Solution: Write the current source in phasor form: Is (phasor) = 8 ∠ 0 A, ω=2rad/s, so: XL = ω L = 2× 0. 125 = 0.25 ω, xc1= xc2 =/kloc-0.

The phasor model of the circuit thus obtained is shown in the above figure. List the node voltage equation according to KCL:

Node 1: V 1 (phasor)/1+[V 1 (phasor) -V2 (phasor) ]/(-j0.5)+ [V 1 (phasor) -V2.

Node 2: [V2 (phasor) -V 1 (phasor) ]/(-j0.5)+ [V2 (phasor) -V 1 (phasor) ]/(j0.25)+V2 (phasor)/(-j0.5).

Solve the equation:

Equation 1 both ends are multiplied by J0.5: (1+J0.5) V 1 (phasor) -V2 (phasor) = 8 ∠ 90.

Multiply both ends of Equation 2 by j0.5: v 1 (phasor) =j2V2 (phasor). Substitute into the above formula:

(1+j0.5)×j2V2 (phasor) -V2 (phasor) =(- 1+j2- 1)V2 (phasor) =(-2+j2)V2 (phasor) = 2 ∠ 2.

Therefore, V2 (phasor) = 8 ∠ 90/2 ∠ 2 ∠ 135 = 2 ∠ 2 ∠-45 (V).

Therefore: V 1 (phasor) = J2× 2 ∠ 2-45 = 4 ∠ 2 ∠ 45 (V).

V 1(t)= 2√2×√2 sin(2t-45)= 4 sin(2t-45)(V).