Basic circuit analysis

Discussion in 'Homework Help' started by markp912, May 6, 2012.

  1. markp912

    Thread Starter Member

    Feb 4, 2009
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    I have some questions about a few basic circuits.
    1. On the circuit below i am looking for I5 and V7. At the top middle node of the image does our voltage go to 36V to each side? Then, to find I5, i need to divide 36v/the total value of our resistors on the left hand side correct? Then, how do we find the voltage at V7? Thank you. [​IMG]


    2. For this one i have to thevinize the circuit to the left of resistor R. So i kill my current source first then take measurements. How do i find the value R for max power transfer to R and the maximum power to R?

    [​IMG]



    Thanks in advance for your help.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    No. Each connecting line represents a wire and it is assumed that the wires are ideal and, hence, every point on a given wire is at the same voltage. Thus, then entire bottom wire is at 0V (because it is connected to the ground symbol) and the entire top line is at a single voltage, namely 72V.

    With this in mind, consider the following (looking at just right hand side):

    1) What is the voltage at the top of R6?

    2) What is the effective resistance of R6, R7 and R8,9?

    3) What is the current flowing in this effective resistance?

    4) How does the answer to Q3 above relate to I6?

    5) How much current is flowing in R6?

    6) What is the voltage drop across R6?

    7) Give the answer to Q1 and Q6, what is the voltage at the bottom of R6?

    8) What is the voltage at the top of R7?

    9) How does the answer to Q8 relate to V7?


    As for the second question, consider the following:

    10) In general, given a function f(x), how do you find the value of x that maximizes f(x)?

    11) Once you have determined the Thevenin equivalent circuit, you will have a Vs, Rs, and R all in series. In terms of these three components, what is the power in R as a function of R, namely P(R)?

    12) What do you get if you apply your answer to Q10 to the answer to Q11?
     
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  3. markp912

    Thread Starter Member

    Feb 4, 2009
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    1) What is the voltage at the top of R6?
    72 volts

    2) What is the effective resistance of R6, R7 and R8,9?
    Does that mean all of the values added together? R7 and R8,9 were in parallel and got 4.5k for their value then added 12k and 4.5k together because they're in series. So i got 16.5k.

    3) What is the current flowing in this effective resistance?
    6mA

    4) How does the answer to Q3 above relate to I6?
    72 total volts over our total resistance of 12k

    5) How much current is flowing in R6?
    6mA

    6) What is the voltage drop across R6?
    72 volts

    7) Give the answer to Q1 and Q6, what is the voltage at the bottom of R6?
    We still have 72v correct?

    8) What is the voltage at the top of R7?
    We need to use a voltage divider so its (9k/9k+9k)72V=36V at R7

    9) How does the answer to Q8 relate to V7?
    Its the divided voltage from the 72v power source.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    CORRECT
    Yes, but saying "added together" is not a good way to say it because it implies "addition", which is not strictly true. It would be better to say something like, "Does that mean all of the values combined together?"
    CORRECT
    How do you get that? You have a voltage of 72V across and effective resistance of 16.5kΩ. Remember, the point of combining a bunch of resistors into a single equivalent (or effective) resistance is that you can replace the bunch of that resistors with the single equivalent resistance and the rest of the circuit can't tell the difference. So exploit that.

    The specific answer is wrong, but the idea is correct. The better, and more useful, answer would have been, "They are the same."

    Wrong, but correct given your previous answers. Correct those and you will probably get this one correct.

    Again, wrong but correct given your previous answers.

    Wrong, period. If you start out at the top of a 72ft tall staircase and drop down 72ft, how far off the ground are you? This is an example of gravitational potential, but the same concepts hold true for electric potential (also known as voltage). If you start out at a node that is at a potential of, say, 28V and you drop through a voltage of 5V, then you are now at a point that has a potential of 23V.

    But you just said that the voltage at the bottom of Q6 is 72V. Now you are saying that the votlage at the top of R7 is 36V. However, these two points are connected by a wire and remember what we said about wires - every point on the wire has the SAME voltage!

    Looking for something much more specific. In fact, I don't know what you mean when you say that V7 is the divided voltage from something. Hint: If I tell you a numerical answer to Q8, what is the numerical answer to V7 (and vice versa)?

    It appears that your main difficulty stems from not grasping the following concepts:

    1) Voltage is a measure of potential energy (just like height is a measure of gravitational potential energy).

    2) All of the points on a wire are at the same voltage (just like all of the points on the same floor of a building are at the same height).

    3) It doesn't matter which path you take to get from one point to another, you have to end up at the same potential (just like it doesn't matter if you take the elevator, the stairs, a fire pole, or scale the outside of the building).

    There may be some more fundamental concepts giving you problems, but you need to rework the set of questions I gave you in light of this new information before I can start seeing if that's the case.

    Good luck.
     
  5. markp912

    Thread Starter Member

    Feb 4, 2009
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    3) What is the current flowing in this effective resistance?
    6mA
    How do you get that? You have a voltage of 72V across and effective resistance of 16.5kΩ. Remember, the point of combining a bunch of resistors into a single equivalent (or effective) resistance is that you can replace the bunch of that resistors with the single equivalent resistance and the rest of the circuit can't tell the difference. So exploit that.

    Okay, i messed up with my calculations. My correct I6 current is 4.36mA.

    5) How much current is flowing in R6?
    My new correct answer is 4.36mA


    6) What is the voltage drop across R6?
    How do we not have 72 volts going across R6 if what you said was that all the spots on the wire have the same voltage of 72V?

    7) Give the answer to Q1 and Q6, what is the voltage at the bottom of R6?
    We still have 72v correct?
    Wrong, period. If you start out at the top of a 72ft tall staircase and drop down 72ft, how far off the ground are you? This is an example of gravitational potential, but the same concepts hold true for electric potential (also known as voltage). If you start out at a node that is at a potential of, say, 28V and you drop through a voltage of 5V, then you are now at a point that has a potential of 23V.

    But arent we still running 72V through the circuit?



    8) What is the voltage at the top of R7?
    We need to use a voltage divider so its (9k/9k+9k)72V=36V at R7
    But you just said that the voltage at the bottom of Q6 is 72V. Now you are saying that the votlage at the top of R7 is 36V. However, these two points are connected by a wire and remember what we said about wires - every point on the wire has the SAME voltage!

    This answer should be our voltage outputted after going thourgh R6. Why do we have a voltage drop after R6? So our voltage at R7 will be the same voltage outputted after R6. I still do not see why it wont be 72V since you say every spot on the wire is the same.
     
  6. markp912

    Thread Starter Member

    Feb 4, 2009
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    Wait, if we have 9K at R7 with 4.36mA of current dont we have 39.24V at V7?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Correct up to this point.

    All the spots on the SAME wire.

    You have one wire across the top connecting the positive terminal of the battery to one side of R1,2,3, R4, and R6. There is a different wire connecting the bottom side of R6, R7, and R8,9.

    You have already determined that I6 is 4.36mA. There is no place for that current to go except through R6. Thus, by Ohm's Law, the voltage that has to exist ACROSS R6 (meaning the voltage on one side minus the voltage on the other) is I6*R6 = 4.36mA*12kΩ = 52.3V. Since I6 is positive, it means that the current is flowing in the direction it is defined on the schematic and therefore is entering the top of R6 and existing the bottom. The side that the current is entering has the higher voltage, so the bottom of R6 is 52.3V lower than the top, which is at 72V.

    Thus, what is the voltage at the bottom of R6?

    That's like saying, "Isn't everything in a ten story building ten stories above the ground?" No, only the things on the roof. Things on the third floor are only three stories above the ground.

    Each node is at a different voltage. This circuit has four nodes: One at the bottom, which is at 0V, one at the top, which is at 72V, and one each at the midpoint of the left and right branches which are at unknown voltages.

    If the voltage at the bottom of R6 was 72V, then the voltage across R6 would be 72V-72V=0V and there would be zero current flowing in it.

     
  8. WBahn

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    You're starting to get the idea, but remember that Ohm's Law relates the voltage ACROSS a particular resistance to the current through THAT SAME resistance and the resistance of THAT SAME resistance. In this case, the 4.36mA is NOT the current flowing in R7, it is the current flowing in R6. My prior post should help you determine the voltage at the bottom of R6, which has to be the same as the voltage at the top of R7 because they are connected to the same node.

    As a sanity check, you know that the 4.36mA has to split up between R7 and R8,9. Since both of these are the same size, symmetry tells you what about how much current is going to go through each?

    Another way to get at it is to consider what the equivalent resistance is between the node (the one between R6/R7) and ground by combining just those two resistors into a single equivalent (which you've already done once before). The entire 4.36mA would be flowing in this equivalent resistance were you to replace those two with their equivalent.

    So you have three ways of getting the answer and they better all agree. So do it all three ways and see if they all agree (to within roundoff error).
     
  9. markp912

    Thread Starter Member

    Feb 4, 2009
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    Thus, what is the voltage at the bottom of R6?

    72V-52.3=19.7V at the bottom of R6.

    As a sanity check, you know that the 4.36mA has to split up between R7 and R8,9. Since both of these are the same size, symmetry tells you what about how much current is going to go through each?

    We now use a current divider so split up our current between the two 9K resistors and we get 1.09mA of current going through R7. Therefore we used 1.09mA*9K to get our V7 equal to 9.81V.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Always ask if the answer makes sense!

    We have already established that the voltage at the bottom of R6 has to be the same as the voltage at the top of R7 (because they are connected to the same wire). But you got a voltage at the bottom of R6 of 19.7V and a voltage at the top of R7 of 9.81V. So you know at least one of them is wrong.

    Now, one thing that I should have asked at the beginning but was willing to make an assumption based on your prior work: In the diagram there is a resistor labeled R8,9 that is given as 9kΩ. What does this mean? Is 9kΩ the equivalent resistance of two resistors, R8 and R9, or is this some kind of unusual shorthand and there are really two 9kΩ resistors there (in parallel with the other 9kΩ resistor, namely R7)?

    Assuming that we have what the diagram shows, namely a total of two 9kΩ resistors (R7 and R8,9) in parallel, what current division rule led you to conclude that there was 1.09mA in R7? How much is flowing in the other 9kΩ resistor?
     
  11. markp912

    Thread Starter Member

    Feb 4, 2009
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    We have already established that the voltage at the bottom of R6 has to be the same as the voltage at the top of R7 (because they are connected to the same wire). But you got a voltage at the bottom of R6 of 19.7V and a voltage at the top of R7 of 9.81V. So you know at least one of them is wrong.

    It should be 19.7V at the top of R7.


    Now, one thing that I should have asked at the beginning but was willing to make an assumption based on your prior work: In the diagram there is a resistor labeled R8,9 that is given as 9kΩ. What does this mean? Is 9kΩ the equivalent resistance of two resistors, R8 and R9, or is this some kind of unusual shorthand and there are really two 9kΩ resistors there (in parallel with the other 9kΩ resistor, namely R7)?

    That is a question i will have to ask my instructor. I do not know what he means by that.

    Assuming that we have what the diagram shows, namely a total of two 9kΩ resistors (R7 and R8,9) in parallel, what current division rule led you to conclude that there was 1.09mA in R7? How much is flowing in the other 9kΩ resistor?

    I used a current divider with (4.5K/18K)(4.36mA) and got 1.09mA. Or should we just divide 4.36mA by two and use that as the current for each resistor?
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    The 4.5kΩ is the equivalent resistance and not the resistors that are there. Think of it this way. I can build two boxes, A and B, each having two terminals coming out of it. One box would have a single 4.5kΩ resistor in it and the other would have two 9kΩ resistors connected in parallel. You could place either one in the circuit (in place of R7 and R8,9) and you wouldn't be able to tell the difference -- that is what makes them equivalent. But you either have one, or the other, in the circuit. Never both. So don't mix and match portions of each during your analysis. One, or the other.

    You need to become comfortable with deriving things like the voltage divider and current divider rules on the fly. For current division, it would go something like this:

    I have a total current, I_T, that splits between R_1 and R_2.

    We know two things:

    1) The individual branch currents, I_1 and I_2 (through R_1 and R_2, respectively, must sum to the total current. Hence

    I_T \ = \ I_1 + I_2.

    2) The voltage across the two resistors is the same, because they are in parallel. Hence

    V_T \ = \ I_1R_1 \ = \ I_2R_2.

    If we want to know the current in R_1 in terms of the total current, just eliminate I_2 from the set of equations. What do you get?
     
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