AP EAMCET · PHYSICS · Current Electricity
In a meter bridge, the null point is located at 20 cm from left end of the wire when resistances R and S are connected in the left and right gaps respectively. If the resistance \(S\) is shunted with \(60 \Omega\) resistance, the null point shifted by 5 cm, then the values of \(R\) and \(S\) are respectively
- A \(24 \Omega, 6 \Omega\)
- B \(6 \Omega, 24 \Omega\)
- C \(5 \Omega, 20 \Omega\)
- D \(20 \Omega, 5 \Omega\)
Answer & Solution
Correct Answer
(C) \(5 \Omega, 20 \Omega\)
Step-by-step Solution
Detailed explanation
\(\frac{R}{S} = \frac{20}{100-20} = \frac{20}{80} = \frac{1}{4}\) \(S = 4R\) Shunted resistance \(S' = \frac{S \times 60}{S+60}\) Null point shifts by 5 cm. Since \(S' \(\frac{R}{S'} = \frac{25}{100-25} = \frac{25}{75} = \frac{1}{3}\) \(S' = 3R\)…
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