A cell of emf 1.1 V and internal resistance \( 0.5\,\Omega \) is connected to a wire of resistance \( 0.5\,\Omega \). Another cell of the same emf is now connected in series with the intention of increasing the current but the current in the wire remains the same. The internal resistance of the second cell is ___________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1\,\Omega \)
For cells in series, the emfs add and the internal resistances add. The current is the total emf divided by the total resistance of the circuit.
With one cell:
\( \displaystyle I = \frac{E}{R + r_1} = \frac{1.1}{0.5 + 0.5} = 1.1\,\mathrm{A} \)
With the second cell (emf 1.1 V, internal resistance \( r_2 \)) added in series:
\( \displaystyle I' = \frac{2E}{R + r_1 + r_2} = \frac{2.2}{1 + r_2} \)
The current stays the same, so \( I' = I \):
\( \displaystyle \frac{2.2}{1 + r_2} = 1.1 \;\Rightarrow\; 1 + r_2 = 2 \;\Rightarrow\; r_2 = 1\,\Omega \)
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2).