The voltage measured across the load resistance \[{{R}_{L}}=100\Omega \]. is 50 volt. Find the value of resistance R in the primary circuit.
A) \[250\Omega \]
B) \[500\Omega \]
C) \[750\Omega \]
D) \[300\Omega \]
Correct Answer: B
Solution :
[b] \[{{V}_{1}}=\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right){{V}_{2}}\,\,\,\,\,\,\,\,\,\,[{{V}_{2}}={{V}_{L}}=50V]\] and \[{{V}_{s}}={{I}_{2}}R+\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right){{V}_{2}}\] Current in secondary circuit is \[{{I}_{2}}=\frac{{{V}_{2}}}{{{R}_{L}}}\] and \[{{I}_{1}}=\left( \frac{{{N}_{2}}}{{{N}_{1}}} \right){{I}_{2}}=\frac{{{N}_{2}}}{{{N}_{1}}}\frac{{{V}_{2}}}{{{R}_{L}}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,{{V}_{s}}\left( \frac{{{N}_{2}}}{{{N}_{1}}} \right)\left( \frac{{{V}_{2}}}{{{R}_{1}}} \right)R+\left( \frac{{{N}_{2}}}{{{N}_{1}}} \right){{V}_{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,R=\frac{{{N}_{1}}{{R}_{L}}}{{{N}_{2}}{{V}_{2}}}\left[ {{V}_{s}}-{{V}_{2}}\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right) \right]\] \[=5\times \frac{100}{50}\,[300-50\times 5]\] \[=10\times 50=500\Omega \]
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