Question

In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The particular integral of the solution of ‘ip’ is?

(a) ip = 0.98cos⁡(1000t+π/2-78.6^o)

(b) ip = 0.98cos⁡(1000t-π/2-78.6^o)

(c) ip = 0.98cos⁡(1000t-π/2+78.6^o)

(d) ip = 0.98cos⁡(1000t+π/2+78.6^o)

Answer 1

The correct option is (a) ip = 0.98cos⁡(1000t+π/2-78.6^o)

For explanation: Assuming particular integral as ip = A cos (ωt + θ) + B sin(ωt + θ). We get ip = V/√(R^2+(ωL)^2) cos⁡(ωt+θ-tan^-1(ωL/R)) where ω = 1000 rad/sec, V = 100V, θ =  π/2, L = 0.1H, R = 20Ω. On substituting, we get ip = 0.98cos⁡(1000t+π/2-78.6^o).