Question

In the circuit shown below steady state is obtained before the switch closes at t = 0. The switch is closed for 1.5 s and is then opened. At t = 1 s, V (t) will be?

(a) – 3.24 V

(b) 1.97 V

(c) 5.03 V

(d) 13.24 V

The question was asked by my college director while I was bunking the class.

Question is taken from Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor in section Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Correct answer is (c) 5.03 V

To explain I would say: V (0^–) = 10 V = V (0^+)

For 0<t≤1.5 s, τ = 4 × 0.05 = 0.2 s

VOC = 5 V

V (t) = 5 + (10 – 5)\(e^{\frac{-t}{0.2}}\) = 5 + 5e^-5t

V (1s) = 5.03 V.