At an instant of time, the current in a part of the circuit is found to be 2A and increasing at the rate of 1 A-s–1. The measured voltage VAB between A and B is 8V. But when the current is 2A, decreasing at the rate of 1 A-s–1, the voltage VAB is found to be 4V. What is the ratio in s

Engineering

Physics

LR Circuit Growth and Decay of Current

Self and Mutual Inductance

Magnetic Flux and Faraday Law and Lenz Law

Question

At an instant of time, the current in a part of the circuit is found to be 2A and increasing at the rate of 1 A-s^–1. The measured voltage V_AB between A and B is 8V. But when the current is 2A, decreasing at the rate of 1 A-s^–1, the voltage V_AB is found to be 4V. What is the ratio $\frac{R}{L}$ in s^–1?

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$v_{A} - iR + 2 - \frac{Ldi}{dt} = v_{B}$

$v_{A} - v_{B} = 8 = iR - 2 + L \frac{di}{dt} = 2 R - 2 + L \times 1$

v_A' – v_B' = 4 = 2R – 2 – L × 1

$L = 2 H, R = 4 Ω \Rightarrow \frac{R}{L} = 2$