A cell of emf 1.1 V and internal resistance 0.5 Ω is connected to a wire of resistance 0.5 Ω. Another cell of the same emf is connected in series but the current in the wire remains the same. The internal resistance of the second cell is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.0 Ω
By Ohm’s law for a complete circuit, the current is the total emf divided by the total resistance. For cells in series both the emfs and the internal resistances add.
With one cell:
\( \displaystyle I = \frac{\varepsilon}{R + r_1} = \frac{1.1\,\mathrm{V}}{0.5\,\Omega + 0.5\,\Omega} = 1.1\,\mathrm{A} \)
With both cells in series, the current is the same:
\( \displaystyle I = \frac{2\varepsilon}{R + r_1 + r_2} \)
\( \displaystyle 1.1\,\mathrm{A} = \frac{2.2\,\mathrm{V}}{1.0\,\Omega + r_2} \)
\( \displaystyle 1.0\,\Omega + r_2 = 2.0\,\Omega \)
\( \displaystyle r_2 = 1.0\,\Omega \)
Shortcut: The emf is doubled, so for the same current the total resistance must also double, from 1.0 Ω to 2.0 Ω; the second cell must add 1.0 Ω.
Note: A similar question was also asked in CUET (UG) 2024 on 16-05-2024 (Slot 1).