In the circuit shown the transformer is ideal with turn Ratio $\frac{N_{1}}{N_{2}} = \frac{5}{1}$ . The voltage of the source is V_s = 300 Volt. The voltage measured across the load resistance R_L = 100 $Ω$ is 50 Volt. Find the value of resistance R in the primary circuit.

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400 Ω

500 Ω

250 Ω

200 Ω

answer is C.

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Detailed Solution

$V_{1} = (\frac{N_{1}}{N_{2}}) V_{2} [V_{2} = V_{L} = 50 V]$ ,
and, $V_{s} = I_{1} R + V_{1} = I_{1} R + (\frac{N_{1}}{N_{2}}) V_{2}$ .
Current in secondary circuit is $I_{2} = \frac{V_{2}}{R_{L}}$ ,
and $I_{1} = (\frac{N_{2}}{N_{1}}) I_{2} = \frac{N_{2}}{N_{1}} \frac{V_{2}}{R_{L}}$ .
$∴ V_{s} = (\frac{N_{2}}{N_{1}}) (\frac{V_{2}}{R_{L}}) R + (\frac{N_{1}}{N_{2}}) V_{2}$
$\begin{array}{l} \Rightarrow R = \frac{N_{1} R_{L}}{N_{2} V_{2}} [V_{s} - V_{2} (\frac{N_{1}}{N_{2}})] \\ = 5 \times \frac{100}{50} [300 - 50 \times 5] \\ = 5 \times 100 = 500 Ω \end{array}$