# NCEES Problem 108



## jdd18vm (Jan 20, 2007)

Sorry if this seems fundamental, but if someone could give me their thoughts.

A Sinusoidal voltage of vt=[50 cos(ωt + 45) + 28.28 cos(ωt)] V is applied across a 5Ω resistor. the Average power (W) deliverd to the Resistor is most nearly?

I see, and can follow the solution, but dont understand the reasoning? Specifically what to do with the two components of the voltage.

John


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## Art (Jan 20, 2007)

convert to phasors

one will have an angle of 45, the other zero

convert to polar and add...

then back to phasor

divide the |V| by sqrt of 2 (avg)

then square it and divide by 5...S = V^2/R

back to polar for the real power

or cos (arctan Q/P) x phasor magnitude...pf x |S|


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## Dark Knight (Jan 20, 2007)

I am trying to remember the specifics of the problem but I had similar concerns. I believe the equation of the sinusoidal signal shows two components. One with a magnitude V1 and an angle of 45 degrees and another component V2 with and angle of 0. If you add the two of them, vectorial or converting it two rectangular, you will get the magnitude of the voltage applied to the resistor.

I will look for my Basic Circuits book, don't know where the heck it is, but I do remember finding the "why" there. If I am not wrong, and I usually am, the issue issue is that the signal it is not a pure sinusoidal wave and has two components. The basic form of the sinusoidal wave is V (sin(omegat + alpha)).

Well...I will stop making a fool of myself




and will get back to you with the basis.



Just trying to give you a heads up.

Ahhhhh...I almost forgot. They are giving you have to use the rms value so be careful with the square root of 2.

Check EERM 27-2 and 27-12


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## Art (Jan 20, 2007)

I come up with

S = 310 + j486 VA

P = 310 W

Q = 486 VAR

or S = 575 ang 58 deg VA

P = cos 58 deg x 575 = 0.53 pf x 575 = 310 W

now I may be off, I did this with a 6" slide rule


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## jdd18vm (Jan 20, 2007)

jdd18vm said:


> Sorry if this seems fundamental, but if someone could give me their thoughts.
> A Sinusoidal voltage of vt=[50 cos(ωt + 45) + 28.28 cos(ωt)] V is applied across a 5Ω resistor. the Average power (W) deliverd to the Resistor is most nearly?
> 
> I see, and can follow the solution, but dont understand the reasoning? Specifically what to do with the two components of the voltage.
> ...


here is the solution and answer the way it appears in NCEES

V=50 Theta 45 / Sqrt2 + 28.2 Theta 0/ sqrt 2 =

(50/sqrt2) (.0707+j0.707) +20 = &lt;-- i dont see what they are doing cos45 is .0707 cos 0 is 1

25+j25+20=

45+j25=

51.5 Theta 29 V

RMS value of the voltage is 51.5 The power is computed by

P=V^2/R = (51.5)^2/5 = 530 W

Art, Luis, thanks for your help and patience

gonna be a LONG 3 months


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## Dark Knight (Jan 20, 2007)

Hey my friend,

This is the purpose of the prep and we are also here to help. Did you check EERM 27-2 and 27-12? It is explained very well.

Don't quit and keep working hard. You will see the reward in the summer.


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## jdd18vm (Jan 20, 2007)

Luis said:


> Hey my friend,
> This is the purpose of the prep and we are also here to help. Did you check EERM 27-2 and 27-12? It is explained very well.
> 
> Don't quit and keep working hard. You will see the reward in the summer.


thanks 

I did actually. the part im confused is when/if to apply the cos (or sin for that matter) to the angle. as they didin the solution above. i did it the way Art said, i got 51.79 for the magnitude which yielded 536 watts...i suppose i would have picked the 530...but i'm missing something?


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## Dark Knight (Jan 20, 2007)

You missed nothing. Nailed it.

The aswer is 530 Watts....I think. Did it on the run here , without writing numbers on a paper and came up close to your number.

Good job my friend...and EERM is a book reference but :"the other board":


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## Art (Jan 20, 2007)

jdd18vm said:


> here is the solution and answer the way it appears in NCEES
> V=50 Theta 45 / Sqrt2 + 28.2 Theta 0/ sqrt 2 =
> 
> (50/sqrt2) (.0707+j0.707) +20 = &lt;-- i dont see what they are doing cos45 is .0707 cos 0 is 1
> ...


I was with you as far as

45+j25= 51.5 Theta 29 V (actually made a small math error...slide rule)

I had the 45 + 25J, but converted it to 53.2 (sqrt 45 x 50, instead of 45 x 45) ang 29

I squared it to 2875 ang 58

divided by 5

575 ang 58 (should be (51.5 x 51.5)/5 ang 58 or 530 ang 58 W)

I considered this a complex power...

it's not, it's a real power phase shifted...that is where I made my mistake...

so it's 575 W or if I did the math right 530 W

we can all learn something...nice problem


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## singlespeed (Jan 21, 2007)

jdd18vm said:


> thanks
> I did actually. the part im confused is when/if to apply the cos (or sin for that matter) to the angle. as they didin the solution above. i did it the way Art said, i got 51.79 for the magnitude which yielded 536 watts...i suppose i would have picked the 530...but i'm missing something?


Hopefully this will help: (I am using &lt;, for the angle notation (not less than))

50 cos(wt + 45) = 50/root2 *&lt;45 = 35.36&lt;45 _converted to rms phasor notation_

35.36&lt;45 = 35.36 * *cos*(45) + j35.36 * *sin*(45) = 25 + j25 _converted to rectangular_

Since both voltages were defined with cosines, you didn't have to worry about the phase difference between sine and cosine which is 90 deg when you converted to phasor notation. Then, the conversion to rectangular always requires the use of sine and cosine as I indicated.

So, 25 + j25 + 20 = 45 + j25 = 51.5&lt;29 _added voltages in rectangular and converted to phasor_

P = V^2/R = 51.5&lt;29^2/5&lt;0 = 530&lt;58 W

My answer was 530W


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