Question

A meter bridge circuit is powered by an ideal battery of emf $5V$ and negligible internal resistance. The bridge wire has a resistance per unit length equal to $0.1\Omega c{m}^{-1}$ . 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

Answer: 1

$\frac{l_1}{l_2}=\frac{2}{3}=\frac{v_1}{v_2}$
Now $v_1+v_2=v=5V$ $=IR$
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 $\frac{0.1\Omega }{cm}=\frac{1\Omega }{m}$
$\therefore \left(x+1\right)=R_{eff}=5$
$\therefore i=\frac{V}{R_{eff}}=\frac{5}{5}=1A$