Question

In the given circuit, where the length AB of the potentiometer wire is $1\text{m}$, the resistors $X$ and $Y$ have values of $5\Omega$ and $2\Omega$, respectively. When $X$ is shunted by a wire, the balance point is found to be $0.625$ $\text{m}$ from $A$. What is the resistance of the shunt?

• A. $10$ $\Omega$
• B. $20$ $\Omega$
• C. $30$ $\Omega$
• D. $40$ $\Omega$

Answer 1

The correct option is A $10$ $\Omega$

Let $\text{R}$ be the resistance of the shunted wire, the effective resistance of $\text{R}$ and $5\Omega$ in parallel $=$ $\frac{5R}{5+R}$

At balance point,

$\frac{\frac{5R}{5+R}}{2}=\frac{0.625}{1-0.625}=\frac{0.625}{0.0375}=\frac{5}{3}$

Upon solving , we get $\text{R}=10\Omega$

Key concept :
If a constant current flows through a wire of uniform area of cross section, then potential drop alog the wire is directly propotional to the length of the wire.

We can compare these type of circuits with the balanced wheatstone bridge, when no current flows through the galvanometer

$\frac{R_1}{R_2}=\frac{R_3}{R_4}=\frac{\left(l\right)cm}{(100-l)cm}$