is: e = L E L e^ -Rt/L = E e^ -Rt/L Substituting the given values E = 2.0 V , R = 20 , L = 1.0 H , and t = 1.0 s : e = 2.0 e^ -20 1.0 / 1.0 = 2.0 e⁻²⁰ To evaluate e⁻²⁰ , taking the base 10 logarithm gives: ₁₀(e⁻²⁰) = -20 ₁₀(e) = -20 0.4343 = -8.686