Four identical cells each of emf \( 2\ \mathrm{V} \) and internal resistance \( 0.5\ \Omega \), are connected in series and then in parallel across an external resistor of \( 2\ \Omega \). The ratio of current in the series to the parallel combination of cells is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 17:8 \)
Each arrangement of cells acts as one equivalent cell, and the current is found from \( I = \varepsilon_{\mathrm{eq}}/(R + r_{\mathrm{eq}}) \).
\( \varepsilon = 2\,\mathrm{V},\ r = 0.5\,\Omega,\ n = 4,\ R = 2\,\Omega \)
Series: the emfs add and the internal resistances add.
\( \varepsilon_{\mathrm{eq}} = 4\times2 = 8\,\mathrm{V},\quad r_{\mathrm{eq}} = 4\times0.5 = 2\,\Omega \)
\( \displaystyle I_{s} = \frac{8\,\mathrm{V}}{(2 + 2)\,\Omega} = 2\,\mathrm{A} \)
Parallel: identical cells give the same emf, and the internal resistance becomes \( r/n \).
\( \varepsilon_{\mathrm{eq}} = 2\,\mathrm{V},\quad r_{\mathrm{eq}} = 0.5/4 = 0.125\,\Omega \)
\( \displaystyle I_{p} = \frac{2\,\mathrm{V}}{(2 + 0.125)\,\Omega} = \frac{2}{2.125}\,\mathrm{A} = \frac{16}{17}\,\mathrm{A} \)
Ratio:
\( \displaystyle \frac{I_{s}}{I_{p}} = \frac{2}{16/17} = \frac{17}{8} \)