' $n$ ' 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

  $\begin{aligned} & \frac{\mathrm{nE}}{\mathrm{nr}} \mathrm{m}+\mathrm{R}\end{aligned}$
• Correct answer
  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

  $\begin{aligned} & \frac{\mathrm{E}}{\frac{\mathrm{nr}}{\mathrm{m}}+\mathrm{R}}\end{aligned}$

Hence, the correct answer is option: 'B'.