Question

Figure shows a meter bridge. Initially null point was achieved at a distance of \(40\ \text{cm}\). When resistance \(16\,\Omega\) is attached in parallel with \(R_2\), new balance point was achieved at \(50\ \text{cm}\). Then find the values of \(R_1\) and \(R_2\):

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

Hint:

For meter bridge problems:
Always write balance condition as \(\frac{R_1}{R_2}=\frac{l}{100-l}\)
Be careful when resistances are modified using series or parallel combinations

Answer 1

The Correct Option is C

Concept:
In a meter bridge at balance condition: \[ \frac{R_1}{R_2}=\frac{l}{100-l} \] where \(l\) is the balance length in cm from the left end.
Step 1: Use the initial balance condition. Given \(l=40\ \text{cm}\): \[ \frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3} \] \[ R_1=\frac{2}{3}R_2 \qquad \cdots (1) \]
Step 2: Find effective resistance when \(16\,\Omega\) is connected in parallel with \(R_2\). \[ R_2'=\frac{16R_2}{16+R_2} \]
Step 3: Apply the new balance condition. New balance length \(=50\ \text{cm}\): \[ \frac{R_1}{R_2'}=\frac{50}{50}=1 \] \[ R_1=R_2' \qquad \cdots (2) \]
Step 4: Substitute from equations (1) and (2). \[ \frac{2}{3}R_2=\frac{16R_2}{16+R_2} \] Cancel \(R_2\neq0\): \[ \frac{2}{3}=\frac{16}{16+R_2} \] \[ 2(16+R_2)=48 \Rightarrow 32+2R_2=48 \Rightarrow R_2=8\,\Omega \]
Step 5: Find \(R_1\). \[ R_1=\frac{2}{3}\times 8=\frac{16}{3}\,\Omega \] \[ \boxed{R_1=\frac{16}{3}\,\Omega,\quad R_2=8\,\Omega} \]