a-determine-the-power-factor-for-an-rc-cir-cuit-with-r40-mathrmk-omega-and-c035-mu-mathrmf-that-is-c

Question

(a) Determine the power factor for an RC cir. cuit with $R=4.0 \mathrm{k} \Omega$ and $C=0.35 \mu \mathrm{F}$ that is connected to an ac
generator with an rms voltage of 24 $\mathrm{V}$ and a frequency of 150 $\mathrm{Hz}$ .
(b) Will the power factor for this circuit increase, decrease, or stay
the same if the frequency of the generator is increased? Explain.

João Gabriel Alencar Caribé · Accepted Answer

the power factor is just the co sign off the face and the co sign of the phase is egos true. The resistance divided by the MP does, and these equals to this square it off r squared plus to pie F. C under one squared is all under the resistance again, so lugging their values that were given leave the problem. We get that the cool sign off. Five Zico&#x27;s troop This square root four time stamped toward there is one key. Loom there squared plus two pi times the frequency 115 times the capacitance 0.25 meat group under one squared square root. This is all under the resistance four times stand to deterred. Finally, disease approximately 0.797 Now we noted that the power factor is proportional the resistance inversely proportional to the impudence on DDE, The impedance is equal to the square root off r squared plus the reactant squared under the resistance again now knows that the reactant is he goes to two pi frequency capacitance under one. So you for you increase the freakin separation. Narrator So by increasing f, we will be decreasing the reactor&#x27;s so we will be decreasing these factor here and by decreasing that this factor here we will be decreasing. They didn&#x27;t need her as a wolf on decreasing the denominator means that the complete fraction will increase, so increasing the frequency implies that the power factor also increases.

João Gabriel Alencar Caribé · Answer

for the first item. Remember that the co sign off the phase Is it close to the resistance divided biting Penis, then the impudence. Easy course, too, the resistance divided by the co sign off the face. The resistance is he goes to 25 homes. The co sign off the face is a course to describe it off. Three Divided by two. This is the co sign off 20 degrees, and this is approximately 29 homes for the second item. Remember that impedance is the question that square it off the resistance squared, plus the difference between the capacity the reactant, says squared. Then we can solve this equation before the capacity reactors. To get the following we first square both sides to get to C squared easy course towards Squared plus Excel, my new sexy squared. Then we know this to get Z squared, minds are squared. He&#x27;s equals two Excel miners XY squared, then the square it off both sides to get pleasure minus the square it off. The Z squared minus R squared is he goes to excel minus X e finally exceeds the capacity reactors is equals. Two Excel plus or minus squared off Z Squared Minds Arctic Squared Dan, we can calculate using the data given in the problem. Exhale is he goes to two pi times the frequency times 160 times 10 to the minus three, which is the doctors, plus or minus the square it off Z squared. So to keep the press even, let&#x27;s do like this 25 divided the way the square root off. Three divided way too squared, which is named P squared minus 25. Squared them. We get the reactive capacitance off approximately 69 point 7 to 6 homes. War 40 point 858 hose. Then we have to choose one off those values. How can he do that? Well, not that the circuit is such that the voltage is leading the current, therefore disease an inductive circuit. Then we must choose. There&#x27;s more capacity reactors. So we go with these value for the capacity reactors. Then we can finally complete One is the capacitance off that circuit. So remember that the capacity reactors is it goes to one divided by two pi times the frequency times the capacitance, then the capacitance is equals to one divided by two pi times the frequency times, the capacity reactors, the reform We get the capacitance off one divided by two pi times 55 times 40.858 And these is approximately 71 me grew for its now for the last item We have to analyze what happens to the impedance when we decrease the capacitance. True knows that, Z because he caused to the square root r squared plus Excel My news XY squared No, that has decreased the capacitance. What happens when we decrease the capacitance? XY is he goes to one If I don&#x27;t bite to pi times the frequency times the capacitance By decreasing the capacitance, we will increase the capacity reacting lt&#x27;s under four. By increasing the capacity reactions thes difference as a wound, we&#x27;ll decrease. Then the impedance will also decrease if we decrease the capacitance.

Ben Nicholson · Answer

31.44. We imagine we have a serious R L C circuit. Yeah, produced by a sea generator with an amplitude of 220 volts, a frequency of 404 100 hertz and then given values of the components we want to find with the passage of react. Ince&#x27;s what the impedance is overall, what the amplitude of the current is. And then the next three parts, as we imagine putting an extra capacitor in Siri&#x27;s with the same value. And then whether these increase, decrease or stay the same so far today, we know that the reactions of a capacitor is one over. Omega finds its value that were given the regular frequency. So accidents. So putting our numbers in, we get 16.6 homes for the impedance. It&#x27;s going to be R squared, plus the inductive minus the capacitive reactor. Ince&#x27;s display aired. We already just found with the capacitive reactor Ince&#x27;s They don&#x27;t need to find that again. R squared plus too high l minus x c cleared baggers route sighing all the way across. So the total impedance is 422 homes. So now the amplitude of the current is equal to the amplitude of the E m f, divided by magnitude of the impedance 0.5 to 1. So now we add an extra acid, Er, um, in Syria&#x27;s with all this. So we have something like to capacitors both of value See a resistor and our conductor Either end of this is connected to the signal generator. So this is proportional to the equivalent capacitance under one. So putting capacitors in Siri&#x27;s decreases the capacitance so see equivalent is decreasing when we add this, uh, so it&#x27;s reactant will increase the recall from the previous screen. The impedance is R squared, minus the difference between the inductive reactant CE and the capacity of reactions spared. So we saw that the all right the acid of reactions increased, which means that this difference here will decrease. And indeed, we can actually just evaluate this and find that it&#x27;s 408 homes which is less than the 422 was originally. So the impedance has decreased and then for the current. So this is inversely proportional to the impedance. So if this decreases, the current will increase

João Gabriel Alencar Caribé · Answer

The power factor is just the co sign off the seas and it is given by the resistance divided by limpy does and these easy coz to square it off the resistance squared plus the reactor squared under the resistance. And this is easy coast through the square it off R squared plus to pie F C under one squared, divided by the resistance No blagging The data given by the problem we got this square root off. 105 squared, plus to pi times 60 times 82.4 me crew This is all under one squared. Um, this is all under the resistance again. 105 and this is approximately 0.96

João Gabriel Alencar Caribé · Answer

by connecting a second capacitor in Siri&#x27;s with the 1st 1 we will end up with a result and capacitance that is smaller done. They need show capacitors and also notice that the IRA mass current is equals two to pi times the frequents times, the capacitance times they are a match voltage. Notice that by increasing the capacitance we will increase the armas voltage and by decreasing the capacitance, we will decrease. They are a master politician. So by including a new capacitor, we are decreasing the capacitance. So we are decreasing the R. M s correct that will flow through that circuit. In the second item, we are connecting the two Capasa pers in Siri&#x27;s these one and he&#x27;s won the result of capacity off The association is given by C. R. Zico&#x27;s too see time see two divided by C plus C two and these years equals two 55.5 times 38.7, divided by 55.5. Close 38.7 and it is approximately 22 0.8 me crew forwards of capacitance. And now we can use the following relation between the ROMs, current and the capacitance before and after, including the new capacitor, The relation is the following I, Ari Mass divided by seat disease before including the new capacitor easy close to I ROMs prime divided by CR disease after including the new capacitor. So by using this relation, we get that the Larry Mize current after including the new capacitor easy goes to CR divided by C times I. R s and these is equals two 22 0.8 divided by 55 0.5 times 1.21 and it is approximately 0.6 to 2. Amperes noticed that these relation holds because we are on Lee, including a new capacity. So we are only changing the capacitance off the circuit. We are not changing the voltage, nor the frequency to change the voltage or the frequency. We have to change the soldiers and we are not doing that. So these is why we can use these relations

Donald Albin · Answer

all right. It&#x27;s showing something. There is describing Teoh something to us. A resistor in doctor and capacitor in series with a generator. Okay, so we&#x27;ve got the generator. I&#x27;m gonna draw this to the voltage source, and then we just have a resistor, a capacitor, and I don&#x27;t know, I&#x27;m just gonna turn it here. An induct er in Siris. The resist er is 400 domes. The reactant of the induct er is 700 OEMs and the circuit impedance is 760. Oh, what can we determine about the react? It&#x27;s of the capacitor. Well, let&#x27;s see. Z is the square root. Uh oh. R squared plus x l minus X c squared. So we know this. We know this and we know this so we can tell the difference between the reactant of the capacitor and the reactant of the in. Dr. Is that what we can tell? But we can&#x27;t tell if the reactant is the reactant of the capacitor is greater than that of the induct er or if it is less than that of the deduct er in doctor. So, yes, we can tell the difference between the reactant practices of the in doctor and capacitor. And I guess the way that I want to show this is that we could square both sides. Okay, We could then subtract r squared on both sides and then we could take the square root of both sides. When we take the square root of both sides, we need to include ah, plus or minus so x l minus. X c is plus or minus the square root of Z squared minus r squared. So this indicates exactly that XL minus X C is a value, but we don&#x27;t know whether that difference is either positive or negative. So we can tell that, um so I guess I could, right? We don&#x27;t know whether the difference is positive or negative, and therefore we don&#x27;t know the exact value of the reactions of the capacitor. All right, so in part B, we find that the power decreases as we raise the frequency. Interesting. Well, power is 1/2 and what do we know here? Do we know V or anything? We don&#x27;t know a whole lot. But ah v squared over easy times. The co sign of Phi, where Fi is the inverse tangent of X l minus X c over our okay. We raised the frequency. So I&#x27;m just going to draw a diagram of what we&#x27;ve got here. The tangent of Fada, His ex l minus x c over. Our tangent is opposite over adjacent. Then this is gonna be easy over here. So the co sign is our over Z. Who sign is adjacent over iPod Innis. Tangent is cheese. So that&#x27;s gonna be 1/2 b squared over Z. Just make sure I&#x27;m doing this right. The tangent is opposite over. Adjacent engine is promising. And the square root of R squared plus ex my ex l minus xy squared is going to be easy. Just using the Pythagorean theory, making sure that you follow this. So what I&#x27;m showing is basically just Pythagorean theorem here. Um inverse tangent is excel of minus x e over r opposite, over adjacent, and then co sign. Wait a minute. Yeah. So I got V squared over Z, and then I&#x27;ve got co sign, which is the adjacent over the high part knees. So now that becomes 1/2 v squared are over Z square. Interesting. And Z is this So let me just write down. What Z is a little more clearly and the power decreases as we raise the frequency. The how are decreases will the voltage The input voltage is independent of the frequency and the resistance is independent of the frequency. So if the power increases, wait a minute. No. The power decreases as the frequency. So power decreases as the frequency increases. So you increase the frequency that decreases the pal and a decrease in power would come with an increase in impedance. So and by the way, as frequency increases than omega also increases. So we&#x27;re saying that we&#x27;re making omega bigger and that&#x27;s causing the impedance to get bigger. Well, let&#x27;s think about this. If x l is greater, then act see that we&#x27;re increasing omega. That would make X l even Mawr excels right here. If omega gets bigger, that would make X l even higher. And so that would cause the impedance to go up. Alright, What about if x L is less than xsi then what if we make omega go up an X C is greater, then it&#x27;s going to make X c smaller. So I&#x27;m thinking that Z would go down, but we&#x27;re saying that Z goes up And so therefore, we now know that X L is greater than xsi. And if Excel is greater than xsi, that means that XL minus X c is a positive number. So x l minus X c is going to be the square root, the positive square root of the squared minus R squared. Okay, let&#x27;s put that in a calculator. Um, we know what X l is, which is 700 so x. See if it&#x27;s if this is the case, then X C is going to be x l minus the square root of Z squared minus r squared. So we put that in a calculator. X Ellis 700 on its a square root Z squared minus R squared. The is 7 60 minus r is 400. That gives me 53 point eight, which is listed is the correct answer. So X c is 53.8 owes. So first we figured out that the react insp the inductive reactions is is greater than the capacitive reactant. Then we know that this has to be positive. All right, let&#x27;s move on to see this is kind of challenging report repeat part a Oh, my goodness. Okay, there&#x27;s a But this time I want to use 200 OEMs instead of 400. Does that has no effect on the answer? Because the question for a was what can you determine about the reactant of the capacitor, but 200 homes again, we can We can tell the difference between the reactions of the induct er and the capacitor, but we can&#x27;t tell whether the difference is positive or negative. I was thinking about whether there would be a different answer, but no, um, I when I was doing part, I didn&#x27;t even really worry about the fact that our was 400. It could have just very well been 200. Um, so this doesn&#x27;t affect the answer to part A. Let&#x27;s just think about this a little bit more. I want to write this down this XL minus X c. Make sure I&#x27;m not making a mistake. We know Z and R 7 60 and 400 and we also know Excel Z and R Z nrz nrz in our 7 60 and 200. Maybe I&#x27;m missing something and we know that Excel 700. So let&#x27;s just put this into the calculator. 700 minus 7 60 squared, minus 200 squared. I&#x27;m thinking maybe I did, miss. If we use the minus here, that would give us a negative XY. The answers are negative. 33.2 or use the positive 1430. We can&#x27;t have negative. So the answer would have to be this and so there.

Problem

The "square-wave" voltage shown in FIGURE $24-31$…

Problem 27 Easy Difficulty

Answer

a) $\cos \phi=0.797$
b) When frequency increases the capacitive reactance decre ases
$\left(\because X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\right)$
As capacitive reactance decreases the impedance of the circuit decreases
$\left(Z=\sqrt{R^{2}+X_{c}^{2}}\right)$

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James S. Walker

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Elyse G.

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a) $\cos \phi=0.797$b) When frequency increases the capacitive reactance decre ases$\left(\because X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\right)$As capacitive reactance decreases the impedance of the circuit decreases$\left(Z=\sqrt{R^{2}+X_{c}^{2}}\right)$

Physics

Elyse G.

Andy C.

Liev B.

Meghan M.

Electric Charge

The Electric Force

James S. WalkerPhysics

Elyse G.

Andy C.

Liev B.

Meghan M.

a) $\cos \phi=0.797$
b) When frequency increases the capacitive reactance decre ases
$\left(\because X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\right)$
As capacitive reactance decreases the impedance of the circuit decreases
$\left(Z=\sqrt{R^{2}+X_{c}^{2}}\right)$

James S. Walker

Physics