When two resistances R1 and R2 are connected in series and introd… — Solution | JEE Physics

NVAMediumJEE 2023Wheatstone Bridge & Meter Bridge

When two resistances $R_1$ and $R_2$ are connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \, \Omega$ is introduced into the right gap, a null point is found at $60 \, \text{cm}$ from the left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3 \, \Omega$ is introduced into the right gap to get the null point at $40 \, \text{cm}$ from the left end. The product of $R_1 R_2$ is _____ $\Omega^2$ :

Answer

Correct answer:30

Step-by-step solution

Standard Method

Given: In the series combination, the left gap contains $R_1 + R_2$ and the right gap contains $10 \, \Omega$ . The null point is at $60 \, \text{cm}$ from the left, so the wire lengths are $60 \, \text{cm}$ and $40 \, \text{cm}$ .

In the parallel case, the solution states:

\frac{R_1 R_2}{R_1 + R_2} = 3

Find: The value of $R_1 R_2$ .

Using the meter bridge principle:

\frac{R_1 + R_2}{10} = \frac{60}{40}

R_1 + R_2 = 15 \, \Omega

For the parallel combination:

\frac{R_1 R_2}{R_1 + R_2} = 3

Substituting $R_1 + R_2 = 15 \, \Omega$ ,

R_1 R_2 = 3 \times 15 = 30 \, \Omega^2

Therefore, the product $R_1 R_2$ is $30$ .

Common mistakes

Using the meter bridge ratio in the reverse order is a common mistake. That gives an incorrect value of $R_1 + R_2$ . Always match the left-gap resistance with the left wire length and the right-gap resistance with the right wire length.
Treating the parallel combination as $R_1 + R_2$ is incorrect. For two resistors in parallel, the equivalent resistance is $\frac{R_1 R_2}{R_1 + R_2}$ , so this expression must be used before substituting values.
Writing the final answer with units in the answer field is wrong for a numerical value answer. The physical quantity is $30 \, \Omega^2$ , but the recorded answer should be only $30$ .

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