Question

A meter bridge with two resistances \( R_1 \) and \( R_2 \) as shown in figure was balanced (null point) at 40 cm from the point \( P \). The null point changed to 50 cm from the point \( P \), when a \( 16\,\Omega \) resistance is connected in parallel to \( R_2 \). The values of resistances \( R_1 \) and \( R_2 \) are 

• A. \( R_2 = 4\,\Omega,\; R_1 = \frac{4}{3}\,\Omega \)
• B. \( R_2 = 16\,\Omega,\; R_1 = \frac{16}{3}\,\Omega \)
• C. \( R_2 = 8\,\Omega,\; R_1 = \frac{16}{3}\,\Omega \)
• D. \( R_2 = 12\,\Omega,\; R_1 = \frac{12}{3}\,\Omega \)

Hint:

For meter bridge problems, always apply the balance condition \( \dfrac{R_1}{R_2} = \dfrac{l}{100-l} \) before and after any modification in the circuit.

Answer 1

The Correct Option is C

To solve the problem of finding resistances \( R_1 \) and \( R_2 \) using a meter bridge, we can utilize the principle of a balanced Wheatstone bridge. The meter bridge consists of a 100 cm wire with a sliding contact or jockey.

Initially, the bridge is balanced at a point 40 cm from \( P \). This means the resistances \( R_1 \) and \( R_2 \) are in proportion to the lengths of the wire on either side of the jockey. The total length of the wire is 100 cm, and the balance length from \( Q \) is 60 cm.

The balance condition for a Wheatstone bridge is given by:

\[\frac{R_1}{R_2} = \frac{40}{60}\]

Simplifying the ratio, we get:

\[\frac{R_1}{R_2} = \frac{2}{3}\]

When a \(16\,\Omega\) resistor is connected in parallel with \( R_2 \), the null point changes to 50 cm from \( P \). The effective resistance \( R_2^\prime \) can be calculated using parallel resistance formula:

\[\frac{1}{R_2^\prime} = \frac{1}{R_2} + \frac{1}{16}\]

At this new balance point, we have:

\[\frac{R_1}{R_2^\prime} = \frac{50}{50}\]

Which implies:

\[R_1 = R_2^\prime\]

Substituting the value of \( R_1 \) in terms of \( R_2 \), we get:

\[\frac{2}{3} R_2 = R_2^\prime\]

From the equation for \( R_2^\prime \):

\[\frac{1}{\frac{2}{3} R_2} = \frac{1}{R_2} + \frac{1}{16}\]

Solving the above equation for \( R_2 \) gives:

\[\frac{3}{2 R_2} = \frac{1}{R_2} + \frac{1}{16}\]

Rearranging and solving:

\[\frac{1}{2 R_2} = \frac{1}{16} \Rightarrow R_2 = 8\,\Omega\]

Then, using \(\frac{R_1}{R_2} = \frac{2}{3}\):

\[R_1 = \frac{2}{3} \times 8 = \frac{16}{3}\,\Omega\]

Thus, the values of the resistances are \( R_2 = 8\,\Omega \) and \( R_1 = \frac{16}{3}\,\Omega \).

The correct answer is:

\( R_2 = 8\,\Omega,\; R_1 = \frac{16}{3}\,\Omega \)