In a meter bridge experiment, the balance point is obtained if the gaps are closed by $2 \Omega$ and $3…
In a meter bridge experiment, the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$. A shunt of $X \Omega$ is added to $3 \Omega$ resistor to shift the null point by 22.5 cm . The value of ' $x$ ' is
$1 \Omega$
$2 \Omega$
$3 \Omega$
$4 \Omega$
Solution
$\begin{array}{ll}
& \text { In case } 1, \\
& \frac{2}{3}=\frac{l}{100-l} \\
\therefore \quad & 200-2 l=3 l \\
\therefore \quad & l=40 \mathrm{~cm}
\end{array}$
In case 2,
$\begin{aligned} & \frac{2}{\frac{3 x}{3+x}}=\frac{40+22.5}{100-(40+22.5)} \\ \therefore \quad & \frac{2(3+x)}{3 x}=\frac{62.5}{37.5}\end{aligned}$
$\begin{array}{ll}\therefore & (6+2 x) \times 37.5=3 x(62.5) \\ \therefore & 225+75 x=187.5 x \\ \therefore & x=\frac{225}{112.5}=2 \Omega\end{array}$