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Basic EE quesiton...

F

FP

Diamond Member
I had a simple question about electricity...

Here is how I understand it...

When you increase the volts in a circuit you also increase the amps (assuming fixed resistance). So if you double the volts you are also doubling the number of electrons flowing through the circuit for a fixed amount of time. Example...

110 Volts = 5 Amps * 22 ohms
220 Volts = 10 Amps * 22 ohms

So in the second example there are twice as many electrons flowing through the circuit in time T.

Since Watts = Volts * Amps we have...

550 Watts = 110 Volts * 5 Amps
2200 Watts = 220 Volts * 10 Amps

The wattage has quadrupled after doubling the volts (assuming a fixed resistance). As stated above we have twice the number of electrons flowing in the circuit. This is confusing me because I can't figure out why power (watts) = volts * amps if amps is the measure of electrons in the circuit.

Are the electrons that are flowing in the circuit in example 2 "more excited" (i.e are they carrying more charge) than the electrons in example 1?

Thanks for any help!
 
Well the equations make since. Watts is voltage * amp, and if you double each one, then you will quadruple the wattage. I think your idea of wattage is just wrong when you are thinking about it in terms of electrons
 
Power (wattage) = (V^2) /R = (I^2)R, so a doubling in current or voltage yields a four-fould increase in power. (2 Squared is 4) 😀
 
oldsmoboat - thanks, your reply was extremely helpful. unfortunately my question wasn't 'where can i pay to have my question answered?'

Jassi - amps = the rate of flow of charges. not necessarily positive ones. and it does represent the number of electrons (charge) flowing in a circuit for a given time period (rate)... at least according to http://en.wikipedia.org/wiki/Amperage.



 
flyboy84 - right. but i am trying to figure out where that exponential increase in power (wattage) is coming from. obviously there are more electrons (amperage) in the system but those only doubled. the only thing i could come up with to explain the quadrupling of power was that each individual electron contains more charge.
 
When you double the voltage across a fixed resistance, you also double the current, as you demonstrated.

Voltage is a measure of energy per charge.
Current is a measure of charges per second through the circuit.
Power = Energy / Sec = Energy/Charge * Charge/Sec = Voltage * Current

...but you already knew all that.

When you double both the voltage and the current, you're not only doubling the number of electrons flowing though the resistor, you're also doubling the energy each charge carries. Thus, the total power is quadrupled. By contrast, if you doubled voltage and resistance, current would remain the same and the total power would only be doubled.

Hope that helps.
 
let me chew on it and check my text books...

Lonely Pheonix seems to have described the phenomena to my satisfaction 🙂
 
Originally posted by: binister
oldsmoboat - thanks, your reply was extremely helpful. unfortunately my question wasn't 'where can i pay to have my question answered?'

Jassi - amps = the rate of flow of charges. not necessarily positive ones. and it does represent the number of electrons (charge) flowing in a circuit for a given time period (rate)... at least according to http://en.wikipedia.org/wiki/Amperage.
Click to expand...

Sorry, you are correct. The charges can be negative as well but it is not a convention that I have had to use. For all intents and purposes, I have used +'ive charge flowing as current. All you have to do is reverse the direction and all is well.
 
the quick answer is that , assuming a purely real load
(1) V = IR
(2) P = VI
rewrite (1) as I = V/R and substitute into (2)
to obtain P = V^2/R, which means that for fixed R, doubling V quadruples P. Simple as that 🙂
 
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