Question

Two resistances are connected in the two gaps of a meter bridge. The balance point is $20cm$ from the zero end. When a resistance $15\Omega$ is connected in series with the smaller of two resistance, the null point get shifted to $40cm$ . What is the smaller of the two resistance values ?

• A. $8\Omega$
• B. $9\Omega$
• C. $10\Omega$
• D. $12\Omega$

Answer 1

Answer: (B)

$\frac{R_1}{R_2}=\frac{20}{80}=\frac{1}{4}$ , It is clear that $R_1$ is smaller resistance
∴ $R_2=4R_1$
When $15\Omega$ is connected in series with $R_1$ , then equivalent resistance is $R_1+15$ , So
$\therefore \frac{R_1+15}{R_2}=\frac{40}{60}=\frac{2}{3}$
$\therefore \frac{R_1+15}{4R_1}=\frac{2}{3}$
$\therefore R_1=9\Omega$