# GATE 2016 EE – SET 2 – Question No:3

Since we need the value of current in the primary side, transfer the secondary parameters to primary side.

$$R_{1}=R\left ( \frac{V_{1}}{V_{2}} \right )^{2}=80\ast 1000\left [ \frac{100}{100\ast 100} \right ]^{2}=8\Omega$$

$$X_{c}^{‘}=X_{c}\ast \left ( \frac{V_{1}}{V_{2}} \right )^{2}=40\ast 1000\left [ \frac{100}{100\ast 100} \right ]^{2}=4\Omega$$

Therefore the current I can be written as an expression of, $$I=\frac{100}{8+j\left ( 10-4 \right )}=\frac{100}{8+6j}$$

Calculating the magnitude, $$I=\frac{100}{\sqrt{8^{2}+6^{2}}}=\frac{100}{10}=10A$$