


Thread Tools  Display Modes 
09212011, 06:59 AM  #31 
Newbie
Join Date: Sep 2011
Posts: 6
Rewards Points: 10

Amp / Watt analysis
Thanks you, that was my point
Advertisement 
09212011, 08:37 AM  #32  
Member
Join Date: Nov 2007
Location: Nashua, NH, USA
Posts: 7,352
Rewards Points: 2,806

Amp / Watt analysisQuote:
The number of volts (not the percentage of volts) dropped in a given piece of wire depends on the amperes flowing. Losing 10 out of 240 volts is less significant compared with losing 10 out of 120 volts. For the typical multivoltage motor the amperes drawn (and the voltage drop in the lines) with the motor set for 240 volts is about half that compared with 120 volts. So we are now talking losing 5 out of 240 volts compared with losing 10 out of 120 volts in the wires. (example only) Quote:
The load (motor, etc.) has a given resistance. Since amperes equals volts divided by resistance at all times, doubling the volts will also double the amps if the resistance stays the same. In practice the resistance usually increases in this situation because the load heats up and the resistance increases so it is not a proportional increase in amperes for an increase in volts. When the motor is reconfigured for the higher voltage, its resistance is greater so the motor delivers about the same output in terms of work performed, the watts consumed is about the same as before under the lower voltage, and the amperes drawn is less. Undervoltage can also burn out motors. The physical load imposed causes the resistance to go down (due to such factors as lowered RPM), the amperes drawn goes up, and the motor can oveheat.
__________________
Do you get coverage you can count on? From the bag of fertilizer? From your lawn sprinklers? Last edited by AllanJ; 09212011 at 09:09 AM. 

09212011, 09:50 AM  #33  
I=E/R
Join Date: May 2010
Location: Minnesota
Posts: 2,052
Rewards Points: 1,000

Amp / Watt analysisQuote:
BTW 1875/220=8.5 amps  needed to make numbers work Ohms Law is a three part equation. Voltage, Current, and resistance (or impedance) The flaw in your above example is that power is constant regardless of the applied voltage to the hair dryer. The only thing constant is the resistance of the hair dryer so the resistance of your hair dryer is 220/8.5=25.9 ohms If you were to plug this hair dryer into a 110 power source: I=E/R so I=110/25.9=4.25 amps as compared to 220/25.9=8.5 amps. If the hair dryer would even operate at 110 volts the power used would be 110*4.25=468 watts Power is also this: P=I²R and therefore R=P/I² P=I²R or 4.25²*25.9=468 watts In regards to the 80 amps for the stove, are you adding the value of two 40 amp breakers? AdvertisementLast edited by a7ecorsair; 09212011 at 10:02 AM. 

The Following User Says Thank You to a7ecorsair For This Useful Post: 
LivingLight (09222011)

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Switching from incandescent to CFL  Mooe  Electrical  8  12252007 11:27 AM 
help me solve this mystery please  lindaberrien  Roofing/Siding  17  06122007 01:06 AM 
100 Watt Max  Big Bill  Electrical  2  03292007 05:34 PM 