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)
*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 Jack-Quick 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.
