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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

 

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