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#13212 - 08/27/02 11:27 PM homework Q  
Eandrew  Offline
Joined: May 2002
Posts: 68
seattle, wa, usa
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?

Work Gear for Electricians and the Trades

#13213 - 08/28/02 12:07 AM Re: homework Q  
elektrikguy  Offline
Joined: May 2002
Posts: 132
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
at 240 volts: 240v/1200=5A

3000/120000=.25 or 1/4 at 240 volts

WHEW!! that has been awhile.

#13214 - 08/28/02 05:43 AM Re: homework Q  
pauluk  Offline
Joined: Aug 2001
Posts: 7,520
Norfolk, England
Don't like to criticize, but I think you're burning a little too much midnight-oil there! [Linked Image]

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

#13215 - 08/28/02 02:43 PM Re: homework Q  
elektrikguy  Offline
Joined: May 2002
Posts: 132
Whoops! Note time and had a long day before that.

#13216 - 08/29/02 06:57 AM Re: homework Q  
Redsy  Offline
Joined: Mar 2001
Posts: 2,056
Bucks County PA
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.

#13217 - 08/29/02 12:26 PM Re: homework Q  
pauluk  Offline
Joined: Aug 2001
Posts: 7,520
Norfolk, England

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.

#13218 - 08/29/02 03:07 PM Re: homework Q  
gpowellpec  Offline
Joined: Dec 2000
Posts: 127
Irving, Tx USA
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

#13219 - 08/29/02 10:46 PM Re: homework Q  
Steve Miller  Offline
Joined: Aug 2001
Posts: 325
Loudoun Cty, VA
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.

#13220 - 09/02/02 02:22 PM Re: homework Q  
Scott35  Offline

Broom Pusher and
Joined: Oct 2000
Posts: 2,707
Anaheim, CA. USA
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 math-wise:

(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)

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 Jack-Quick when subjected to higher than rated Voltages (AKA: 120 Volt lamp connected to 240 Volt circuit).

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!

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