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The resistance of each arm of the Wheatstone bridge is 10 ohm. A resistance of 10 ohm is connected in series with galvanometer then the equivalent resistance across the battery will be :
A
10 ohm
B
15 ohm
C
20 ohm
D
40 ohm
Detailed Solution
All four arms are equal (10 $\Omega$ each), so the bridge is balanced: $\dfrac{P}{Q} = \dfrac{R}{S}$.
No current flows through the galvanometer branch, whatever its resistance (10 $\Omega$ added in series makes no difference). So that branch can be removed.
Then there are two branches of $10 + 10 = 20\ \Omega$ in parallel.
$R_{eq} = \dfrac{20 \times 20}{20 + 20} = 10\ \Omega$
No current flows through the galvanometer branch, whatever its resistance (10 $\Omega$ added in series makes no difference). So that branch can be removed.
Then there are two branches of $10 + 10 = 20\ \Omega$ in parallel.
$R_{eq} = \dfrac{20 \times 20}{20 + 20} = 10\ \Omega$