Question

In a meter bridge, the gaps are closed by resistance $2\Omega$ and $3\Omega$. The value of shunt to be added to $3\Omega$ resistor to shift the balancing point by $22.5\text{cm}$ is

• A. $5\Omega$
• B. $2.5\Omega$
• C. $2\Omega$
• D. $1\Omega$

Answer 1

The correct option is C $2\Omega$

Given, Shunt resistors $R_1=2\Omega ;R_2=3\Omega$

The initial balancing length $l$ of the bridge is,

$\frac{R_1}{R_2}=\frac{l}{100-l}.....\left(1\right)$

$\frac{2}{3}=\frac{l}{100-l}$

$3l=200-2l$

$5l=200\Rightarrow l=40\text{cm}$

When, adding a unknown resistance $x$ parallel to $3\Omega$ makes $25\text{cm}$ shift in the balancing length of the bridge.

The new balancing length is,

$l_1=l+22.5=40+22.5=62.5\text{cm}$

Using (1) we get,

$\frac{2}{\left(\frac{3x}{3+x}\right)}=\frac{l_1}{100-l_1}$

$\frac{6+2x}{3x}=\frac{5}{3}$

$6+2x=5x$

$6=3x$

$x=\frac{6}{3}=2\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.