Question

A current 0.8 A is drawn from a cell of e.m.f. 1.5 V, when internal resistance of cell is 1.6 Ω. The drop in potential across the terminal of cell will be:

पर्याय

• 1.24 V

• 1.28 V

• 1.20 V

• 1.16 V

Answer 1

1.28 V

Explanation:

Drop in potential $$= Ir$$

$$= 0.8 \times 1.6$$

$$= 1.28$$ V

This is the potential lost inside the cell due to its internal resistance.