Circuit Analysis Question -- Solving for a multi-resistor circuit

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Homework Statement:
In a circuit R1=R2, these are in series linked in parallel to R3=150 ohms and R4=200 ohm which are in series The current flowing through R1 is 4 amperes. The source has 15V in the total circuit. Solve for I3 and R1 and R2
Relevant Equations:
Vparalell=V1=V2.. Iseries=I1=I2.. Rseries=R1+R2... 1/Rpara=1/R1 + 1/R2..
From what I understand is that you can sum up R3 and R4 to get 350 ohms
Then you can can a parallel circuit with the missing resistors, 4amp and the new 350 ohm resistor and that's about it.

My textbook says the answer is 55 ohms for the resistors and the current is 0.073 but I got 3.75 ohms (15v/4 amp) and that's about it. I'm very lost
 

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  • #2
berkeman
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Welcome to the PF. :smile:

Can you upload a picture or diagram of the circuit? That will help us to help you solve this. Use the Attach Files link below the Edit window to upload a PDF or JPG file. Thanks.
 
  • #3
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Heres the question
It's number 63
 
  • #4
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You said "I got 3.75 ohms (15v/4 amp)". I agree with you. There may be an error in the write-up of the problem or in what your book said the right answer should be.

For (b), can you explain what the (11.9) means? That might be a clue.

Can you ignore a for now and answer b?
 
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  • #5
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You said "I got 3.75 ohms (15v/4 amp)". I agree with you. There may be an error in the write-up of the problem or in what your book said the right answer should be.

For (b), can you explain what the (11.9) means? That might be a clue.

Can you ignore a for now and answer b?
The 11.9 is the section of my textbook to which circuit analysis is located but this is very different from the problems there
 
  • #6
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I got 3.75 ohms (15v/4 amp)". I agree with you.
I should make certain that you know - that is the sum R1+R2. So what are the R1 and R2 values?

About part b, do you have an answer? Since you understood that 3.75 ohms = 15v/4 amp, I expect that you can find an answer for the current thru R3+R4. Or does does the fact that R3 and R4 have different values cause you uncertainty?
 
  • #7
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I should make certain that you know - that is the sum R1+R2. So what are the R1 and R2 values?

About part b, do you have an answer? Since you understood that 3.75 ohms = 15v/4 amp, I expect that you can find an answer for the current thru R3+R4. Or does does the fact that R3 and R4 have different values cause you uncertainty?
Yes, It does but in my textbook, it states that R1 and R2 are 55 ohms and I3 is 0.073 A so trying to figure out how is extremely confusing
 
  • #8
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The answer in your book and the statement of the problem that you posted are not compatible. To be sure that you know how to solve those questions, please answer this: Just from the statement of the problem, what would your answers be?
 
  • #9
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Ok, since 4 amps is flowing through r1, then r2 must have 4 amps, but the 15volts must be split across the 2 sections as the incoming voltage must equal the outgoing. So I would get a paralell circuit. 1/x (since r1 and r2 equal) + 1/200+150 so 1/x +1/350=1/r then I'd multiply that by 4 to get 150 and solve for x?
 
  • #10
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Your book's answers are wrong on this question!!!
Your last post showed significant confusion caused by the error in your book saying what the answers to these 2 parts should be. I will show you how close you were and hopefully eliminate some confusion.

In your initial post, you said
I got 3.75 ohms (15v/4 amp)
. You almost had the answer to part a then. (One simple step is yet to be done.) The battery's 15 V is across R1 and R2 and that 4 A is running thru them. Ohm's Law applies to the R1&R2 branch. And that was how you calculated the 3.75 Ohms. So R1+R2=3.75 Ohms. The problem says that R1=R2, so one simple step will give you the resistance of both.

The R3&R4 branch does not change how Ohm's Law applies to the voltage across a series group of resistors, the current thru them, and the equivalent resistance of the group. Let me explain the significance of calculating equivalent resistance, Req, when you analyse part of a network. You could replace the group of resistors (R1+R2+...) with a single resistor of the value Req and nothing would change.

Now, about part b. As you said,
you can sum up R3 and R4 to get 350 ohms
. So you have the Req of that branch. And since this series pair is in parallel with R1&R2, this applies
Vparalell=V1=V2
. So you know the voltage across this series pair. If you use Ohm's Law on this equivalent resistance you will find the current thru these 2. And then you can use Iseries=I1=I2.

So you see that Ohm's Law applies to each of the parallel branches as if the other branch were not even there. It is more work for the battery since it has to put out more current, but the current is divided at the node at the "Y" in the path according to the needs of each branch. The 2 parallel branches are not confused by each other.
 
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