Voltages in two branch networks

Discussion in 'Homework Help' started by Markafano, May 28, 2016.

  1. Markafano

    Thread Starter New Member

    May 28, 2016

    I have the following task, I've managed to do the first part but was needing some help for the second.

    The circuit can be seen below (please note a switch exists between the supply and L1):
    Via the use of Mesh analysis, determinents and patial fractions I have found the currents in each branch and therefore the total current as well, my workings can be seen in the attached PDF.

    The question I'd like some help with is:
    The second part asks for the voltage to be calculated across the centre branch at the following times (s) after the switch has been closed:
    0, 1, 10, 100, and at 't' infinity.

    Now I understand that in a DC circuit the inductor will offer no impedance after a certain length of time, and in an AC circuit the impedance of an inductor is 2*pi*f*l but how is it dealt with in the example above?
  2. RBR1317

    Active Member

    Nov 13, 2010
    Suppose instead that the circuit would be configured with all resistors instead of a mixture of coils & resistors. Could you use a DC voltage divider to find the voltage across the center branch? No difference in this problem in the complex frequency domain except that the 'resistance' of the coils is replaced by sL. Then take the result from the voltage divider and use a table of inverse Laplace transforms with some partial fraction expansion as necessary to get back to the time domain.
  3. WBahn


    Mar 31, 2012
    If you have the current in the center branch, what is giving you trouble in finding the voltage across the center branch. Hint: It's a resistor!

    Your work keeps referring to things like "I1 branch current," but I don't see anywhere that you identify which branch that refers to. That forces your readers to guess, and engineering is not about guessing.

    You final line indicates that you are claiming the total current in steady state is 7.86 A. Does that make any sense? You know that, in DC steady state, an inductor looks like a short circuit. That means that the battery only sees the two resistors connected in parallel. What will that current be?
  4. Markafano

    Thread Starter New Member

    May 28, 2016
    PBR thank you for your reply.

    WBahn, apologies I1 refers to that part of the circuit which contains the components V1, L1 and R1, plus the switch, not shown.
    The branch I2 refers to that part of the circuit containing the components R1, R2 and L2.

    I understand that when looking at just the resistors, the total resistance is 22.4 ohms ((42*48)/(42+48)). When dividing the voltage (120V) by this I get a value of 5.36 A, which is my calculated value as per the attached PDF, for branch 1.

    However what I need help with is that at 1 second I have a current of 4.68A running through this branch which means a resistance (or impedance) of 25.6 ohms (120/4.68). Aren't the inductors having an effect right after the switch is turned on?

    It is this effect that I am trying to figure out how to deal with in this exercise.

    I've run a simulation of this circuit in Multisim:
    Without inductors:

    With the inductors included:

    With regards to the total current being 7.86A I have no reason to believe that by adding the currents in the two branches together would not give anything other than the correct toal current.
    Although looking at a value of 7.86A and 120V gives a resistance of 15.26 ohms which does seem low.
  5. WBahn


    Mar 31, 2012
    Part of the confusion is that you are using the term 'branch' when you should be using 'mesh' or 'loop'. A branch consists of elements that are all in series.

    Your circuit has three branches: {V1, L1}, {R1}, and {L2, R2}

    A loop is a complete circuit while a mesh is a special kind of loop that doesn't encompass any other components.

    Even with your additional descriptions, we have to guess which direction those currents flow.

    Instead of describing parts of your circuit and your current assignments verbally, why not just annotate your schematic using Paint or any other free graphic editor to make things clear?

  6. Markafano

    Thread Starter New Member

    May 28, 2016

    Sorry but I am only using the terminology given to me by my lecturer, now it may be a case of me misunderstanding something he has said but I don't know until someone points it out.

    But your diagram is effectively what I am working with.

    I was told that the first mesh is:

    and that the 2nd mesh is:

    My understanding is that the use the above of determinants helps to identify the unknown values, i.e. I1 (the current in mesh 1) and I2 (the current in mesh 2).

    Now if instead I1 is the total current and I2 the current in Mesh 2 then this would make a bit more sense with regards to the values that I have got but means that I am missing some piece of knowledge somewhere as why would I take the current away that is in mesh 2 from the current in mesh 1 as we were told that this was not the total current, sorry if rambling.

    But sticking with the calculated value of I1 being in actuality the total current then at say 10 seconds I get:

    5.36-2.5=2.86 A

    120/2.86=41.95 ohms (approx 42 ohms) the vlaue of R1
    As expected the inductor is not offering any impedance.

    and at say 1 second:
    4.68-2.09=2.59 A

    so 120/2.59=46.33 ohms
    From this I can say that at 1 second the inductor is offering:
    46.33-42=4.33 ohms impedance

    and in Mesh 2:
    120/2.09=57.4 ohms total.
    So can I also say that the inductor is offering
    57.4-48=9.4 ohms impedance

    If so then I can redraw the circuit for time frame 1 second replacing L1 with a 4.33 ohm resistor and L2 with a 9.4 ohm resistor.

    Now of course I can simply change out a 9.4 ohm resistor along with the 48 ohm resistor to one 57.4 ohm resistor.

    Presumably I use Thevenins to work this out

    At t=10 s and higher I can just ignore the inductors because as I have already stated earlier they offer zero resistance at this point so I am dealing with just two resistors in parallel.

    If the above makes sense or is total gibberish please let me know, thanks.
  7. WBahn


    Mar 31, 2012
    Which only underscores the value of a properly annotated schematic instead of relying on verbal descriptions -- which is not to say that misunderstandings can't happen with schematics, but it is far less likely and generally far easier to spot and correct.

    The first problem is that you are trying to add things that can't be added. L1 is an inductance while R1 is a resistance. Adding them is like trying to add feet to gallons -- simply can't be done.

    I'm going to take this opportunity to harp on one of my biggest pet peeves -- in part because I think you are in a position to benefit nicely from it.

    You need to track your units throughout your work -- not just tack the units onto the final result that you want that result to have. Had you done that, you would have saved a LOT of time.

    What would have happened had you done that? When it came time to evaluate (L1+R1) you would have had (8.4 H + 42 Ω). What would you have done then? You would have been faced with the reality that you simply can't add them. 1 H is equal to 1 V·s/A which is also equal to 1 Ω·s. You can't add that to something that has units of Ω, which would have let you see that something was wrong right at that point before you proceeded to waste time and effort pushing through to an answer that is guaranteed to be wrong.

    Recall that the impedance of an inductor in the complex frequency domain is

    Z = Ls

    The units on the domain variable 's' is the same as radian frequency, namely 1/second. So when we multiply that by inductance we are left with units of resistance -- and hence something that CAN be added to resistance.
  8. Markafano

    Thread Starter New Member

    May 28, 2016
    Thanks for your help, I apologies unreservedly for the stupid mistake I made and wasting some of your time, I see that I've not correctly typed up my original Mesh equations and they should read:

    For Mesh 1:

    and for Mesh 2:

    The thing is that when I have my three determinants, they are:


    DI2 = 5040/s

    So in order to work out the current in Mesh 1 (I1):

    and the current in Mesh 2 (I2):

    So why in order to get the caclulated values to fit the circuit I have to use:

    I would have thought I would have to use:
    I1+I2=Total current; but as you have already pointed out the values for this don't add up.

    Surely if I1 was the total current I'd have to include resistor R2 and inductor L2 within the Mesh 1 equation?

    Edited for clarity
  9. WBahn


    Mar 31, 2012
    Look at the diagram. I1 is a current flowing downward in R1 while I2 is a current flowing upward in R1. Thus the net current flowing downward in R1 is (I1-I2). If want the current flowing upward out of the source or left-to-right in L1, then you just use I1. Similarly, if you want the current flowing downward in R2 or left-to-right in L2, then that is just I2. The sum of the two currents is meaningless since there is nowhere that the two currents for flowing through a branch in the same direction.
  10. Markafano

    Thread Starter New Member

    May 28, 2016
    OK so it is the direction of current flow is why I have to carry out a subtraction rather than adding them together.
  11. WBahn


    Mar 31, 2012
    Yes. As I noted much earlier, voltages and currents have both a magnitude and a polarity. You can't ignore either.