Show that the power delivered by a three-phase ac source equal...

Question

Show that the power delivered by a three-phase ac source equals a constant P = 3Vo²/2R, by combining the four equations in Section 30–11.

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with ac circuits. This problem says in a balanced three phase ac circuit, the phase voltages are given by VA is equal to V not multiplied by the sine of Omega TV. B is equal to V knot multiplied by the sine of Omega T minus two pi divided by three and VC is equal to V knot multiplied by the sine of Omega T plus two pi divided by three. Given that the resistive load in each phase is R an expression for the total power delivered by the three phase AC source. We're given four possible choices as our answers or choice A we have P is equal to three V knot squared divided by the quantity of two R. And Choice BP is equal to three VNO squared divided by the quantity of five R. For choice C we have P is equal to three are divided by the quantity of five V knots squared. And for choice DP is equal to three are divided by the quantity of two V knots squared. Now we're gonna need to um calculate the total power for this AC source and to do that, since we were given the voltage, we need to write our power in terms of our voltage and our resistance. So we call for power P is equal to V squared divided by R repeat. Here is gonna be the power V is the voltage and R is the resistance. And so in order to get our total power, we'll call that P, that's just gonna be the power from each of the phases. So we'll have va squared divided by R plus VB squared divided by R plus the C squared divided by R. And so we can plug in our formulas for each of the voltages. For each phase. We'll have P is equal to the quantity of V not multiplied by the sign of omega T. And that quantity is squared and that's gonna be divided by R and we'll have plus or VB will have the quantity of be not multiplied by the sine of the quantity of Omega T minus two pi divided by three. And that whole quantity is squared divided by R. It'll have plus the quantity of P not multiplied by the sine of omega T plus two pi divided by three, that quantity is squared and divided by R. So in each term, I'll have a V not squared divided by R. So I can factor it out. And so I have P equal to are V not squared divided by R multiplying the quantity of P squared of omega T plus the sign squared of omega T minus two pi divided by three. Then we'll have plus sign squared of the quantity of omega T plus two pi divided by three. So then we want to do is combine all those trick functions into possibly one trick function. So we have to do that through some trig identities. The first one we're gonna need to use is to write the sign squared in terms of cosine, recall from your trig identities, that sign squared of X is equal to one minus cosign of two X. That whole quantity is divided by two. So we're gonna use this to replace the sign squares with just cosine. That means that our power P gonna be equal to VNO squared divided by R multiplying the quantity of. And here I'm gonna go ahead and factor out the one half out of everything. So that VNO um squared is gonna be actually divided by two R. And what will be left with is one minus cosine of two omega T don't have plus one minus cosine of two omega T minus four pi divide by three. Then we have plus one minus the cosine of two Omega T plus four pie divided by three. So I can combine some more terms in this case. So I have P is equal to V not squared divided by two R. And that's gonna be multiplying the quantity I have three ones here. So I have three, I have minus cosine of two omega T. And then if I look at the next two terms, they're of the form of cosign of X minus Y plus the cosine of X plus Y. And that's gonna be equal to two multiplied by the cosine of X multiplied by the cosine of Y. That's another trigger, identity that we're going to need to use. And so the next term in the sum is actually going to be minus two, multiplied by the cosine of two omega T multiplied by the cosine of war pi divided by three. So we can actually factor out a cosine of two omega T out of the last two terms in the sum. And so we'll have P is equal to VK not squared divided by two R multiplying the quantity of three will have minus cosine of two omega T. And that's gonna be multiplying the quantity of one plus two cosign of or pie divided by three. However, cosine of four pi divided by three is just negative one half. And so that means that term in the parentheses will actually be equal to zero. So that removes the time dependence out of our equation. And so what I'm left with is P is equal to three V not squared divided by two R. So that's gonna be my final answer here. And if I look at my answer choices, that actually corresponds to answer choice A, so just a quick little recap here of what we've done. We started off by using our definition of power in terms of the voltage and the resistance. And then we added the powers together from each of the three phases. We then had to simplify the expression using some trig identities. But ultimately, we arrived at the solution that we were looking for. So I hope that this has been useful and I'll see you in the next video.

Show that the power delivered by a three-phase ac source equals a constant P = 3Vo²/2R, by combining the four equations in Section 30–11.

Key Concepts

Three-Phase AC Power

Voltage and Resistance Relationship

Power Calculation in AC Circuits

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