AP EAMCET · PHYSICS · Current Electricity
Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at \(20 \mathrm{~cm}\). When a resistance of \(15 \Omega\) is connected in series with the smaller resistance of the two, the balancing point shifts to \(40 \mathrm{~cm}\). The value of smaller resistance is
- A \(9 \Omega\)
- B \(12 \Omega\)
- C \(6 \Omega\)
- D \(3 \Omega\)
Answer & Solution
Correct Answer
(A) \(9 \Omega\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{L}=20 \mathrm{~cm} ; \mathrm{L}^{\prime}=40 \mathrm{~cm} \\ & \text { Using } \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}}{100-\mathrm{L}}=\frac{20}{100-20}=\frac{20}{80} \\ & \Rightarrow \mathrm{R}_2=4 \mathrm{R}_1 \\ &…
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