A cell sends a current of 2 A through an external resistor. Its terminal voltage is then 1.3 V. Later the same cell is charged with a current of 1 A. Its terminal voltage during charging is 1.6 V. The emf of the cell is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.5 V
While the cell gives current, the current inside it flows from the negative to the positive terminal. Its terminal voltage is then less than the emf, \( V=\varepsilon-Ir \). During charging the current inside flows the other way, so \( V=\varepsilon+Ir \).
The two readings give
\( \begin{aligned} \varepsilon-2r &= 1.3 \\ \varepsilon+r &= 1.6 \end{aligned} \)
Subtracting the first relation from the second,
\( \begin{aligned} 3r &= 0.3 \\ r &= 0.1\ \Omega \end{aligned} \)
\( \begin{aligned} \varepsilon &= 1.6-r \\ &= 1.6-0.1 \\ &= 1.5\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Kirchhoff’s Rules – “… This will tell us the potential difference, V = V (P) – V (N) = ε – I r … between the positive terminal P and the negative terminal N; I here is the current flowing from N to P through the cell. If, while labelling the current I through the cell one goes from P to N, then of course V = ε + I r”