First slide

meter bridge

Question

In the given circuit of fig.no current is passing
through the galvanometer connected to a meter bridge. Distance AC corresponding to null deflection of galvanometer is x. What would be its value if the radius of the wire AB is doubled ?

Easy

A

X / 2
B

2X
C

X
D

None of these

Solution

When Wheatstone bridge is balanced, then
$\frac{R_{1}}{R_{2}} = \frac{resistance of AC part}{resistance of BC part}$
When the diameter of the wire is doubled, the ratio of R.H.S. remains the same. Hence the value of ,AC (= x) remains the same,

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