The filament of a bulb takes a current 100 mA when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 400 mA. Calculate the resistance of filament in each case and account for the difference.

Case 1 —

Current (I) = 100 m A = 0.1 A

Potential Difference (V) = 0.2 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

$0.2 = 0.1 \times R \\[0.5em] \Rightarrow R = \dfrac{0.2}{0.1} \\[0.5em] \Rightarrow R = 2 \varOmega \\[0.5em]$

Hence, resistance of filament of bulb = 2 Ω

Case 2 —

Current (I) = 400 m A = 0.4 A

Potentia Difference (V) = 1.0 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

$1.0 = 0.4 \times R \\[0.5em] \Rightarrow R = \dfrac{1.0}{0.4} \\[0.5em] \Rightarrow R = 2.5 \varOmega \\[0.5em]$

Hence, resistance of filament of bulb = 2.5 Ω

Therefore, we observe that with increase in temperature resistance of the wire increases.

Hence, resistance of filament increases with the increase in temperature.