What are the factors of heat dissipated by a conductor depends?
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When current flows through a conductor, heat energy is generated in the conductor. The heating effect of an electric current depends on three factors:
The resistance, R of the conductor. A higher resistance produces more heat.The time, t for which current flows. The longer the time the larger the amount of heat producedThe amount of current, I. the higher the current the larger the amount of heat generated.
Hence the heating effect produced by an electric current, I through a conductor of resistance, R for a time, t is given by H = I2Rt. This equation is called the Joule’s equation of electrical heating.
Electrical energy and power
The work done in pushing a charge round an electrical circuit is given by w.d = VIt
So that power, P = w.d /t = VI
The electrical power consumed by an electrical appliance is given by P = VI = I2R = V2/R
Example
An electrical bulb is labeled 100W, 240V. Calculate:a)The current through the filament when the bulb works normally
b)The resistance of the filament used in the bulb.
Solution
I = P/V = 100/240 = 0.4167AR = P/I2 = 100/ 0.41672 = 576.04Ω or R = V2/P =2402/100 = 576ΩFind the energy dissipated in 5 minutes by an electric bulb with a filament of resistance of 500Ω connected to a 240V supply. { ans. 34,560J}
The resistance, R of the conductor. A higher resistance produces more heat.The time, t for which current flows. The longer the time the larger the amount of heat producedThe amount of current, I. the higher the current the larger the amount of heat generated.
Hence the heating effect produced by an electric current, I through a conductor of resistance, R for a time, t is given by H = I2Rt. This equation is called the Joule’s equation of electrical heating.
Electrical energy and power
The work done in pushing a charge round an electrical circuit is given by w.d = VIt
So that power, P = w.d /t = VI
The electrical power consumed by an electrical appliance is given by P = VI = I2R = V2/R
Example
An electrical bulb is labeled 100W, 240V. Calculate:a)The current through the filament when the bulb works normally
b)The resistance of the filament used in the bulb.
Solution
I = P/V = 100/240 = 0.4167AR = P/I2 = 100/ 0.41672 = 576.04Ω or R = V2/P =2402/100 = 576ΩFind the energy dissipated in 5 minutes by an electric bulb with a filament of resistance of 500Ω connected to a 240V supply. { ans. 34,560J}
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