Q 12-03-033NEETJEE MainMedium
Four identical cells, each of emf $1.5$ V and internal resistance $0.5\ \Omega$, are connected in series across a $4\ \Omega$ resistor. The current through the resistor is
Answer: (B) $1$ A
$E = 6$ V, internal resistance $= 2\ \Omega$. $I = \dfrac{6}{4 + 2} = 1$ A.
Solution by Sreeraj P, M.Sc Physics