A) \[25\Omega \]
B) \[22\Omega \]
C) \[5\Omega \]
D) \[12\Omega \]
Correct Answer: B
Solution :
Initially, let \[{{I}_{1}}\]be the current through G. then \[{{I}_{1}}=\frac{V}{220+G}.\] After the key \[{{K}_{2}}\]is closed, the circuit is shown as Apply KVL on loop 1, \[5{{I}_{3}}=G{{I}_{4}}\Rightarrow {{I}_{3}}=\frac{G{{I}_{4}}}{5}\] Also,\[{{I}_{3}}+{{I}_{4}}={{I}_{2}}\]\[\Rightarrow \]\[\left( \frac{G}{5}+1 \right){{I}_{4}}={{I}_{2}}\] \[\Rightarrow \]\[{{I}_{4}}=\frac{V}{\left( {{R}_{1}}+\frac{5G}{G+5} \right)\left( \frac{G+5}{5} \right)}\] For a galvanometer, \[I\propto \theta \] So,\[\frac{{{I}_{1}}}{{{I}_{4}}}=\frac{{{\theta }_{0}}}{{{\theta }_{0}}/5}\Rightarrow 5=\frac{V}{220+G}\frac{{{R}_{1}}(G+5)+5G}{5V}\] \[\Rightarrow \]\[25(220+G)=(220)(G+5)+5G\] \[\Rightarrow \]\[25(220)+20G=220G+1100\] \[\Rightarrow \]\[200G=4400\Rightarrow G=22\Omega \]You need to login to perform this action.
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