# gate 2017 EE(Set-B) q42-solution

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$$X’_1\frac{20}{60}\times 0.41=0.13666\Omega$$
$$X’_2\frac{20}{60}\times 0.41=0.13666\Omega$$
$$I_{st}=I_{ph}=I_{sc}=\frac{V_{ph}}{z_{eq}}$$
$$=\frac{80/\sqrt{3}}{[(0.3+0.3)+j(0.1366+0.1366)]}=\frac{46.18}{0.65}$$
$$=71.4\angle -24.46^0Amps$$