AP EAMCET · PHYSICS · Current Electricity
' \(n\) ' identical cells each of emf \(E\) and internal resistance ' \(r\) ' are joined in series to form a row. ' \(m\) ' such rows are joined in parallel across a load resistance \(R\). The current in each cell is
- A \(\begin{aligned} & \frac{\mathrm{nE}}{\mathrm{nr}} \mathrm{m}+\mathrm{R}\end{aligned}\)
- B \(\begin{aligned} & \frac{\mathrm{nE}}{\mathrm{nr}+\mathrm{mR}} \end{aligned}\)
- C \(\begin{aligned} & \frac{\mathrm{mE}}{\frac{\mathrm{mr}}{\mathrm{n}}+\mathrm{R}}\end{aligned}\)
- D \(\begin{aligned} & \frac{\mathrm{E}}{\frac{\mathrm{nr}}{\mathrm{m}}+\mathrm{R}}\end{aligned}\)
Answer & Solution
Correct Answer
(B) \(\begin{aligned} & \frac{\mathrm{nE}}{\mathrm{nr}+\mathrm{mR}} \end{aligned}\)
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