A cell of internal resistance 0.5 Ω is connected across an external resistor R. The potential difference across R is 80% of the emf of the cell. The value of R is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2 Ω
The cell drives a current \( I=\varepsilon/(R+r) \) through the circuit. So the voltage across the external resistor is \( V=IR \), which equals \( \varepsilon R/(R+r) \). The rest of the emf is used up across the internal resistance.
Here \( V=0.8\,\varepsilon \) and \( r=0.5\ \Omega \).
\( \begin{aligned} \frac{R}{R+r} &= 0.8 \\ R &= 0.8R+0.8r \\ 0.2R &= 0.8\times0.5 \\ R &= 2\ \Omega \end{aligned} \)
NCERT (2026-27):
Summary – “11. When a source of emf ε is connected to an external resistance R, the voltage Vext across R is given by”
\( V_{ext}=IR=\frac{\varepsilon}{R+r}R \)
“where r is the internal resistance of the source.”