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Gate 2016 EE - Set 2 - Question 35

Given parameters, Z_{11}=40\Omega; Z_{12}=60\Omega; Z_{21}=80\Omega; Z_{22}=100\Omega

From the figure,

V_{2}=-20I_{2} __________(1)

V_{1}=Z_{11}I_{1}+Z_{12}I_{2}

V_{1}=40I_{1}+60I_{2} _______ (2)

V_{2}=Z_{21}I_{1}+Z_{22}I_{2}

V_{2}=80I_{1}+100I_{2} _________(3)

Substituting (1) into (3), -20I_{2}=80I_{1}+100I_{2}

-120I_{2}=80I_{1}

I_{2}=\frac{-80}{120}I_{1}=\frac{-2}{3}I_{1}

Substituting the above value of I_{2} into (2) V_{1}=40I_{1}+20\left ( \frac{-2}{3} \right )I_{1}

V_{1}=40I_{1}-40I_{1}\Rightarrow V_{1}=0

From the figure, loop 1 gives us the equation, 20=10I_{1}+V_{1}

I_{1}=2A

I_{2}=\frac{-2}{3}(2)=\frac{-4}{3}A

Power dissipated in R_{L}=I_{2}^{2}R_{L}

\Rightarrow \left ( \frac{-4}{3} \right )^{2}\times 20\Rightarrow 35.55W

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