AP EAMCET · PHYSICS · Current Electricity
\(n\) identical resistance are taken in which \(\frac{n}{2}\) resistors are joined in series in the left gap and the remaining \(\frac{n}{2}\) resistances are joined in parallel in the right gap of a metre bridge. Balancing length in \(\mathrm{cm}\) is
- A \(100 \cdot \frac{n^2}{n^2+4}\)
- B \(100 \cdot \frac{n^2}{n^2+1}\)
- C \(400 \cdot \frac{1}{n^2+4}\)
- D \(400 \cdot \frac{1}{n^2+1}\)
Answer & Solution
Correct Answer
(A) \(100 \cdot \frac{n^2}{n^2+4}\)
Step-by-step Solution
Detailed explanation
Meter bridge is shown in the figure below, When \(\frac{n}{2}\) resistances are joined in series in left gap each of resistance \(R_1\), then equivalent resistance in left gap.…
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