Two identical immersion heaters are to be used to heat water,in a large container.Wich one of the following arrangement would heat the water faster: 1)connecting the heaters in series with the maiin supply, 2)connecting the heaters in parallel with the main supply?
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If two immersion heaters (resistance R each) are connected in series, then the total resistance = 2 R Ohms
Let the Mains voltage supply be V Volts.
Power dissipated = V² / (2R) Watts
If the heaters are connected in parallel with the mains:
Effective resistance = R *R /(R+R) = R/2
Power dissipated = V² / (R/2) = 2 V²/R
This is four times the power dissipated in a series connection.
Let the Mains voltage supply be V Volts.
Power dissipated = V² / (2R) Watts
If the heaters are connected in parallel with the mains:
Effective resistance = R *R /(R+R) = R/2
Power dissipated = V² / (R/2) = 2 V²/R
This is four times the power dissipated in a series connection.
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