Two cells with internal resistances \( r_1 \) and \( r_2 \) are connected across the load resistance R as shown in the figure. The values of \( r_1, r_2 \) and R are \( (1/2)\,\Omega \), \( (1/3)\,\Omega \) and \( (9/5)\,\Omega \), respectively. The current through R is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 7/10 A
The two cells are joined in parallel with like terminals together (the longer, positive plates face the same way). Such a pair acts as a single cell of
\( \displaystyle E_{\mathrm{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \qquad r_{\mathrm{eq}} = \frac{r_1 r_2}{r_1 + r_2} \)
\( E_1 = 2\,\mathrm{V},\; r_1 = 1/2\,\Omega \)
\( E_2 = 1\,\mathrm{V},\; r_2 = 1/3\,\Omega \)
\( \displaystyle E_{\mathrm{eq}} = \frac{2 \times (1/3) + 1 \times (1/2)}{1/2 + 1/3}\,\mathrm{V} = \frac{7/6}{5/6}\,\mathrm{V} = \frac{7}{5}\,\mathrm{V} \)
\( \displaystyle r_{\mathrm{eq}} = \frac{(1/2)(1/3)}{5/6}\,\Omega = \frac{1}{5}\,\Omega \)
The load \( R \) is in series with this equivalent cell.
\( \displaystyle I = \frac{E_{\mathrm{eq}}}{R + r_{\mathrm{eq}}} = \frac{7/5}{9/5 + 1/5}\,\mathrm{A} = \frac{7/5}{2}\,\mathrm{A} = \frac{7}{10}\,\mathrm{A} \)