Alternating current added a new complication to Ohms law. If we are working with only resistance, there is no problem as long as we use RMS values for volts and current. The math works great including the calculation for power. However, as soon as we added inductance or capacitance things started getting complicated because we have the additional problems of phase shift. Tonight we are going to take a small step in dealing with the problem. Most technician books do not go past where we will end up tonight.

Imagine we have a series RL circuit and after we calculate the inductive reactance for the inductor at the frequency of our input we have an X_{L} of 50 Ω. We also have a 100Ω resistor in series with it. Because the inductor wants to shift the phase by 90 degrees when we calculate the total impedance we have to imagine a right triangle as shown in the picture. The Pythagorean theorem allows us to calculate the impedance, Z. Z = √(R^{2} + X_{L}^{2}) which in this case comes out to be 111.8Ω.

Because the voltage and current are out of phase, we cannot calculate the power by simply multiplying the Volts X Amps. That will give us a number, but not the power. The easiest way to calculate the power is to first find the current flow. Lets say the voltage supplied to the whole circuit is 100 Volts rms. Then the current flowing in the circuit is 100 V / 111.8 = 0.894 A.

When we talked about power in DC circuits we presented 3 formulas for finding power P= V x A in DC. P = V^{2}/R, and P=I^{2} x R. The last two formulas work for AC or DC because no phase shift is involved since we are squaring values. Since the circuit we are dealing with is a series circuit, the current is the same throughout the loop so the last equation is the one most useful to us. However we will have to keep in mind that the phase shifting occurs.

To make life a little easier for both of us I will square the current one time and we will use it for all three sides of a new triangle. I^{2} = 0.894 x 0.894 = .799.

To calculate the power consumed: P=I^{2} x R = .799 x 100 = 79.9 Watts.

Now let’s take I^{2} x X_{L} = .799 x 50 = 39.96. However, we cannot call this Watts. The term used for it is VAR for Volt Amps Reactive. All of this energy is stored during one part of the AC cycle and then dumped back into the resistor during another part of the cycle so it is not power consumed.

Now let’s take I^{2} x Z = .799 x 111.8 = 89.4 VA. This is not power, and it is not just VAR it is a combination of both of those. We could have got the same value by simply multiplying V x I = 100 x 0.894. This also forms a triangle.

The power triangle is drawn with the real power on the horizontal line and the VAR drawn on a vertical line. The VA is the hypotenuse of the right triangle.

The Pythagorean theorem still works and VA^{2} = VAR^{2} + P^{2}.

In the previous post about transformers, I said “old smokey” had a rating of 70 VA. VA is the rating used by all power supplying devices that I can think of. Why is this? If you think of the wires and the heat being produced in them or in other words the power being lost. It is related to the I^{2} x R value. It does not matter when the phase of the current happens because the power lost is related to the the current squared. Similarly, when a transformer is designed the voltage is determined by the windings on the primary and the secondary windings. However the current is reactive to the load. This current is limited both by the amount of heat the wire can dissipate, but also the amount of magnetic flux the iron core as been designed to accommodate. So a transformer is rated in VA (usually KVA for one of any real size).

There is one more important ratio to discuss. The ratio of P to VA. P/VA is called the Power Factor. If you look at the power triangle P/VA is also the cosine between those two sides. This is the amount of phase shift of the current compared to the voltage. This is a very important number to power companies when they sell power to industrial users. Most residential loads are considered to be mostly resistive so residential customers buy power based upon the watts used throughout the month. However, industrial users have a lot of motors operating and they get charged a penalty the they have a power factor that is very much less than unity.

So far all I have talked about in this is inductive loads. What happens if it is capacitive reactance? The equations are exactly the same. The only difference is I would draw the power triangle with the vertical line pointing upward. This would make the VA line leading the power line. Remember ELI the ICE man. The current and therefore the VA would lead the voltage in a capacitive circuit.

Also the next question you probably have is what happens if both a capacitor and an inductor is present? if it is a simple RLC series circuit, then X_{L} and X_{C} are subtracted from each other and the largest determines if the circuit is leading or a lagging circuit.

Probably the next question is: What happens if it is a parallel RLC circuit? At which time I would say, don’t you get tired of asking questions? Seriously, the equation for X and R in parallel is Z = (R x X)/√(R^{2} + X^{2}). In this case also X_{L} and X_{C} are subtracted from each other to determine the X type.

At this time we have completed all of the normal way it is taught to most technicians. However, some of you may be wanting a little more high-powered way of calculating so you can have L, R, and C in various series parallel combinations. There is a way to do that, it will require probably about 3 posts to get there and two of those will be pure math. I am still thinking about it. Please e-mail me or comment on this post your opinion.

Gary.

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This was rather simply put for ease of understanding (Y)Quite brief but informative.