Question

In a meter bridge consisting of $R_1$ and $R_2$ is balanced at point J, which is 40 cm from A. When 10 $\Omega$ resistor is connected in series with $R_1$, then balance point shifts by 10 cm towards right. How much resistance should be connected in parallel with $(R_1+10)$ so that balance point regain it previous position?

• A. 60 $\Omega$
• B. 80 $\Omega$
• C. 120 $\Omega$
• D. 10 $\Omega$

Answer 1

The correct option is A 60 $\Omega$

From meter bridge principle, we get that,
$\frac{R_1}{R_2}=\frac{40}{60}$
$\Rightarrow R_1=\frac{2}{3}{R}_2$

Now, according to question,
$\frac{R_1+10}{R_2}=\frac{50}{50}$
$\Rightarrow R_1+10=R_2$

Again if we connect a resistor $R_3$ in parallel to $R_1+10$, we have :
$\frac{(R_1+10)||R_3}{R_2}=\frac{40}{60}$

Combining the above equations, we get $R_3=60\Omega$