A cell of emf 3 V and internal resistance 0.1 Ω is connected to a 3.9 Ω external resistance. What will be the potential difference across the terminals of the cell ?
Answer & explanation
Correct answer: option 2
Electric Current $I = \frac{E}{r+R} = \frac{3}{0.1+3.9} = 0.75A$
Potential difference across terminal of cell is $ V = E - Ir = 3 - 0.75\times 0.1 = 2.925V$