When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is

When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is

A
1.0 V
B
1.5 V
C
2.0 V
D
2.5 V

Solution:

Since $i = (\frac{E}{R + r}) . we get$

$0.5 = \frac{E}{2 + r}$ ….(i)

$0.25 = \frac{E}{5 + r}$ ….(ii)

Dividing (i) by (ii), we get $2 = \frac{5 + r}{2 + r} \Rightarrow r = 1 Ω$

$∴ 0 .5 = \frac{E}{2 + 1} \Rightarrow E = 1 .5 V$

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