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?

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? Correct Answer ip = (4.99×10-3) cos⁡(100t+π/4+89.94o)

Assuming particular integral as ip = A cos (ωt + θ) + B sin (ωt + θ) we get ip = V/√(R2+(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+89.94o).

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