Resistor Between Two Parallel Circuits

Discussion in 'Homework Help' started by Shoutymatt, Dec 14, 2012.

  1. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    [​IMG]

    Uploaded with ImageShack.us

    Hi,

    I'm just wondering how you would go about finding the current across each resistor?


    I just don't have a clue :/ I've changed the numbers from the question I've been given as I'm more after the theory behind how you go about the solution as opposed to someone doing the problem for me.

    I've just never seen a resistor added between two parallel circuits before

    Does anyone have any clues?

    Thanks if you can help.
    [​IMG]
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,429
    3,360
    Use KVL and KCL to write out the equations.
     
  3. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    but i dont really know where to start

    if 5 amps goes into that node then 5 amps must come out. But in the question I've been given, when i do

    I = v / r on r1 and the same again for r2 to find the current across each resistor

    it adds up to more than the total amps going into the node :/
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,429
    3,360
    You have changed the numbers. Ignore the 5 amps.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    If you've changed the numbers, then you don't know if there is actually 5A flowing out of the battery.

    When you used I = v/r on r1, which v did you use? You can't just grab any v that happens to be handy. Ohm's Law related the resistance of a resistor to the voltage across THAT resistor and the current flowing in THAT resistor.

    See the attached diagram. You have four nodes in this problem. You get to pick one and define it to be any voltage you want it to be. By convention, we pick a node the looks like it will make the analysis easier (not always easy to tell) or one that just seems like an obvious reference node and call it "ground" and assign it a voltage of 0V. In the diagram, I have picked the bottom node. I've also labeled the other three nodes A, B, and C.

    Notice also that I have removed the numbers altogether. If you want to learn circuit analysis, then learn to do it symbolically.

    I have arbitrarily assigne a current direction for each resistor. It's safe to assume that the current directions are correct for the vertical resistors, but who knows about the horizontal one -- may I picked right and maybe I didn't. If the value of that current comes out to be positive, then I picked right. If it comes out negative, then I picked wrong but now I know the current is actually flowing the other direction with the same magnitude.

    It's fundamentally important to understand that I could have flipped a coin in assigning the current directions.

    Having assigned the directions of the current for a given resistor, the voltage polarity, as shown by the + and - marks, is not arbitrary. The "passive sign convention" dictates that, for a passive component (which a resistor is the classic example), current enters the positive terminal and exits the negative terminal. The reverse is true for an active device, such as a battery, and then current exits the positive terminal and reenters the negative terminal.

    Note that while I have not labeled the currents and voltages associated wiith each resistor, all of the information to do so is there. For instance, V2 is the voltage across R2 and I2 is the current through R2. Thus, by Ohm's Law, we have

    V2 = I2*R2

    I2 is the current flowing downward through R2 and V2 is the voltage at the top of R2 relative to the voltage at the bottom. This last statement means that

    V2 = Vc - Vb = Vcb

    The notation Vxy means the voltage on node 'x' minus the voltage on node 'y', or saying it differently, the voltage on node 'x' relative to the voltage on node 'y'.

    With all of this in mind, try writing equations for the relationships between the various node voltages and the currents in terms of supply voltage and the resistances. See if you can solve for each of them and, having done so, see if they do, in fact, satisfy KVL and KCL.
     
  6. Engr

    Member

    Mar 17, 2010
    114
    5
    Use KCL and KVL to solve this circuit. Since you change the numbers, I dont think the current is still 5Amps, it could be a different number. That is another variable that you have to solve.
     
  7. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    Thanks for the help everyone.

    I realised the circuit is essentially a bridge circuit. So I've redesigned how the problem is presented so that it resembles some of the online help available. I'm trying to apply the mesh method to the problem.

    [​IMG]

    Uploaded with ImageShack.us

    Loop one:

    V2 + V5 + V1 = 0
    R2(I1) + R5(I1+I2) + R1(I1+I3) = 0


    Loop two:
    V5 + V3 + V4 = 0
    R5(I1+I2) + R3(I2-I3) + R4(I2) = 0

    Loop Three:
    Vtotal - V1 - V3 = 0
    100V - R1(I1+I3) - R3(I2-I3) = 0

    I'm a bit weary of whether the negatives and positives are right. Is this a possible solution or am I on the road to nowhere :/

    Thanks again for any help
     
  8. MrChips

    Moderator

    Oct 2, 2009
    12,429
    3,360
    WBahn already pointed out. The sign of the current will indicate the direction.
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    You are pretty close, but are making some mistakes here and there.

    You can't assign two different polarities to the same voltage, like you do for R3. Consider this - once you have a value for V3, which polarity applies?

    You have to be very careful to get the polarity of the currents to match the polarities of the voltage drops at each resistor in each equation. You have done that incorrectly in a few places.

    You seem to have some of the mechanics of mesh analysis in your grasp, but have missed some of the key concepts. The basic idea behind mesh analysis is that it is a highly systematic application of KVL that is specifically designed to make it so that the mesh equations can be written down by inspection will little probability of error. To make this happen, it is critical that all of the mesh currents be in the same direction -- either all clockwise or all counterclockwise. You have a mixture. If they are all the same direction (let's pick counterclockwise), then in any given component the actual current is the difference between the mesh current in that mesh and the mesh current in the adjacent mesh. So, for instance, the equation for Mesh 2 (remember, Loop 1 is now going the opposite direction to your picture) becomes:

    R5(I2-I1) + R4(I2) + R3(I2-I1) = 0

    Which becomes

    -I1(R5) +I2(R3+R4+R5) -I3(R3) = 0

    The pattern is that we can multiply that mesh's current (I2, in this case) by the sum of all the resistance in the mesh and then, for each adjacent mesh, subtract that the product of that mesh's current multiplied by the sum of the resistances shared by those two meshes. That's the left hand side. The right hand side is simply the sum of the explicit voltage gains due to supplies (which, for mesh 2, is zero).
     
  10. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    I've changed the current flow. I'm just wondering when two currents "mesh" how do you know which current to take away from the other in these:

    R5(I2-I1) and R3(I2-I3) and R1(I1-I3)

    I.e why is it I2-I1, I2-I3 I1-I3 and not the other way around I1-I2 or I3-I2 or I3-I1?


    [​IMG]

    Uploaded with ImageShack.us


    My New Calculations would be:

    Loop One:
    V1 + V5 + V2 = 0
    R1(I1-I3) + R5(I2-I1) + R2(I1) = 0

    2(I1-I3) + 10(I2-I1) + 4(I1) = 0

    2(I1) – 2(I3) +10(I2) -10(I1) +4(I1) = 0

    10(I2) -2(L3) -4(I1) = 0

    Loop two:
    V5 + V3 + V4 = 0
    R5(I2-I1) + R3(I2-I3) + R4(I2) = 0

    10(I2-I1) + 6(I2-I3) + 8(I2) = 0

    10(I2) – 10(I1) + 6(I2) – 6(I3) + 8(I2) = 0

    24(I2) -10(I1) – 6(I3) = 0

    Loop Three:
    Vtotal + V3 + V1 = 0

    Vtotal+ R3(I2-I3) + R1(I1-I3) = 0

    100V + 6(I2-I3) + 2(I1-I3) = 0

    100v + 6(I2) – 6(I3) + 2(I1) -2(I3) = 0

    100V + 6(I2) +2(I1) – 8(I3) = 0


    Though I'm not sure about Loop three because its going from the negative of the battery to the negative of the resistor. If I put the figures in would that work? :/
     
    Last edited: Dec 16, 2012
  11. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    Ergh, I"ve tried to edit the above but they won't let me.

    I realise my above calculations were wonky.

    I'm assuming that when calculating meshing Loops, you deduct the current of the loop you aren't calculating from the loop you are. E.g for Loop one it would be R1(I1-I3)


    I've ammended the calculations accordingly but i still think i'm a bit wonky on loop 3 with the 100v

    Loop One:
    V1 + V5 + V2 = 0
    R1(I1-I3) + R5(I1-I2) + R2(I1) = 0

    2(I1-I3) + 10(I1-I2) + 4(I1) = 0

    2(I1) – 2(I3) + 10(I1) – 10(I2) + 4(I1)

    16(I1) – 2(I3) – 10(I2)


    Loop two:
    V5 + V3 + V4 = 0
    R5(I2-I1) + R3(I2-I3) + R4(I2) = 0

    10(I2-I1) + 6(I2-I3) + 8(I2) = 0

    10(I2) – 10(I1) + 6(I2) – 6(I3) + 8(I2) = 0

    24(I2) -10(I1) – 6(I3) = 0

    Loop Three:
    Vtotal + V3 + V1 = 0
    -100V + 6(I3-I2) + 2(I3-I1) = 0

    -100v + 6(I3) – 6(I2) + 2(I3) -2(I1) = 0

    8(I3) -2(I1) – 6(I2) -100V = 0

    Add 100v to each side =

    Loop Three:

    8(I3) -2(I1) – 6(I2) = 100V
     
  12. Shoutymatt

    Thread Starter New Member

    Dec 14, 2012
    9
    0
    Many thanks to all those who helped. I managed to work it all out in the end :)
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    Glad to hear it and sorry I didn't get back to you. It dropped of my radar screen.
     
Loading...