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Old 27th October 2009, 06:00 PM   #1
Default Calculating power in a series-parallel circuit

Hello,

I'm trying to verify that I correctly calculated the power in R5 shown in the circuit below.



I first determined that the voltage over R5 is 4.91 v. Then I calculated power as P = (V^2)/R5 = 4.3 mW. I would greatly appreciate it if anyone could verify my calculations. Thanks!
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Old 27th October 2009, 07:55 PM   #2
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Show us your calculations........
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Old 27th October 2009, 08:04 PM   #3
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Quote:
Originally Posted by Chippie View Post
Show us your calculations........
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Old 27th October 2009, 08:12 PM   #4
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R5 isnt in parallel with R3/R4......See your third line
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Old 27th October 2009, 08:38 PM   #5
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Quote:
Originally Posted by Chippie View Post
R5 isnt in parallel with R3/R4......See your third line
Thanks for your help. It looks like I should have had R5 in parallel with R2.

So then VR5= (R2||R5 + R3||R4) / (R1 + R2||R5 + R3||R4).

Does that seem reasonable?
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Old 27th October 2009, 08:56 PM   #6
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Calc the value of R3 in parallel with R4 (514.29 ohms)....add R2 to that ( 144.29) , then calc that result in parallel with R5...(950.14)...call that Rz
Add that to R1...

Calc the current Iw thru R1+Rz...

then Pdiss in R5 = Iw squared*R5
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Old 28th October 2009, 12:10 PM   #7
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Why don't you use mesh analysis? you do the three meshes find the current thru R5, and you're done, it will be very fast if you use a graphing calculator.
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Old 29th October 2009, 08:09 PM   #8
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Quote:
Originally Posted by Karkas View Post
Why don't you use mesh analysis? you do the three meshes find the current thru R5, and you're done, it will be very fast if you use a graphing calculator.
I'm afraid that I haven't studied mesh analysis at all. But I'll look into that; thanks.

Last edited by Emil09; 29th October 2009 at 08:51 PM.
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Old 29th October 2009, 08:51 PM   #9
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Actually, what I'm really trying to verify is the following:

for the circuit I originally posted.

Thank you.
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Old 29th October 2009, 09:07 PM   #10
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deleted my answer
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Last edited by Chippie; 29th October 2009 at 09:12 PM.
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Old 30th October 2009, 12:08 AM   #11
1 out of 1 members found this post helpful.
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As I see it, it's good, you'll get a the result, as long as you don't make a mistake with the numbers.
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Old 30th October 2009, 12:47 AM   #12
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Thanks. I'm studying for an exam and this is one of the things that was causing some confusion for me.
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Old 30th October 2009, 12:42 PM   #13
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As you see, there's a bunch of people here that can help you, hope you do it good.
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