Q165 / 170 · Manipal (MET) 2017 · Physics
In the adjoining circuit, the resistances are given in ohm, $X$ and $Y$ are unknown resistances. The current through the $10 \Omega$ resistance is 3A while that through the resistance $X$ is 1 A . No current passes through the galvanometer. The values of the unknown resistances $X$ and $Y$ are respectively
- A$12 \Omega$ and $12 \Omega$
- B$12 \Omega$ and $6 \Omega$
- C$6 \Omega$ and $12 \Omega$
- Correct answerD$6 \Omega$ and $6 \Omega$
Solution
When no current passes through the galvanometer, the bridge is balanced hence,
$\begin{aligned}
& \frac{P}{Q}=\frac{R}{S} \\
\Rightarrow & \frac{24+X}{10}=\frac{84+Y}{30}
\end{aligned}$
Also, applying Kirchhoff's law
$\begin{aligned}
\sum i R & =\Sigma i^{\prime} R^{\prime} (parallel)\\
(X+24)+(84)+Y & =3(10+30) \\
\Rightarrow X+Y & =12 .....(i)\\
\text { and } 3 X-Y & =12 ....(ii)
\end{aligned}$
From Eqs. (i) and (ii), we get
$X=6, Y=6 \text { (neglecting negative value) }$
Hence, the correct answer is option: 'D'.