In a metre bridge experiment, the balance point is obtained if the gaps are closed by 2 and 3. A shunt of X is added to the 3 resistor to shift the balancing point by 22.5 cm. The value of X is —.

Given: The two gaps are closed by 2 and 3. The initial balance point is determined by the metre bridge relation, and after adding a shunt X across the 3 resistor, the balance point shifts by 22.5 cm. Find: The value of X. For the initial balance condition: l1l2 = R1R2 with l1 + l2 = 100 cm Here, R1 = 2, R2 = 3 So, l1l2 = 23 Hence, l1 = 22+3 100 = 40 cm After adding the shunt, the effective resistance of the 3 resistor becomes R2' = R2 XR2 + X = 3X3 + X The new balance point shifts by 22.5 cm, so l1' = 40 + 22.5 = 62.5 cm Therefore, l2' = 100 - 62.5 = 37.5 cm Using the new balance condition: l1'l2' = R1R2' Substituting the values: 62.537.5 = 2 3X3 + X So, 53 = 2(3 + X)3X Cross-multiplying, 15X = 18 + 6X Thus, 9X = 18 and X = 2 Therefore, the value of X is 2.

In a metre bridge experiment, the balance point is obtained if th… — Solution | JEE Physics

NVAMediumJEE 2023Wheatstone Bridge & Meter Bridge

In a metre 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 the $3 \, \Omega$ resistor to shift the balancing point by $22.5 \, \text{cm}$ . The value of $X$ is — $\Omega$ .

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: The two gaps are closed by $2 \, \Omega$ and $3 \, \Omega$ . The initial balance point is determined by the metre bridge relation, and after adding a shunt $X \, \Omega$ across the $3 \, \Omega$ resistor, the balance point shifts by $22.5 \, \text{cm}$ .

Find: The value of $X$ .

For the initial balance condition:

\frac{l_1}{l_2} = \frac{R_1}{R_2}

with

l_1 + l_2 = 100 \, \text{cm}

Here,

R_1 = 2 \, \Omega, \qquad R_2 = 3 \, \Omega

So,

\frac{l_1}{l_2} = \frac{2}{3}

Hence,

l_1 = \frac{2}{2+3} \times 100 = 40 \, \text{cm}

After adding the shunt, the effective resistance of the $3 \, \Omega$ resistor becomes

R_2' = \frac{R_2 X}{R_2 + X} = \frac{3X}{3 + X}

The new balance point shifts by $22.5 \, \text{cm}$ , so

l_1' = 40 + 22.5 = 62.5 \, \text{cm}

Therefore,

l_2' = 100 - 62.5 = 37.5 \, \text{cm}

Using the new balance condition:

\frac{l_1'}{l_2'} = \frac{R_1}{R_2'}

Substituting the values:

\frac{62.5}{37.5} = \frac{2}{\frac{3X}{3 + X}}

So,

\frac{5}{3} = \frac{2(3 + X)}{3X}

Cross-multiplying,

15X = 18 + 6X

Thus,

9X = 18

and

X = 2 \, \Omega

Therefore, the value of $X$ is $2 \, \Omega$ .

Common mistakes

Using the shunted resistance as $3 + X$ instead of the parallel combination is incorrect because the shunt is connected across the $3 \, \Omega$ resistor. Use $R_2' = \frac{3X}{3 + X}$ instead.
Taking the new balance length as $40 - 22.5$ without checking the stated shift direction is wrong here because the extracted solution uses the shifted point as $40 + 22.5 = 62.5 \, \text{cm}$ . Follow the balance relation used in the working.
Forgetting that the metre bridge wire has total length $100 \, \text{cm}$ leads to an incorrect second segment length. After finding one balance length, always use $l_1 + l_2 = 100 \, \text{cm}$ to get the other.

Practice more Wheatstone Bridge & Meter Bridge questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Wheatstone Bridge & Meter Bridge

Answer

Step-by-step solution

Common mistakes

Practice more Wheatstone Bridge & Meter Bridge questions

Related questions