GATE 2015 EE – SET 2 – Question 44

Solution :

$$V_{1}$$ = 100$$\angle 0^{0} $$ = 100 + 0.00 j
$$V_{2} $$= 100 $$\angle -120^{0}$$ = -50 – 86.60j
$$V_{3}$$ = 100 $$\angle 120^{0}$$ = -50 + 86. 60 j

$$I_{1}$$ = $$\frac{(-50+86.60j)-(100+0j)}{-j1}$$

$$I_{1}$$=$$\frac{-150 + 86.60 j}{-j}$$

$$I_{1}$$ =$$ \frac{173.20\angle 150^{0}}{1.\angle -90} $$
$$I_{1}$$ = 173.20$$\angle 150$$ + 90
$$I_{1}$$ = 173.20 $$\angle 240$$
$$I_{1}$$ = 173.20$$\angle -120$$

$$I_{2}$$ = $$\frac{V_{3}-V_{2}}{j}$$ = $$\frac{(-50 + 86.60j)-(-50-86.60j)}{j1}$$

$$I_{2}$$ = $$\frac{173.2j}{1j}$$ = 173.2

i = $$I_{1} + I_{2}$$

i = 173.2 + 173.2 $$\angle -120$$

i = 173.2 + ( -86.60-150j)

i  =  86.60 – 150j

i = 173.2  $$\angle 300$$

i  = $$\angle -60 $$