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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.

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Question

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.

Short/Brief Note
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Solution

According to Ohm's law,

V = IR

∴ R = `"V"/"I"`

`therefore "R"_1 = "V"_1/"I"_1`

= `0.2/0.1 = 2  Omega`

Similarly ,

`"R"_2 = "V"_2/"I"_2`

= `1/0.4`

= 2.5 Ω

Resistance of the wire increases with increase in temperature. So the difference arises because the temperature of the filament increased.

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Chapter 8: Current Electricity - EXERCISE - 8(A) [Page 187]

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Selina Concise Physics [English] Class 10 ICSE
Chapter 8 Current Electricity
EXERCISE - 8(A) | Q 13. | Page 187

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