Question

When two resistances 𝑅₁ and 𝑅₂ connected in series and introduced into the left gap of a meter bridge and a resistance of 10Ω is introduced into the right gap, a null point is found at 60 cm from left side. When 𝑅₁ and 𝑅₂ are connected in parallel and introduced into the left gap, a resistance of 3Ω is introduced into the right-gap to get null point at 40 cm from left end. The product of 𝑅₁𝑅₂ is _______ Ω².

Answer 1

Correct Answer: 30

To determine the product \(R_1R_2\), we need to use the principles of a meter bridge. By applying the formula for the meter bridge in both scenarios, we can create equations to solve for \(R_1\) and \(R_2\).

For the first scenario, where \(R_1\) and \(R_2\) are in series, the total resistance \(R_s=R_1+R_2\), and we have a null point at 60 cm from the left side. Using the meter bridge principle:

\(\frac{R_s}{10}=\frac{60}{40}\)

\(\Rightarrow R_s = \frac{60}{40} \times 10 = 15 \, \Omega\)

For the second scenario, where \(R_1\) and \(R_2\) are in parallel, the equivalent resistance \(R_p=\frac{R_1R_2}{R_1+R_2}\), and the null point is at 40 cm from the left. Thus:

\(\frac{R_p}{3}=\frac{40}{60}\)

\(\Rightarrow R_p = \frac{40}{60} \times 3 = 2 \, \Omega\)

Now, \(R_p=\frac{R_1R_2}{R_1+R_2}=2\) and \(R_1+R_2=15\). Substituting \(R_1+R_2=15\) into the equation for \(R_p\):

\(\frac{R_1R_2}{15} = 2\)

\(\Rightarrow R_1R_2 = 2 \times 15 = 30 \, \Omega^2\)

Thus, the product \(R_1R_2\) is \(30 \, \Omega^2\), which falls within the given range [30,30].