# Why does current lag voltage by 90 degrees?

Discussion in 'General Electronics Chat' started by foolios, Jan 23, 2010.

1. ### foolios Thread Starter Senior Member

Feb 4, 2009
163
1
Edit:
http://www.pcguide.com/ref/power/ext/basicsPower-c.html

But where it reads:
where "cosine" is the trigonometric function. "cosine(phase)" is also called the power factor of the load. Let's try an example. Let's suppose we are trying to run a power supply and the power supplied is 115V voltage and 2A of current. The apparent power is 115 * 2 = 230 VA. If the nature of the power supply is that its voltage and current are out of phase by 50 degrees, then the power factor is cosine(50º) = 0.642 (sometimes expressed as 64.2%) and the power used by the load is 148 W.

Wasn't the apparent power 230W?
So the true power used is only 148W.
Why is this a problem for power utility companies? I thought that poor power factor meant that power was being wasted, but instead these calculations seem to show that less power is actually needed.
This is very confusing.

Last edited: Feb 14, 2010
2. ### foolios Thread Starter Senior Member

Feb 4, 2009
163
1
I think I have found enough information to answer my questions.

But I do have one more new one.

If a circuit is purely inductive and it results in no power because too much power is coming back and negating it.
What about a purely capacitive circuit? Is the power also cancelled out completely in this kind of circuit too?
How would that be?

3. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
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For power usage, the problem with poor power factor is that when the peak current is drawn out of phase with the peak voltage, higher current is needed to achieve the same amount of work (watts or horsepower).

This means machines need thicker cables, or conversely the existing distribution cables cannot supply as much useful power if the power factor is poor.

The electricity companies typically surcharge industrial users based on how bad the power factor is.

A large factory with mostly inductive loads (ie. motors and transformers rather than just electric heating) can reduce their electricity costs by sometimes 30% by adding their own power factor correction equipment, so the power factor at the meters is then good.

A capacitor on an AC supply will draw current (depending on it's value).

The peak current is when the voltage is changing fastest, around the zero crossing of the sine wave. Half each current peak is while the voltage is positive, half negative (& vice versa).

This is drawing power, and an electricity meter that is 'power factor aware' will read this. The old aluminium disc type meters do not read, as the 90' out-of-phase current and voltage cancel out in this mechanical system.

4. ### foolios Thread Starter Senior Member

Feb 4, 2009
163
1
I don't see why the power is dipping just that amount below zero in this image. It appears that voltage and current are both still moving along that length of the graph in the same direction yet the curve in power begins to change direction. The current and voltage are both still dropping all along that point where power decides to change direction in it's wave.

::image removed due to host server image bandwidth constraints

Last edited: Jan 23, 2010
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,667
1,311
When capacitor is full charged then voltage on capacitor is Vin but current is 0A.
And when capacitor is discharge and then start to charge-upthen voltage on capacitor is 0V but current is at max value. And that is why we say that capacitor voltage lags capacitor current by 90. We have similar situations with coil.
When we connect the coil to a voltage source then at the beginning there is non current flow through inductance but voltage on the coil is equal Vin.
In steady state there is no induce voltage on the coil but flow max current through coil. So in inductor cause current to lag voltage.