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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?

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? Correct Answer ip = 0.98cos⁡(1000t+π/2-78.6o)

Assuming particular integral as ip = A cos (ωt + θ) + B sin(ωt + θ). We get ip = V/√(R2+(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/2-78.6o).

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