Question

In a meter bridge experiment to determine the value of unknown resistance, first the resistances \(2\,\Omega\) and \(3\,\Omega\) are connected in the left and right gaps of the bridge and the null point is obtained at a distance \(l\) cm from the left end. Now, when an unknown resistance \(x\,\Omega\) is connected in parallel to \(3\,\Omega\), the null point is shifted by \(10\,\text{cm}\) to the right. The value of \(x\) is ________ \(\Omega\).

Hint:

Start by writing the balance condition for the first setup, $\dfrac{2}{3} = \dfrac{l}{100-l}$, and solve it to fix the value of l first. Then work out the new balance length after the shift, and use the new balance condition to write an equation connecting x with the combined resistance of 3 ohm and x in parallel. Remember that connecting x in parallel with the 3 ohm resistor always makes the combined resistance smaller than 3 ohms, which tells you which direction the null point should move.

Answer 1

Correct Answer: 6

In the meter bridge experiment, we start by using the formula for the Wheatstone bridge balance: \(\frac{R_1}{R_2}=\frac{l}{100-l}\), where \(R_1=2\,\Omega\) and \(R_2=3\,\Omega\). Therefore:
\( \frac{2}{3}=\frac{l}{100-l} \)
Cross-multiplying gives:
\(200-2l=3l\)
This simplifies to:
\(5l=200\)
Thus, \(l=40\,\text{cm}\).

Now, when \(x\) is connected in parallel with \(3\,\Omega\), the equivalent resistance \(R_2'\) is:

\(R_2'=\frac{3x}{3+x}\, \Omega\)
The new null point shifts 10 cm to the right, so \(l=50\,\text{cm}\).
The new balance equation is:
\( \frac{R_1}{R_2'}=\frac{50}{50}\Rightarrow R_1=R_2'\)
Therefore:
\(2=\frac{3x}{3+x}\)
Cross-multiplying yields:
\(2(3+x)=3x\)
Expanding gives:
\(6+2x=3x\)
Thus:
\(x=6\,\Omega\)
Verification confirms \(x\) is within the range \(6,6\). Therefore, the value of the unknown resistance \(x\) is 6 \(\Omega\).

Answer 2

Concept:

• For two resistors in parallel, the reciprocal formula $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ is often quicker to solve than the product-over-sum form, especially when solving for an unknown resistor.

Step 1: Find the first null point $l$.
$\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow l = 40\ \text{cm}$

Step 2: Find the new null point.
$l' = l + 10 = 50\ \text{cm}$

Step 3: Write the new balance condition.
$\dfrac{2}{R} = \dfrac{l'}{100-l'} = \dfrac{50}{50} = 1 \Rightarrow R = 2\ \Omega$

Step 4: Apply the reciprocal formula for the parallel combination.
$\dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{x}$
$\dfrac{1}{2} = \dfrac{1}{3} + \dfrac{1}{x}$

Step 5: Isolate and solve for $x$.
$\dfrac{1}{x} = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{3-2}{6} = \dfrac{1}{6}$
$x = 6\ \Omega$

Final Answer: $x = 6\ \Omega$