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In Wheatstone bridge, 4 resistors $\mathrm{P}=10 \Omega$, $\mathrm{Q}=5 \Omega, \mathrm{R}=4 \Omega, \mathrm{S}=4 \Omega$ are connected in cyclic order. To ensure no current through galvanometer
Options:
Solution:
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Verified Answer
The correct answer is:
$5 \Omega$ resistance is connected in series with $\mathrm{Q}$
For no current through the galvanometer, the wheatstone bridge should be balanced. For this, we must have
$$
\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{S}}{\mathrm{R}}
$$
This condition is satisfied with only option (a).
When a $5 \Omega$ resistor is connected in series with $Q$, the equivalent resistance in the P-arm becomes $10 \Omega$.
$$
\therefore \quad \frac{\mathrm{P}}{\mathrm{Q}}=\frac{10}{10}=1
$$
and $\frac{\mathrm{S}}{\mathrm{R}}=\frac{4}{4}=1$ $\Rightarrow \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{S}}{\mathrm{R}}$
$$
\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{S}}{\mathrm{R}}
$$
This condition is satisfied with only option (a).
When a $5 \Omega$ resistor is connected in series with $Q$, the equivalent resistance in the P-arm becomes $10 \Omega$.
$$
\therefore \quad \frac{\mathrm{P}}{\mathrm{Q}}=\frac{10}{10}=1
$$
and $\frac{\mathrm{S}}{\mathrm{R}}=\frac{4}{4}=1$ $\Rightarrow \frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{S}}{\mathrm{R}}$
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