Three cells with internal resistance are connected as shown in the figure. The equivalent emf of the combination of three cells is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 V
Two cells in parallel act as one cell of emf
\( \displaystyle E_{p} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \)
where an emf is taken as negative if that cell faces the other way. Cells in series then add their emfs with signs. In a cell symbol the longer plate is the positive terminal.
\( E_1 = 4\,\mathrm{V},\; r_1 = 3\,\Omega \;(\text{positive terminal on the left}) \)
\( E_2 = 10\,\mathrm{V},\; r_2 = 3\,\Omega \;(\text{positive terminal on the right}) \)
\( 5\,\mathrm{V},\; 1.5\,\Omega \;(\text{positive terminal on the left}) \)
The 4 V and 10 V cells face opposite ways. Taking the direction of the 10 V cell as positive,
\( \displaystyle E_{p} = \frac{10 \times 3 + (-4) \times 3}{3 + 3}\,\mathrm{V} = \frac{18}{6}\,\mathrm{V} = 3\,\mathrm{V} \)
with its positive terminal on the right. The 5 V cell, in series with this pair, has its positive terminal on the left, so it opposes the 3 V.
\( \displaystyle E_{\mathrm{eq}} = 5\,\mathrm{V} - 3\,\mathrm{V} = 2\,\mathrm{V} \)