Two cells of emf E and 2E with internal resistances r and 3r, respectively are connected in parallel. The emf and internal resistance of the combination is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{5}{4}E, \frac{3}{4}r \)
For two cells in parallel (like terminals joined), the equivalent internal resistance is that of the two internal resistances in parallel, and the equivalent emf is the resistance-weighted average of the emfs.
\( \displaystyle \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}} = \frac{E_1}{r_1} + \frac{E_2}{r_2}, \qquad \frac{1}{r_{\mathrm{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} \)
\( \displaystyle r_{\mathrm{eq}} = \frac{r \times 3r}{r + 3r} = \frac{3}{4}r \)
\( \displaystyle E_{\mathrm{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{E \times 3r + 2E \times r}{4r} = \frac{5}{4}E \)