Series-Parallel Circuits

Discussion in 'Homework Help' started by Fidd, Mar 11, 2016.

  1. Fidd

    Thread Starter New Member

    Mar 11, 2016
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    Hello everyone, I'm quite new to Electrical Engineering. My background is mostly in computer science and i'm having issues figuring out how to setup the correct equations for series-parallel circuits. #13.png This is my problem. I've attempted to get the correct equations from this but always come out with two unknowns. I know that RT1 (total resistance within the two resistors in parallel at the top of the circuit) will be (47)(R2)/(47+R2), I also know that we have 220V coming from the source and there must be a voltage drop across each resistor. I know the voltage drop across the resistors in parallel are going to be equivalent. So i'm guessing V1 = V2 = Vs-V3? Also, there is a 1A current flowing through R2 (the resistance of R2 is what i'm trying to determine). I fell like I would be pretty quick to pick up on this problem if I could just be pointed in the right direction. Thanks in advance.
     
  2. WBahn

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    If you were given the value of R2 but not the value of the current flowing in R2, could you find the value of the current flowing in R2?
     
  3. Fidd

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    Mar 11, 2016
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    I believe if that was the case I could use I = V/R, but I still don't know our voltage across the resistor.
     
  4. WBahn

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    If you were just given the circuit layout with the value of the voltage source and all three resistor values, could you find the total current flowing in the circuit?
     
  5. Fidd

    Thread Starter New Member

    Mar 11, 2016
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    Yes, I could use Is=220V/RT1+R3?
     
  6. WBahn

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    What the math.

    What you wrote is equivalent to

    Is = (220V/RT1) + R3

    What you meant was

    Is = 220 V / (RT1 + R3)

    Okay, so now that you have Is, what is the current through R2?
     
  7. Fidd

    Thread Starter New Member

    Mar 11, 2016
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    The current through R2 is 1A, that is a given.
     
  8. WBahn

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    Look at what I said in Post #2 and Post #4 -- pretend that you have NOT been given the current in R2!

    I'm trying to lead you along a path that will help you solve the problem by using what you already know -- how to analyze a circuit in which the component values are given but the currents are not -- to solve a problem in which some of the currents are given but not all of the component values are. The idea is to use what you know to find an expression for the current in R2 in terms of the value of R2. With such an equation if you are given one value you can find the other.
     
  9. Fidd

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    Mar 11, 2016
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    Okay.. To answer the previous question again, would I2 = IT * (RT1/R2)?
     
  10. WBahn

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    Correct. So now expand RT1 in terms of R1 and R2 and also expand IT in terms of R1, R2, R3 and Vs so that you have I2 = a function of Vs and the three resistances and nothing else. Leave the terms as Vs and R1, R2, R3 for now. The work is best left completely symbolic right up to the end at which point you substitute in all the values to get the final answer.
     
  11. Fidd

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    Mar 11, 2016
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    Okay, I'm getting seriously confused now. This is what I got from that and I don't think it makes sense. I2 = IT * (R1*R2)/R2... I'm sorry but i'm having trouble understanding what you mean when you say "expand IT in terms of R1, R2, R3 and Vs so that you have I2 = a function of Vs and the three resistances and nothing else."
     
  12. WBahn

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    Your starting point of

    I2 = IT * RT1/R2

    Is correct (since the voltage across R2 is IT*RT1).

    But think about

    I2 = IT * (R1*R2)/R2

    What are the units of the right hand side? It's a current multiplied by a resistance-squared divided by a resistance. The end result is a voltage. You KNOW that a current cannot be equal to a voltage, so you KNOW that this equation is incorrect.

    What is RT1? It is the parallel combination of R1 and R2, right?

    RT1 = (R1*R2)/(R1+R2)

    So substitute THAT into

    I2 = IT * RT1/R2

    and what do you get?

    At that point you will have I2 in terms of IT, R1, and R2. The problem is that IT itself is a function of R2, so we need to express IT in terms of R2 (and the other circuit parameters).

    The goal is to end up with an equation that has I2 on the left and an expression on the right that ONLY has Vs, R1, R2, and R3 in it. Nothing else (other than, perhaps, constants).
     
  13. Fidd

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    Mar 11, 2016
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    Okay, So that gives me I2 = IT *((R1 R2)/(R1+R2))/R2?
     
  14. WBahn

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    Yes, although that can be simplified a bit -- look at the R2 factors.
     
  15. Fidd

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    Mar 11, 2016
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    I'm guessing I2 = IT *((R1 R2)/(R1) ?
     
  16. WBahn

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    No. Once again, look at the units of the expression on the right hand side. You have a current multiplied by a resistance (since resistance-squared over resistance is just resistance) so the right hand side has units of voltage while the left hand side has units of current. You therefore KNOW that this is WRONG.

    By now you should be LONG past having to GUESS about basic algebra concepts.

    You are starting with

    <br />
I_2 \; = \; I_T \, \frac{\( \frac{R_1 R_2}{R_1 \, + \, R_2}\)}{R_2}<br />

    Is this not the same as

    <br />
I_2 \; = \; I_T \, \( \frac{R_1 R_2}{R_1 \, + \, R_2}\)\( \frac{1}{R_2} \)<br />

    Is this not the same as

    <br />
I_2 \; = \; I_T \, \frac{R_1 R_2}{\(R_1 \, + \, R_2\) R_2}<br />

    Is this not the same as

    <br />
I_2 \; = \; I_T \, \frac{R_1}{\(R_1 \, + \, R_2\)} \( \frac {R_2}{R_2} \)<br />

    Until your math skills come up to speed, don't shortcut things -- take it in as small of steps as is needed to be sure that you are doing it right. And then CHECK your results to see if they make sense -- if the units don't work out, the answer is wrong. Period.
     
  17. Fidd

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    Mar 11, 2016
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    Right, No I simply just typed this in wrong. I have R1/R1+R2 Sorry.
     
  18. WBahn

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    Again, R1/R1 + R2 is the same as 1 + R2. You need to stop being sloppy with order of operations and relying on the "you know what I meant" approach. If you don't, you will type those kinds of equations into programs, simulators, spreadsheets, etc, and get wrong results and have a hard time tracing down the cause because "you knew what you meant". The problem is that the computer only knows what you actually wrote.

    Engineering is about attention to detail.
     
  19. manson1972

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    Sep 14, 2015
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    Hello, I know this thread is kind of old, but I wanted to check my answer since I've been trying to learn this stuff. Anyway, I got 110 ohms for R2, I checked it out via voltage drops and current dividing, and seems to check out. Can someone tell me if that is correct?
     
  20. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Looks good, the exact value is 109.8625Ω
     
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