5

Write an expression for the instantaneous voltage V(t) delivered by an ac generator supplying 120 V rms at 60 Hz. Assume the voltage is 0 at time = 0,and leave the ...

Question

Write an expression for the instantaneous voltage V(t) delivered by an ac generator supplying 120 V rms at 60 Hz. Assume the voltage is 0 at time = 0,and leave the factors in exact form; do not round their values. Write the argument of the sinusoidal function to have units of radians, but omit the unitsV(t)

Write an expression for the instantaneous voltage V(t) delivered by an ac generator supplying 120 V rms at 60 Hz. Assume the voltage is 0 at time = 0,and leave the factors in exact form; do not round their values. Write the argument of the sinusoidal function to have units of radians, but omit the units V(t)



Answers

In the United States, a standard electrical outlet supplies sinusoidal electrical current with a maximum voltage of $V=120 \sqrt{2}$ volts $(\mathrm{V})$ at a frequency of 60 cycles per second. Write an equation that expresses $V$ as a function of the time $t,$ assuming that $V=0$ if $t=0$

Hi In the given problem, RMS voltage is given as we are. MS is equal 220 fold. So peak voltage will be given by route to times of Farmers voltage means this is 120 into route to world And frequency is given as 60 heart. So using the equation, the standard equation of alternating voltage, which varies as a function of time and given as VT is equal to we not into sign to buy ft. So here it will be VT for we know this is 120 route to wall into sign to buy into frequency, which is 60 B. So finally it to me were given ass we D. is equal 220 route to hold into sign. 120 by P. Which is the answer for this given problem here. Thank you.

In this question were given some information about a function for voltage. We have V of tea equals 120. Sign of 120. Hi, three. So we're finding out the amplitude. Well, we see the amplitude is going to be 1 20 So this is question A So amplitude is 1 20 the period we're looking at this horizontal compression. So we will divide to pie by 120 pie, which gives us 1 60 year. So a period is going to be 1/60 and now we're gonna graph two periods of this function. So we're starting at zero, and we're going down to negative 1 20 and up to positive 1 20 And then this would be 1/60. And here we have one 30th and then we're gonna graph this like, usual sine function. So we go up and down and then come up through 1/60 and then repeat. So we go up down on up through 1/30 and then this would continue. But here are two periods of this graph V. Now the next thing we're looking at is homes Law, where we have V equals I. You are where I is, Theo. The current and are is resistance. And in this example we have here we have V and are is 20. So in order to find the function for I were going to be dividing be by 20 so we can scroll down. So the divided by 20 is I of tea, which equals six. Because we have won 20 redivide by 20 we are left with six sign of 1 20 Hi t And now we're going to draw this function. So we have the amplitude, which is six. So the amplitude of six and a period of 1/60 because we have the same horizontal compression. So the period is one 60th Now when we here, I'll screw up a little bit. Now when we graph this, it's gonna look pretty similar to what we have above adjust with. Our new guidelines are new amplitude. We're just six instead of 1 20 And then we have 1/60 and one 30th they were gonna be drawing. That's pretty much the same just with these as our, uh, guidelines. So we come up and down and then through zero at 1/60. So that's the first period. And then we repeat up 3 1/30 on this would continue. But here we have two periods of our function. I of tea, which is six. Sign 1 20 high tea. So we've graft both I of t and V f t. So there are your answers.

So we first have the given output voltage. We can say beasts of Al is gonna be equaling 1.20 times 10 to the second bolts. Or better yet, we can simply say 120 volts multiplied by sine of 30 pie t where the induct Ince's equaling half of 1.500 Henry's And so we can say for party by inspection, Omega is equaling 30 tie radiance per second. And so we find that then the frequency F is equaling omega divided by two pi and 30 pie. Divide by two pi 15 0.0 hurts. This would be your answer for part A for part B. Then we find that the maximum voltage is equaling 120 volts. So we can say that the R M s voltage felt visa. L are a mess is equaling 120 volts that maximum voltage divided by radical too. And so this is equaling 84.9 volts. This would be our answer for part B for part C. Then we want to find the inductive reactant ce This would be equaling to pi multiplied by the frequency times Al the induct INTs. This would be simply omega times l. So this would be equaling 30 pie radiance per second multiplied by point 500 Henry's And this is giving us 47.1 comes. So this would be our answer for part C for part D. Then we can say that the Army's current will be equaling the arm s voltage divided by the inductive reactant CE. And so this would be equaling 84.9 volts. Divided by 47.1 comes. This is equaling 1.80 and peers or amps, and we find that then the maximum current would simply be equaling. The arm s current won't supplied by radical too. So this would be 1.80 amps multiplied by radical, too giving us 2.55 m's. This would be our answer for Part E four part F. We know that here the face difference between the voltage across the induct er and the current through the induct er would be because they fisa bell equaling 90 degrees. So here the average power delivered to the induct er would be equaling the army's current multiplied by the R M s voltage multiplied by co sign of FISA bell. And as you can see, this is gonna be zero. So this would be equaling zero watts. This would be our answer for part F four parts G. Now, excuse me when I sign your soul, you Saudi sold all my apologies. Voltage with a peak value is applied two in Dr the current through the induct er also varies sign of soldierly with with respect to time And so with the same frequency as the applied voltage, we know that the again the maximum current is gonna be equaling the maximum voltage divided by the inductive reactant ce. However, we can say current lags behind the voltage in phase by 1/4 cycle or pie over to radiance. And so we can say that if the voltage is given as dealt to the max, the maximum current multiplied by sine of omega t, the current as a function of time would be I equaling the maximum current times sign of omega T minus pi over too. And so, in the case of the conductor that they're giving us here, we could say that the current I would be equaling 2.55 amps multiplied by sine of 30. Hi t minus. Pi over too. This would be our final answer for part G for part h. We can say when eyes equaling positive 1.0 and peers. We can say that. Then we have that sign of 30 pie t minus pi over too is gonna be equaling 1.0 and peers divided by 2.55 amps. And so we can then say that 30 pie t equaling Rather 30 pie t minus pi over to is gonna be equaling arc sine of 1.0 divided by 2.55 And this is giving us Excuse me. 0.403 radiance. And now T is gonna be equaling pie over to radiance plus 0.403 radiance. And this would be divided by 30 pie radiance per second. And we find that this is equaling 2.9 times 10 to the negative second seconds. This would be our final answer for part H. That is the end of the solution. Thank you for watching

And a C. Source that has RMS value. We are a mess equals to 1 20 volts and it will taste goes from zero to its maximum value in time. T equals to 4.20 milliseconds. Okay, so now first of all calculating peak value from here so peak value we're not equals to two times of RMS value. So we get underway to MacLeod by 1 20. So from here we get 1 70 Wolf. No, This time from 0 to peak maximum value or peak value is always equals to T by 4 30 by four is equal to 4.20 milliseconds. Where capital is time? Period of time, Is he? So so from here we get time period equals two 16.8 million seconds. Okay now the angle of frequency omega can be calculated as two by by capital T. So substituting the day of T. Here. So we get 16.8 manipulated by million second, which is 10 to the power minus three seconds. Okay, So from here, after solving We get angular frequency omega equals to 374 radiant per second. Okay, so this is the angular frequency of the source. Okay. No, the expression for the world is will be equals to be not sign omega. Okay. So now substituting all the values so we get equals to be not, which is equals to 1 70 world. So 1 17 year old Sine Omega which is 374 and time T. So this is the expression for the voltage at any time. Okay, so this is the expression for the required questions. Okay?


Similar Solved Questions

5 answers
Q6. A particle of mass m describes & complete vertical circle on the end of a light inextensible string of length T Given that the speed of the particle at the lowest point is twice the speed at the highest point, show that:rg a) the speed of the particle at lowest point is 2_ 3 (b) the tension in the string when the particle is at the highest point is mg 3
Q6. A particle of mass m describes & complete vertical circle on the end of a light inextensible string of length T Given that the speed of the particle at the lowest point is twice the speed at the highest point, show that: rg a) the speed of the particle at lowest point is 2_ 3 (b) the tension...
5 answers
Secure https / /session mastenngengineeni S Preblereo Problem 10.70Part AThe and axes puss through the centroid C of the rectangular area . Suppose that & 160 IL (Figure_1)Deterrnin ExpressSubrFigurePart BDeternExpre30 MMPart €Dete
Secure https / /session mastenngengineeni S Preblereo Problem 10.70 Part A The and axes puss through the centroid C of the rectangular area . Suppose that & 160 IL (Figure_1) Deterrnin Express Subr Figure Part B Detern Expre 30 MM Part € Dete...
5 answers
1) Solte each of the following ordinary differential equations O initial value problems (if possible} When possible; find the explicit solution of the equation. Provide all the steps in each problem points each)2Vr4 Vi -4? Uv (u" + I) cos(r) 2yy 9(5) = 2 T2 16 tan(z) y F 4 9(w/2) =*/2 6e2r-" 9(o) = 0 y =1-2+42 Ty?
1) Solte each of the following ordinary differential equations O initial value problems (if possible} When possible; find the explicit solution of the equation. Provide all the steps in each problem points each) 2Vr4 Vi -4? Uv (u" + I) cos(r) 2yy 9(5) = 2 T2 16 tan(z) y F 4 9(w/2) =*/2 6e2r-&qu...
3 answers
Question 182 Point-A process in statistical control with X = 108 , $ = 1.35 and n = 5 . The process specifications are 85 * 13What is the potential process capability Cp ?Add your answer
Question 18 2 Point- A process in statistical control with X = 108 , $ = 1.35 and n = 5 . The process specifications are 85 * 13 What is the potential process capability Cp ? Add your answer...
5 answers
The mean weight of male aerobics instructors in certain city is more than 173 lbs Express the null and alternative hypotheses in symbolic form for this claimHo :Hi : pUse the following codes to enter the following symbols: enter enter enter _=
The mean weight of male aerobics instructors in certain city is more than 173 lbs Express the null and alternative hypotheses in symbolic form for this claim Ho : Hi : p Use the following codes to enter the following symbols: enter enter enter _=...
5 answers
22+3t-4 Find lim +71 02 _ 1+3t - 4 lim t71 22 _ 1 (Type an integer or a simplified fraction )
22+3t-4 Find lim +71 02 _ 1 +3t - 4 lim t71 22 _ 1 (Type an integer or a simplified fraction )...
5 answers
Find f ' (x).flx) = e Nx+12(x+ 12)2 e f"(x) = 2 2(x + 12)
Find f ' (x). flx) = e Nx+12 (x+ 12)2 e f"(x) = 2 2(x + 12)...
4 answers
Ine Economk edley Inititute reparts that the average untr-level age Jemale standard cevialan Vleze gutcuatos 520.55 prr hojr id Ma & Er cyaies 53.94and Frduates Is 53.07, Assumt wares far ferslc coltege graduates (s 517.76 per hour Jnr plodem nomaly Ustrbuted (Scn eacicisc 'On page 427 0fyouf [Ealbook Ir gLetb"Atennethutin
Ine Economk edley Inititute reparts that the average untr-level age Jemale standard cevialan Vleze gutcuatos 520.55 prr hojr id Ma & Er cyaies 53.94and Frduates Is 53.07, Assumt wares far ferslc coltege graduates (s 517.76 per hour Jnr plodem nomaly Ustrbuted (Scn eacicisc 'On page 427 0fyo...
5 answers
[A Cat aEpioD o thc *Y planc thc elecui potcnlal rnlcn dccnclicld cqua [0 Zcio04 Un 2DIEm 0 8 0c2 - mn 2s0.7 9 m 00
[A Cat aEpioD o thc *Y planc thc elecui potcnlal rnlcn dccnclicld cqua [0 Zcio 04 Un 2DIEm 0 8 0c2 - mn 2s0.7 9 m 00...
5 answers
Refer to the following experiment: Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a heart given that the second card is a heart?
Refer to the following experiment: Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a heart given that the second card is a heart?...
1 answers
Show for the natural numbers $\mathbb{N}$ that $$|P(N)|>|\mathbb{N}|$$
Show for the natural numbers $\mathbb{N}$ that $$|P(N)|>|\mathbb{N}|$$...
5 answers
If a spinning body of mass initially has a moment of inertia of457kg*m2 and an angular velocity of 1.24rad/s, then the angularvelocity becomes 3.18rad/s. Then what is the object's new moment ofinertia?r I2 = ____unit ___
If a spinning body of mass initially has a moment of inertia of 457kg*m2 and an angular velocity of 1.24rad/s, then the angular velocity becomes 3.18rad/s. Then what is the object's new moment of inertia? r I2 = ____unit ___...
5 answers
A block of mass m = 0.5 kg is kicked so that it slides up a rampinclined at an angle θ = 30 . At the bottom of the ramp, the blockhas speed v0 = 4.0 m/s. There is a coefficient of kinetic frictionbetween the block and ramp μk = 0.18. (a) How far up the ramp doesthe block slide? (b) Assume that the block comes to rest and doesnot slide back down the ramp. What must be the minimum coefficientof static friction between the block and ramp?
A block of mass m = 0.5 kg is kicked so that it slides up a ramp inclined at an angle θ = 30 . At the bottom of the ramp, the block has speed v0 = 4.0 m/s. There is a coefficient of kinetic friction between the block and ramp μk = 0.18. (a) How far up the ramp does the block slide? (b) Assume ...
5 answers
1) Type I error rate (probability of making type I error) isequal to the significance level alpha. True or False2) We may make a type I error when we conclude to fail toreject H0. True or False 3) We can change the signficance level alpha of a testto favor the conclusion we want to reach. True or False 4) The smaller the significance level we chose to use, the moreprotect H0 (the harder to reject H0). True or Fasle
1) Type I error rate (probability of making type I error) is equal to the significance level alpha. True or False 2) We may make a type I error when we conclude to fail to reject H0. True or False 3) We can change the signficance level alpha of a test to favor the conclusion we want to reach. True ...
5 answers
If $5000 is invested at 6% annual simple interest, how longdoes it take to be worth $7100?
If $5000 is invested at 6% annual simple interest, how long does it take to be worth $7100?...
5 answers
Find the points of intersection of the graphs ofand~28x y = -84.Each point in your answer should be written as an ordered pair (1,y) , and points should be separated by semicolons.y = 722
Find the points of intersection of the graphs of and ~28x y = -84. Each point in your answer should be written as an ordered pair (1,y) , and points should be separated by semicolons. y = 722...
5 answers
Find the given limit:lim (-x+5x-9) X7-4Iim (-x+5x-9) = 77-4(Simplify your answer:)
Find the given limit: lim (-x+5x-9) X7-4 Iim (-x+5x-9) = 77-4 (Simplify your answer:)...

-- 0.020991--