Two cells of emf 3 V and 1.5 V, each of internal resistance 0.5 Ω, are joined in series and connected across a 2 Ω resistor. The current is \(I_1\) when the negative terminal of one cell is joined to the positive terminal of the other. The current is \(I_2\) when the two negative terminals are joined together. The ratio \(I_1:I_2\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3:1\)
In both cases the internal resistances add, so the total resistance of the circuit is the same. Only the equivalent emf changes. When the two negative terminals are joined, the current leaves the 1.5 V cell from its negative terminal. So its emf is taken with a negative sign.
\( \begin{aligned} R+r_1+r_2 &= 2+0.5+0.5 \\ &= 3\ \Omega \end{aligned} \)
\( \begin{aligned} I_1 &= \frac{\varepsilon_1+\varepsilon_2}{3} \\ &= \frac{3+1.5}{3} \\ &= 1.5\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} I_2 &= \frac{\varepsilon_1-\varepsilon_2}{3} \\ &= \frac{3-1.5}{3} \\ &= 0.5\ \mathrm{A} \end{aligned} \)
Therefore, \( I_1:I_2=1.5:0.5=3:1 \).
NCERT (2026-27):
Cells in Series and in Parallel – “The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and (ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.”
Cells in Series and in Parallel – “This is so, when the current leaves each cell from the positive electrode. If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for εeq with a negative sign, as in Eq. (3.47).”
\( \varepsilon_{eq}=\varepsilon_1-\varepsilon_2\ \ (\varepsilon_1 \gt \varepsilon_2) \)\( \qquad (3.47) \)