A cell of emf 1.1 V and internal resistance 0.5Ω is connected to a wire of resistance 0.5Ω. 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 resistances of the second cell is ___________.

Fill in the blank with the correct answer from the options given below.

$1 \Omega$

$ I = \frac{1.1}{0.5+0.5} = 1.1A$

After connecting the anothe cell the current become again 1-1 A.

$\Rightarrow 1.1 = \frac{1.1+1.1}{1+R} \Rightarrow R+1 = 2 \Rightarrow R = 1\Omega$

The correct answer is Option (1) → $1 \Omega$