Two cells are joined to form a closed loop. One cell has emf 6 V and internal resistance 1 Ω. The other has emf 3 V and internal resistance 2 Ω. The positive terminal of each cell is joined to the negative terminal of the other. There is no other resistor in the loop. The potential difference across the terminals of the 6 V cell is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 V
Both cells drive current the same way round the loop. The only resistances in the loop are the internal resistances of the cells. By the loop rule, the changes in potential round the loop add up to zero.
\( \begin{aligned} 6+3-I\times1-I\times2 &= 0 \\ I &= \frac{9}{3} \\ &= 3\ \mathrm{A} \end{aligned} \)
This current flows through the 6 V cell from its negative to its positive terminal. So the terminal voltage of this cell is less than its emf.
\( \begin{aligned} V &= \varepsilon-Ir \\ &= 6-3\times1 \\ &= 3\ \mathrm{V} \end{aligned} \)
For the 3 V cell, the terminal voltage is \( 3-3\times2 \), which is \( -3\ \mathrm{V} \).
NCERT (2026-27):
Kirchhoff’s Rules – “Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero …”
Cells, Emf, Internal Resistance – “If however R is finite, I is not zero. In that case the potential difference between P and N is V = V++ V– – I r = ε – I r”