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/4-89.94o)
• B. ip = (4.99×10-3) cos⁡(100t-π/4-89.94o)
• C. ip = (4.99×10-3) cos⁡(100t-π/4+89.94o)
• D. ip = (4.99×10-3) cos⁡(100t+π/4+89.94o)