First slide

Ohm's law

Question

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

Easy

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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