The resistance ofeach arm of the Wheatstone's bridge is $10\; ohm$. A resistance of $10 \;ohm$ is connected in series with a galvanometer then the equivalent resistance across the battery will be
AIPMT 2001, Easy
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It is a balanced Wheatstone Bridge because $\frac{P}{Q}=\frac{R}{S}$$\Rightarrow\frac{10}{10}=\frac{10}{10}=1$
Equivalent resistance is given by $R=\frac{(10+10)(10+10)}{(10+10)+(10+10)}$
$=\frac{20 \times 20}{40}=10 \Omega$
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