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In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is :
A
$\dfrac{R}{4}$
B
$\dfrac{R}{2}$
C
R
D
2R
Detailed Solution
All four arms are equal, so $\dfrac{P}{Q} = \dfrac{R}{S}$ and the bridge is balanced. No current flows through the galvanometer arm, so it can be removed.
The circuit then has two branches in parallel, each of $R + R = 2R$.
Equivalent resistance $= \dfrac{2R \times 2R}{2R + 2R} = R$
So the resistance seen by the battery, i.e. the equivalent resistance between A and B, is R.