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In the figure given below, the current source is 1 ∠ 0 A, R = 1 Ω, the impedances are
ZC = – jΩ
and Z1 = 2j Ω.
The Thevenin equivalent looking into the circuit across X-Y is
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- √2 ∠0 V , (1 + 2j) Ω
- 2 ∠ 45° V , (1 - 2j) Ω
- 2 ∠ 45° V , (1 + j) Ω
- √2 ∠ 45° V , (1 + j) Ω
- √2 ∠0 V , (1 + 2j) Ω
Correct Option: D
Zeq = 1 + (2j – j) = 1 + j = √2 ∠45°
V = IZeq = (1 ∠ 0°)( √2 ∠ 45° ) = √2 ∠45° 
