Why two 1000-W heaters in series draw less than 2000 W

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
16 replies · 7K views
chantalprince
Messages
54
Reaction score
0

Homework Statement


The power rating of a 1000-W heater specifies the power consumed when the heater is connected dto an AC voltage of 120 V. Explain why the power consumed by two of these heaters connected in series with a voltage of 120 V is not 2000-W.



Homework Equations


P= (V)^2/R

(?) P = IVsin^2(2 pi ft)

f= frequency
t= time



The Attempt at a Solution


Is it because there are 2x the resistance? Therefore lowering P? In my text it starts to explain something about sinusoidal fluctuation but I don't really get how it pertains to this question, however I feel it may...
 
Physics news on Phys.org

Discussion

chantalprince said:
That the watts used doesn't change.
:bugeye:

So you think that if you have a device that uses 1000W when it is hooked up to 120V it will still use up 1000W if you turned the voltage down by half? :wink:
 
(blush) Well, I thought that each unit was turned down by half. There are 2 heaters connected to the in the series.
 
chantalprince said:
(blush) Well, I thought that each unit was turned down by half. There are 2 heaters connected to the in the series.
Yes.

The only way that each heater can produce 1000W (for a total of 2000W) is if each heater is given the full 120V. But we just showed that each heater gets only half the voltage, thus the total power must be less than 2000W.
 
Ok :) So is the total power between the 2 heaters 1000-W?
 
chantalprince said:
Ok :) So is the total power between the 2 heaters 1000-W?
Not necessarily. You can answer that by applying one of the equations from your first post. What happens to the total resistance?
 
Last edited:
What?! I am so lost here! Ok, so if the total watts for one heater connected to 120 V is 1000 watts, and if 2 heaters are in a series with 120 V and each heater has 60 V, then wouldn't the total watts add back up to 120 and leave us where we started?
 
Ok, I think I got it. So P = V^2/R that gives me R = 14.4 When the voltage is cut in half then it becomes P = 60^2/14.4 so P = 250...is this correct?!
 
The total voltage (not wattage) would add up to 120 V. But that voltage is now spread out over both heaters. Use your power equation: P = V^2/R. If the total volts remains the same, but the resistance doubles, what happens to the total power consumed?
 
So R = 14.4 when voltage is 120 for ONE heater. When there are 2 R doubles. So, the total watts for the 2 heaters is 500 W?
 
Right. Since the total resistance doubles while the voltage remains the same, the power drops to half.
 

Similar threads

Power Rating of 570 W Heaters in Series: 1140 W
  • GreenDinos
  • · Replies 2 ·
  • Introductory Physics Homework Help
Replies
2