Consider a series combination of 5 cells, each of internal resistance 2.0 Ω and e.m.f. 6.0 V, connected across an external load of 50 Ω. The current through the load is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5 A
For cells in series, both the emfs and the internal resistances add; the current is the total emf divided by the total resistance of the circuit.
\( \varepsilon_{\mathrm{eq}} = 5 \times 6.0\,\mathrm{V} = 30\,\mathrm{V} \)
\( r_{\mathrm{eq}} = 5 \times 2.0\,\Omega = 10\,\Omega \)
\( \displaystyle I = \frac{\varepsilon_{\mathrm{eq}}}{R + r_{\mathrm{eq}}} = \frac{30\,\mathrm{V}}{50\,\Omega + 10\,\Omega} = 0.5\,\mathrm{A} \)