prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt-x))  TOPIC_SOLVED

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prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt-x))

Postby vgreen74 on Sat May 22, 2010 3:07 am

prove that P=VI where V=5sinwt and I=10sin(wt-x) and p=25(cosx-cos(2wt-x))
am I correct so far and where do I go from here?
Let wt=A and x=B
P=(5sinA)(10sin(a-b))
P=50(sinA)(sinAcosB-cosAsinB)
P=50(sin^2AcosB-sinAsinBcosA)

P=50(cosB+(sin^2AcosB-sinAsinBcosA-cos^2AcosB))
P=25(cosB-(2sinAsinBcosA+cosBcos^2A-sin^2AcosB))
P=25(cosB-cos(2A-B))
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Postby stapel_eliz on Sat May 22, 2010 12:31 pm

How are ''p'' and ''P'' related?

Thank you! :wink:
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Re: prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt

Postby vgreen74 on Sat May 22, 2010 9:46 pm

Sorry P and p are the samething I just missed the shift key. I have to prove that starting with the equations for V and I I end up with the equation for P.
using the cos(A+B) and cos(A-B) rules
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Postby stapel_eliz on Sat May 22, 2010 10:43 pm

vgreen74 wrote:where do I go from here?

I'm not sure what you mean by this, since you seem to have arrived at the final result...? :confused:

vgreen74 wrote:P=50(cosB+(sin^2AcosB-sinAsinBcosA-cos^2AcosB))
P=25(cosB-(2sinAsinBcosA+cosBcos^2A-sin^2AcosB))
P=25(cosB-cos(2A-B))

I was able to figure out some of the missing steps in your displayed work, but I'm afraid I'm drawing a blank here. Please reply with the rest of what you did. Thank you! :wink:
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Re: prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt

Postby vgreen74 on Sun May 23, 2010 12:42 am

I have tired to work from both ends of the equation as I was drawing a blank on working from the begining however I may not have the thrid line correct when tring to work back from the equation for P
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Postby stapel_eliz on Mon May 24, 2010 1:27 am

I could be wrong, but I think the proof isn't working because the equality isn't true.

What did you get when you graphed the left- and right-hand sides of the equations? Did the lines match up? Mine aren't.
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Re: prove P=VI for V=5sinwt, I=10sin(wt-x),p=25(cosx-cos(2wt  TOPIC_SOLVED

Postby vgreen74 on Thu Jun 10, 2010 10:06 pm

sorry my first mistake was with my grouping it should have been let A=wt and B=wt-x
so P=(5sinA)(10sinB)
P=50sinAsinB
using product to sum
2sinAsinB=cos(A-B)-cos(A+B)
therefore
P=25(2sinAsinB)
P=25(cos[wt-{wt-x}]-cos[wt+{wt-x}])
P=25(cos[wt-wt+x]-cos[wt+wt-x])
P=25(cosx-cos[2wt-x])
which was what I had to prove
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