The filament of a bulb takes a current 100 mA when the 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.

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Solution

Case I:

$V_{1} = 0.2 V I_{1} = 100 m A = 0.1 A$

$R_{1} = \frac{V_{1}}{I_{1}} = \frac{0.2}{0.1} = 2 Ω$

Case II:

$V_{2} = 1 V I_{2} = 400 m A = 0.4 A$

$R_{2} = \frac{V_{2}}{I_{2}} = \frac{1}{0.4} = 2.5 Ω$

The difference is on account of heating, which increases the resistance of the filament.

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The filament of a bulb takes a current 100 mA when the 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 I: V1=0.2 VI1=100 mA=0.1 A R1=V1I1=0.20.1=2 Ω Case II: V2=1 VI2=400 mA=0.4 A R2=V2I2=10.4=2.5 Ω The difference is on account of heating, which increases the resistance of the filament.

Case I:

$V_{1} = 0.2 V I_{1} = 100 m A = 0.1 A$

$R_{1} = \frac{V_{1}}{I_{1}} = \frac{0.2}{0.1} = 2 Ω$

Case II:

$V_{2} = 1 V I_{2} = 400 m A = 0.4 A$

$R_{2} = \frac{V_{2}}{I_{2}} = \frac{1}{0.4} = 2.5 Ω$

The difference is on account of heating, which increases the resistance of the filament.