Question

The circuit shown below is at steady state before the switch closes at t=0. The switch is closed for 1.5 s and is then opened. At t = 2 s, V (t) will be?

(a) 5.12 V

(b) 6.43 V

(c) 8.57 V

(d) 9.88 V

The question was posed to me in a job interview.

The above asked question is from Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor in portion Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

Right choice is (c) 8.57 V

The best explanation: V (1.5s) = 5.002 V

For t > 1.5 s, τ = 8 × 0.05 = 0.4

V (t) = 10+ (5-10)\(e^{\frac{-(t-1.5)}{0.4}}\) = 10 – 5e^-2.5(t-1.5)

For t≥1.5 s, V (2 s) = 8.57 V.