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homework Q
#13212
08/27/02 11:27 PM

Joined: May 2002
Posts: 68
OP
Member

it askes
If a 240 volt electric heater is used on a 120 volt ckt., the amount of heat produced will be _____. (=1/4 on 120 volts)
Can anyone prove this mathematically?



Re: homework Q
#13213
08/28/02 12:07 AM

Joined: May 2002
Posts: 132
Member

ok here goes:
Remamber that the resistance never changes but the voltage applied will.
I will use 1200 ohms for example
at 120 volts: 120v/1200=10A (10Ax10A)x1200=120000W at 240 volts: 240v/1200=5A (5Ax5A)x1200=3000W
3000/120000=.25 or 1/4 at 240 volts
WHEW!! that has been awhile.



Re: homework Q
#13214
08/28/02 05:43 AM

Joined: Aug 2001
Posts: 7,520
Member

Don't like to criticize, but I think you're burning a little too much midnightoil there! 120V / 1200 ohms = 0.1A (0.1 x 0.1) x 1200 = 12W 240V / 1200 ohms = 0.2A (0.2 x 0.2) x 1200 = 48W 12 / 48 = 0.25



Re: homework Q
#13215
08/28/02 02:43 PM

Joined: May 2002
Posts: 132
Member

Whoops! Note time and had a long day before that.



Re: homework Q
#13216
08/29/02 06:57 AM

Joined: Mar 2001
Posts: 2,056
Member

Well, You know that Watts = V X A. And you know that Amps = V/R If you half the voltage on a fixed resistance, you also cut the current flow in half. So, you have half the voltage and half the current which is equal to 1/4 the wattage.
240 V. x 6 A. = 1,440 W. 120 V. x 3 A. = 360 W.
1,440 x 1/4 = 360.



Re: homework Q
#13217
08/29/02 12:26 PM

Joined: Aug 2001
Posts: 7,520
Member

Erik, You might find it useful to compare this situation with this thread from a couple of months ago. The 60W bulb example I used there shows how the resistance of the filament in a 240V bulb has to be four times greater than that of a 120V bulb if the power is to remain at 60W. Your present problem is just looking at this same basic equation from the opposite direction, i.e. what happens to power if the resistance is kept constant.



Re: homework Q
#13218
08/29/02 03:07 PM

Joined: Dec 2000
Posts: 127
Member

To prove this mathematically without creating voltage, resistance and current values, and to eliminate variables, use the formula Power = the square of the Voltage divided by the Resistance.
Equation 1 P = E Square/R
Equation 1A P = E X E / R This is the same as the following equation because of the understood coefficients of the components.
Equation 1B 1P = 1E X 1E / 1R
Equation 2 1P = 0.5E X 0.5E /1R
Equation 2A P = 0.25E / R
Both equations 1 and 2 are valid and the value for E and R in the equations have not changed. Just the coefficients of E has changed. The original voltage value is halved. By halving the coefficients of V the value of Equation 2 is 0.25 the value of Equation 1.
Since equation 2A is proven mathematically it can be used in the following form: P = (New E/Original E) X (New E/Original E) X Original E / R
If the value for E is unchanged the ratio of new to original is one and there is no change in power. If new is larger the ratio is greater than 1 and the power is increased; if original is larger the ratio will be less than 1 and the power is decreased
Gerald Powell



Re: homework Q
#13219
08/29/02 10:46 PM

Joined: Aug 2001
Posts: 329
Member

As an apprenticeship instructor I only have one comment .... I wish my students were resourceful enough to ask for help in here. Heck I'd answer their questions if I saw them posted here. After all, electricity is an open book career.



Re: homework Q
#13220
09/02/02 02:22 PM

Joined: Oct 2000
Posts: 2,721
Broom Pusher and Member

Here's a simple proof formula:
Figure a certain Heater [AKA Resistor] having a Resistance of 200 Ohms will draw a True Power of 200 Watts [dissipate that much heat energy], when connected to a power supply with an EMF [Voltage] of 200 Volts  thus driving one Ampere through the Resistor.
Looks like this mathwise:
(E= Voltage / EMF, I= Amperes / Intensity, R= Resistance in Ohms, P= True Power in Watts)
I = E / R [or E x R = I] R=200 Ohms E=200 Volts I=1 Ampere ( 200 / 200 = 1 )
E x I = P [True Power across a linear or pure resistance load] E=200 Volts I=1 Amperes P=200 Watts
Now Reduce the Voltage by 50% [200 x .5 = 100]
R=200 Ohms E=100 Volts I=0.5 Ampere (100 / 200 = 0.5)
*ExI=P* E=100 Volts I=0.5 Ampere P= 50 Watts.
Total Power drawn by the load is 25% [1/4] when applied to 50% [1/2] the voltage.
Works the other way too! This is why Incandescent Lamps blow out JackQuick when subjected to higher than rated Voltages (AKA: 120 Volt lamp connected to 240 Volt circuit). Example:
100 Volt, 100 Watt Incandescent Lamp. R=100 Ohms (at full brightness/ operating Temperature) E=100 Volts I=1 Ampere P=100 Watts
Double the Voltage; R=100 Ohms [at design limits shown above] E=200 Volts I=2 Amperes P=400 Watts
This should be conclusive Math Proof
Scott S.E.T.
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!




