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Q.
If there is no deflection in the galvanometer connected in a circuit shown in figure, then the ratio of lengths AC/CB is
MGIMS WardhaMGIMS Wardha 2009
Solution:
Let $ AC={{l}_{1}},CB={{l}_{2}} $ and r the resistance per unit length of wire AB. As there is no current in galvanometer G, the resistance 60 Q, 15 0, AC and CB form the arms of a balanced Wheatstone's bridge $ \therefore $ $ \frac{P}{Q}=\frac{R}{S} $ $ \Rightarrow $ $ \frac{60}{15}=\frac{r\times {{l}_{1}}}{r\times {{l}_{2}}} $ Or $ \frac{{{l}_{1}}}{{{l}_{2}}}=\frac{4}{1} $