A battery of 2 V emf and internal resistance of \( 0.5\,\Omega \) is connected across a resistance of \( 9.5\,\Omega \). The number of electrons passing through the resistance in 1 second.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.25 \times 10^{18} \)
The current is fixed by the emf and the total resistance (external plus internal). The charge passing in time \( t \) is \( q = It = ne \).
\( \displaystyle I = \frac{E}{R + r} = \frac{2\,\mathrm{V}}{9.5\,\Omega + 0.5\,\Omega} = 0.2\,\mathrm{A} \)
In \( t = 1\,\mathrm{s} \), with \( e = 1.6\times10^{-19}\,\mathrm{C} \),
\( \displaystyle n = \frac{I t}{e} = \frac{0.2 \times 1}{1.6\times10^{-19}} = 1.25\times10^{18} \)