Eight cells each having emf 2 V and the internal resistance of \( 0.02\,\Omega \) are connected in series with an external resistance of \( 9\,\Omega \). The current flowing through the external resistance is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.75 A
For cells in series the emfs add and the internal resistances add; the same current flows through the cells and the external resistor.
\( E_{\mathrm{total}} = 8 \times 2\,\mathrm{V} = 16\,\mathrm{V} \)
\( r_{\mathrm{total}} = 8 \times 0.02\,\Omega = 0.16\,\Omega \)
\( \displaystyle I = \frac{E_{\mathrm{total}}}{R + r_{\mathrm{total}}} = \frac{16\,\mathrm{V}}{9\,\Omega + 0.16\,\Omega} = \frac{16}{9.16}\,\mathrm{A} \)
\( \displaystyle I \approx 1.75\,\mathrm{A} \)