Question

A meter bridge circuit is powered by an ideal battery of emf $5 \, V$ and negligible internal resistance. The bridge wire has a resistance per unit length equal to $0.1 \, \Omega \, cm^{- 1}$ . Unknown resistance $X$ is connected in the left gap and $6 \, \Omega $ in the right gap. The null point divides the wire in the ratio $2:3$ . What is the current (in $A$ ) drawn from the battery?

Answer 1

$\frac{l_{1}}{l_{2}}=\frac{2}{3}=\frac{v_{1}}{v_{2}}$
Now $v_{1}+v_{2}=v=5 \, V$ $=IR$
For potentiometer wire $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}=\frac{2}{3}$
$\frac{x}{6}=\frac{2}{3}\Rightarrow \frac{x}{l}=4 \, \frac{\Omega }{m}$
The resistance per unit length of the wire is $\frac{0.1 \Omega }{c m}=\frac{1 \Omega }{m}$
$\therefore \, \left(x + 1\right)=R_{e f f}=5$
$\therefore \, i=\frac{V}{R_{e f f}}=\frac{5}{5}=1 \, A$