In a meter bridge experiment, the balance point is obtained if the gaps are closed by 2Ω and 3Ω . A shunt of X Ω is added to 3Ω resistor to shift the null point by 22.5 cm . The value of ' x

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In a meter bridge experiment, the balance point is obtained if the gaps are closed by 2Ω and 3Ω . A shunt of X Ω is added to 3Ω resistor to shift the null point by 22.5 cm . The value of ' x ' is

TestHub · Accepted Answer

For a meter bridge, the balancing condition is given by $\frac{R _{1}}{R _{2}} = \frac{l}{100 - l}$

Initially, $R_{1} = 2Ω$ and $R_{2} = 3Ω$ . Let the initial balance point be $l_{1}$ .

$\frac{2}{3} = \frac{l _{1}}{100 - l _{1}}$

$200 - 2 l_{1} = 3 l_{1} \Rightarrow 5 l_{1} = 200 \Rightarrow l_{1} = 40 cm$

When a shunt of $X Ω$ is connected across the $3Ω$ resistor, the equivalent resistance in the right gap becomes $R_{2}^{'} = \frac{3 X}{3 + X}$ .

Since the resistance in the right gap decreases, the balance point shifts to the right. The new balance point is $l_{2} = 40 + 22.5 = 62.5 cm$ .

Applying the balancing condition again:

$\frac{2}{R _{2}^{'}} = \frac{62.5}{100 - 62.5}$

$\frac{2}{R _{2}^{'}} = \frac{62.5}{37.5} = \frac{5}{3}$

$R_{2}^{'} = \frac{6}{5} = 1.2Ω$

Substituting $R_{2}^{'}$ into the equivalent resistance formula:

$\frac{3 X}{3 + X} = 1.2$

$3 X = 1.2 (3 + X)$

$3 X = 3.6 + 1.2 X$

$1.8 X = 3.6 \Rightarrow X = 2Ω$

Answer: $2Ω$

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