Two cells, each of emf 2 V, have internal resistances 1 Ω and 0.25 Ω. They are joined in series, and the combination is connected across an external resistor \(R\). The potential difference across the terminals of the cell of internal resistance 1 Ω is found to be zero. The value of \(R\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.75 Ω
In a series combination, the emfs add and the internal resistances add. The potential difference across the terminals of the first cell is \( V_1=\varepsilon-Ir_1 \). It becomes zero when the lost voltage \( Ir_1 \) equals the emf.
\( \begin{aligned} \varepsilon-Ir_1 &= 0 \\ I &= \frac{\varepsilon}{r_1} \\ &= \frac{2}{1} \\ &= 2\ \mathrm{A} \end{aligned} \)
The same current flows through the whole series circuit.
\( \begin{aligned} I &= \frac{2\varepsilon}{R+r_1+r_2} \\ 2 &= \frac{4}{R+1.25} \\ R+1.25 &= 2 \\ R &= 0.75\ \Omega \end{aligned} \)
From the above, \( R=0.75\ \Omega \). Hence, the terminal voltage of the first cell is zero for this value of \( R \).
NCERT (2026-27):
Cells in Series and in Parallel – “Then V (A) – V (B) is the potential difference between the positive and negative terminals of the first cell. …”
\( V_{AB}\equiv V(A)-V(B)=\varepsilon_1-Ir_1 \)\( \qquad (3.41) \)
“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.”