Q. Two resistances are connected in the two gaps of a meter bridge. The balance point is $20cm$ from the zero end. When a resistance $15\Omega$ is connected in series with the smaller of two resistance, the null point get shifted to $40cm$ . What is the smaller of the two resistance values ?
NTA AbhyasNTA Abhyas 2022
Solution:
$\frac{R_{1}}{R_{2}}=\frac{20}{80}=\frac{1}{4}$ , It is clear that $R_{1}$ is smaller resistance
∴ $R_{2}=4R_{1}$
When $15\Omega$ is connected in series with $R_{1}$ , then equivalent resistance is $R_{1}+15$ , So
$\therefore \frac{R_{1} + 15}{R_{2}}=\frac{40}{60}=\frac{2}{3}$
$\therefore \frac{R_{1} + 15}{4 R_{1}}=\frac{2}{3}$
$\therefore R_{1}=9\Omega$
∴ $R_{2}=4R_{1}$
When $15\Omega$ is connected in series with $R_{1}$ , then equivalent resistance is $R_{1}+15$ , So
$\therefore \frac{R_{1} + 15}{R_{2}}=\frac{40}{60}=\frac{2}{3}$
$\therefore \frac{R_{1} + 15}{4 R_{1}}=\frac{2}{3}$
$\therefore R_{1}=9\Omega$