9 cells each of emf 3V and internal resistance \( 1\,\Omega \) are connected in 3 rows, each having 3 cells in series. The current flowing in the external resistance of \( 5\,\Omega \) connected across this arrangement of the cells is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.5 A
This is a mixed grouping of cells. Cells in series add their emfs and internal resistances; identical rows in parallel keep the emf of one row and divide its internal resistance by the number of rows.
For one row of 3 cells in series:
\( E_{\mathrm{row}} = 3 \times 3 = 9\,\mathrm{V} \)
\( r_{\mathrm{row}} = 3 \times 1 = 3\,\Omega \)
For 3 such rows in parallel:
\( E_{\mathrm{eq}} = 9\,\mathrm{V} \)
\( r_{\mathrm{eq}} = 3/3 = 1\,\Omega \)
The current in the external resistance is
\( \displaystyle I = \frac{E_{\mathrm{eq}}}{R + r_{\mathrm{eq}}} = \frac{9\,\mathrm{V}}{5\,\Omega + 1\,\Omega} = 1.5\,\mathrm{A} \)