Alternating Current Reverse Of Polarity

Thread Starter

hitmen

Joined Sep 21, 2008
161
In an alternating current (AC) circuit the two poles alternate between negative and positive and the direction of the current (electron flow) reverses periodically.

Question: If the 2 poles reverse periodically, wont they cancel each other out.
If the electrons go in a certain direction and then suddenly go into another direction, wont the effect be net?
 

Ian0

Joined Aug 7, 2020
6,272
You are correct. If the current alternated, but the voltage stayed the same, then it would be like charging then discharging a battery, and the net result would be no change in the state of the charge of the battery.

However, when we say “alternating current” we generally mean that the voltage alternates as well.
Power is voltage x current, so if both voltage and current are positive, then power is positive; and if both voltage and current are negative, then power is also positive. Work done is power x time, so the work done with alternating voltage/current is not zero.
 

dcbingaman

Joined Jun 30, 2021
720
What you are saying is also true if the circuit is 100% capacitive or 100% inductive. This causes the current to be out of phase with the voltage by 90 degrees. In this situation power is both positive and negative thus cancelling itself out. Most circuits are always a mixture of reactive and resistive and some power is consumed because the phase angle between the voltage and the current is somewhere between 0 and 90 degrees. For a reactive only circuit the AC waveform does not have to be a sine wave, it can be any periodic wave, as long as the current is going in opposite directions for each phase, no power will be consumed by a reactive circuit. For example, a rectangular wave with equal and opposite currents will create a triangular voltage because V(t)=Integral(current)/C for a capacitive only circuit.
 
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crutschow

Joined Mar 14, 2008
30,754
a rectangular wave with equal and opposite currents will create a triangular voltage because V(t)=Integral(current)/C for a capacitive only circuit.
It will be a triangular wave only if there is some other impedance in the circuit.
For a capacitor only, the rectangular wave will generate a high pulse of current that will in fully charge/discharge the capacitor during the rise and fall time of the wave.
Simulation example below:

1655951572238.png
 
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dcbingaman

Joined Jun 30, 2021
720
It will be a triangular wave only if there is some other impedance in the circuit.
For a capacitor only, the rectangular wave will generate a high pulse of current that will in fully charge/discharge the capacitor during the rise and fall time of the wave.
Simulation example below:

View attachment 270016
What was 'rectangular' was the current waveform, not the voltage waveform. I should have clarified that better. Keep the waveform the same and change it to a current source instead of a voltage source.

From my quote:
"rectangular wave with equal and opposite currents"
A rectangular wave of equal an opposite current implies a current waveform is applied to the capacitance. For example: apply 1uA for 5 microseconds then apply -1uA for 5 microseconds and repeat indefinitely. The result will be a triangular voltage waveform.
 
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boostbuck

Joined Oct 5, 2017
239
If the electrons go in a certain direction and then suddenly go into another direction, wont the effect be net?
The net current and voltage over time is zero, but since power = voltage x current it is always positive so power has a net positive value.
 

dcbingaman

Joined Jun 30, 2021
720
Okay, I misread your post.
I thought we were talking about AC voltages and the currents they generate.
No worries. I could have worded it better to avoid confusion that's my fault. Most engineers when mentioning a 'waveform' are almost always referring to a 'voltage' waveform.
 

dcbingaman

Joined Jun 30, 2021
720
Errr electrons carry electrical energy?
Isnt that so?
They carry voltage and not current right?
Pls help to unconfuse me.
Not sure if this will help but keep in mind voltage is just joules of work per coulomb of charge. If it takes x joules of energy (work) to transfer 1 coulomb of charge from point A to point B, it is said that x volts of potential difference then exists between points A and B. The same is true going the other way, if x joules of energy (work) are 'used' that is performed on the system when 1 C of charge moves between A and B, then x volts of potential difference is between points A and B. The reason energy is being used: if a force is applied to an object over a distance, that is work (force times distance). If you take an electron and move it through an electric field going against that field requires a force over a distance which is what 'energy' is. Physics typically uses 'electron volts'. The is the energy (work) required to move 1 electron through a potential difference of 1 Volt. Being that is 1V= 1 Joule/C and C= 6.241 x 10^18 electrons that is the same as 1eV=1.6e-19 joules.
 
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