# Question 1.SGPYQ.3: In the figure given below, Z1  = 10∠−60°, Z2 = 10∠60°, Z3 = ......

In the figure given below, $Z_1=10 \angle-60^{\circ}$, $Z_2=10 \angle 60^{\circ}, Z_3=50 \angle 53.13^{\circ}$. Thevenin impedance seen from X-Y is

(a) 56.6∠45°            (b) 60∠30°

(c) 70∠30°                  (d) 34.4∠65°

Step-by-Step
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Given that

$Z_1=10 \angle-60^{\circ}, Z_2=10 \angle 60^{\circ}, Z_3=50 \angle 53.13^{\circ}$

To obtain Thevenin impedance, short circuit the source. Therefore,

$Z_{ Th }=Z_3+\left(Z_1 \| Z_2\right)$

Convert polar co-ordinates into rectangular co-ordinates.

\begin{aligned} Z_1 & =5(1-\sqrt{3} j) Z_2=5(1+\sqrt{3} j) Z_3=10(3+4 j) . \\ Z_{\text {th }} & =10(3+4 j)+\left[\frac{5(1-\sqrt{3} j) \times 5(1+\sqrt{3} j)}{5(1-\sqrt{3} j)+5(1+\sqrt{3} j)}\right] \\ & =40+40 j \end{aligned}

$Z_{ Th }$ in polar co-ordinates $=40 \sqrt{2} \cdot \angle 45^{\circ} \Omega$

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