Question

In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The particular integral of the solution of ‘ip’ is?

(a) ip = (4.99×10^-3) cos⁡(100t+π/4-89.94^o)

(b) ip = (4.99×10^-3) cos⁡(100t-π/4-89.94^o)

(c) ip = (4.99×10^-3) cos⁡(100t-π/4+89.94^o)

(d) ip = (4.99×10^-3) cos⁡(100t+π/4+89.94^o)

I have been asked this question during an internship interview.

This intriguing question originated from Sinusoidal Response of an R-C Circuit in portion Transients of Network Theory

Answer 1

The correct answer is (d) ip = (4.99×10^-3) cos⁡(100t+π/4+89.94^o)

For explanation I would say: Assuming particular integral as ip = A cos (ωt + θ) + B sin (ωt + θ)

we get ip = V/√(R^2+(1/ωC)^2) cos⁡(ωt+θ-tan^-1(1/ωRC))

where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get ip = (4.99×10^-3) cos⁡(100t+π/4+89.94^o).