A cell is a source of electric current in the electrical circuit. The Potential difference between terminals of a cell in an open circuit (when no current is drawn) is called electromotive force (emf) of the cell. When electrical circuit is closed and current is drawn from the terminal potential difference between two terminals is called terminal potential difference (v) of the cell. The cells can be connected in series as well as in parallel combinations. Like resistor cell also offers opposition to the flow of current. This opposition offered by cell is called internal resistance of the cell.
Two cells of emf's \( \varepsilon_1 \) and \( \varepsilon_2 \) and respective internal resistances \( r_1 \) and \( r_2 \) are connected in parallel as shown in figure. The effective emf will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \)
In the figure both cells have their positive terminals on the same side, so they are in parallel with like terminals joined. Both cells have the same terminal potential difference \( V \), and the total current \( I \) is the sum of the currents through the two cells.
\( V = \varepsilon_1 - I_1 r_1 \)
\( V = \varepsilon_2 - I_2 r_2 \)
\( \displaystyle I = I_1 + I_2 = \frac{\varepsilon_1 - V}{r_1} + \frac{\varepsilon_2 - V}{r_2} \)
Multiplying by \( r_1 r_2 \) and solving for \( V \):
\( \displaystyle I\,r_1 r_2 = \varepsilon_1 r_2 + \varepsilon_2 r_1 - V\,(r_1 + r_2) \)
\( \displaystyle V = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} - I\,\frac{r_1 r_2}{r_1 + r_2} \)
Comparing with \( V = \varepsilon_{\mathrm{eq}} - I r_{\mathrm{eq}} \) for a single equivalent cell:
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \)
Shortcut: Put \( \varepsilon_1 = \varepsilon_2 = \varepsilon \): two identical-emf cells in parallel must give emf \( \varepsilon \). Option (2) gives \( \varepsilon(r_1 + r_2)/(r_1 + r_2) = \varepsilon \). Option (4) gives \( \varepsilon(r_1 - r_2)/(r_1 + r_2) \) and Option (3) gives \( -\varepsilon \), so both fail. Option (1) also gives \( \varepsilon \), but put \( r_2 \to 0 \): an ideal cell \( \varepsilon_2 \) fixes the voltage, so the answer must be \( \varepsilon_2 \); Option (2) gives \( \varepsilon_2 \), while Option (1) gives \( \varepsilon_1 \).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Cells – “… If we want to replace the combination by a single cell, between B1 and B2, of emf εeq and internal resistance req, we would have
\( \displaystyle V = \varepsilon_{eq} - I r_{eq} \qquad (3.53) \)
The last two equations should be the same and hence
\( \displaystyle \varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \qquad (3.54) \)”