Two known resistances are connected in two gaps of a meter bridge. The null point is obtained at $20…
Two known resistances are connected in two gaps of a meter bridge. The null point is obtained at $20 \mathrm{~cm}$ from zero end. A resistance of $15 \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \mathrm{~cm}$. The smaller resistances is
$9 \Omega$
$7 \Omega$
$3 \Omega$
$5 \Omega$
Solution
Given $l=20 \mathrm{~cm} \quad l^{\prime}=40 \mathrm{~cm}$
Using $\frac{R_1}{R_2}=\frac{l}{100-l}$
$\begin{aligned} & \therefore \frac{R_1}{R_2}=\frac{20}{100-20} \\ & \Rightarrow R_2=4 R_1\end{aligned}$
Again using $\frac{R_1}{R_2}=\frac{l^{\prime}}{100-l^{\prime}}$, where $l^{\prime}=40 \mathrm{~cm}, R_1=R^{\prime}{ }_1+15$ and
$\begin{aligned} & R_2^{\prime}=R_2=4 R_1 \\ & \therefore \frac{R_1+15}{4 R_1}=\frac{40}{100-40} \\ & \Rightarrow R_1=9 \Omega\end{aligned}$