Question

Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length $l_1$ is $40\, cm$. Now an unknown resistance $x$ is connected in series with $P$ and new balancing length is found to be $80 \,cm$ measured from the same end. Then the value of $x$ will be _____ $\Omega$

Answer 1

Initially, $\frac{ P }{ Q }=\frac{40\, cm }{60\, cm }=\frac{2}{3} ....$(1)
Finally, $\frac{ P + x }{ Q }=\frac{80 \,cm }{20 \,cm }=\frac{4}{1}.....$(2)
Divide (2) by (1)
$ \frac{P+x}{P}=4 \times \frac{3}{2}=6$
$\Rightarrow 1+\frac{x}{P}=6 \Rightarrow \frac{x}{P}=5$
$ \therefore x=5 P=5 \times 4=20 \Omega$