A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. The current and terminal voltage are

(C) 0.5 A and 1.75 V. I = (2)/(3.5 + 0.5) = 0.5 A. V = E - Ir = 2 - 0.25 = 1.75 V.

Current Electricity question for NEET / JEE Main, with solution

Q 12-03-030NEETJEE MainEasy

A cell of emf $2$ V and internal resistance $0.5\ \Omega$ is connected to a $3.5\ \Omega$ resistor. The current and terminal voltage are