Joe wants to heat his 12\'X20\' workshop with electric heat. He has hired the HA
ID: 1899893 • Letter: J
Question
Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1380.0W 240V baseboard heaters to provide a total heating capacity of 4140.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram: Joe wants to heat his 12'X20' workshop with electric heat. He has hired the HACME electrician company to build the system. They propose to use three 1380.0W 240V baseboard heaters to provide a total heating capacity of 4140.0W. (A heater is basically a resistor. This is not quite true, because there is a thermostatic switch incorporated into the heater and because the resistance of a heater varies a bit with its temperature. But we will use a linear resistor as a model of a heater.) In the proposed system the heaters are connected in parallel with the 240V 60Hz AC power line (modeled by a voltage source) as shown in the diagram:
Remember that AC power-line voltages and currents are specified as RMS values. So 120V AC heats a given resistance exactly as much as 120V DC would heat that same resistance.
How much current is expected to be drawn from the power line by this heating system when all three heaters are on?
If instead, HACME chose to implement the system with 120V heaters, how much current would have been needed?
Notice that this would require much heavier and more expensive wire to distribute the power.
Back to the original plan with 240V power.
Unfortunately, Sparky, who works for HACME, was a little sleepy that day. He accidentally connected the heaters as shown below:
As a consequence, Joe found his workshop too cold. H1 was weak; H2 and H3 barely warmed up.
What power was being dissipated in H1?
What power was being dissipated in H2 (or in H3)?
So the total heating power in Joe's shop was:
Explanation / Answer
P = EI so if P = 1380 W and E = 240 V, I = P/E = (1380 W)/(240 V) = 5.75 A which is the current each heater draws; since you have three of them, the total current drawn is 17.25 A, which is the answer to your first question.
The answer to the second question, if this were implemented with 120V instead of 240V, is twice as much current (since you're using half as much voltage) so 34.5 A.
For the last series of questions, we know the heaters are resistors; their resistance, from Ohm's Law, is R = E/I = (240 V)/(5.75 A) = 41.739 . In the configuration shown, we have H2 and H3 in parallel, making a resistance of 20.870 , and H1 in series, for a total of 62.609 . This fact allows us to calculate the total current leaving the voltage source I = E/R = (240 V)/(62.609 ) = 3.833 A.
Knowing this, we know the power dissipated in H1 is I2R = (3.833 A)2(41.739 ) = 613.33 W. Since H2 = H3, we know the current divides equally through those two heaters, so the power dissipated in each is (3.833 A/2)2(41.739 ) = 153.33 W.
The grand total of power dissipated is from H1, H2, and H3 together = 920 W.
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