Calculating power in a series-parallel circuit

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Emil09

New Member

• Oct 27, 2009

• #1

Hello,

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

**broken link removed**

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!

 

Chippie

Member

• Oct 27, 2009

• #2

Show us your calculations........

 

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Emil09

New Member

• Oct 27, 2009

• #3

Chippie said:

Show us your calculations........

Click to expand...

**broken link removed**

 

Chippie

Member

• Oct 27, 2009

• #4

R5 isnt in parallel with R3/R4......See your third line

 

E

Emil09

New Member

• Oct 27, 2009

• #5

Chippie said:

R5 isnt in parallel with R3/R4......See your third line

Click to expand...

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?

 

Chippie

Member

• Oct 27, 2009

• #6

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

 

K

Karkas

Member

• Oct 28, 2009

• #7

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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Emil09

New Member

• Oct 29, 2009

• #8

Karkas said:

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.

Click to expand...

I'm afraid that I haven't studied mesh analysis at all. But I'll look into that; thanks.

 

Last edited: Oct 29, 2009

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Emil09

New Member

• Oct 29, 2009

• #9

Actually, what I'm really trying to verify is the following:
**broken link removed**
for the circuit I originally posted.

Thank you.

 

Chippie

Member

• Oct 29, 2009

• #10

deleted my answer

 

Last edited: Oct 29, 2009

K

Karkas

Member

• Oct 29, 2009

• #11

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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Emil09

New Member

• Oct 29, 2009

• #12

Thanks. I'm studying for an exam and this is one of the things that was causing some confusion for me.

 

K

Karkas

Member

• Oct 30, 2009

• #13

As you see, there's a bunch of people here that can help you, hope you do it good.