An electric bulb is connected to a 220 V
genera-
tor. The current is 0.5 A. What is the power of
the bulb? Calculate the total cost of heat gener-
ated by it for 2 hours when 1 joule cost Rs 3.
Answers
Answer:
(a) Joule's law of heating states that the power of heating generated by an electrical conductor is proportional to the product of its resistance and the square of the electric current passing through the conductor.
Electric bulbs are usually filled with argon or krypton.
In Light Bulb, tungsten filament (due to its high melting temperature) is heated to extreme temperature so as to emit enough amount of visible light. However to prevent tungsten from catching fire (combustion) at these high temperatures, tungsten filament is sealed in oxygen free chamber. But creating vacuum, i.e., lack of surrounding air pressure to avoid tungsten filament combustion leads to evaporation of tungsten at extreme high temperatures which gets collected inside portion of the glass of electric bulb and reduces light output. Hence in electric bulb inert gases typically argon is filled to avoid loss of tungsten due to evaporation. Also as argon is inert gas there is no chance of combustion of tungsten filament.
(b) The power input is (P)=VI
Thus, the current, I=
V
P
When heat is at the maximum rate, I=
220
840
=3.82A
The resistance of the electric iron, R=
I
V
==
3.82
220
=57.59Ω
Explanation:
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