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Find current using Mesh Analysis

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Use the mesh-current method to find the branch currents ia, ib, and ic in the circuit in figure.(Figure 1) And find currents i_a, i_b, and i_c if the polarity of the 77V source is reversed. The values of v1 and v2 are Figure_P04.33.jpg 77V and 16V , respectively.

    2. Relevant equations
    Mesh Analysis
    Mesh 1: R's in mesh 1 x i1 - R's in common between Mesh 1 and 2 x i2 = Voltage source in Mesh 1
    Mesh 2: -R's in common between Mesh 1 and 2 x i1 + R's in Mesh 2 x i2 = Voltage source in Mesh 2


    3. The attempt at a solution

    I know this problem can be solved by simply using kirchoffs current and voltage laws, but the problem specifically asked to use mesh current. I designated the left mesh as mesh 1 and the right as mesh 2. Mesh 1 goes clockwise while mesh 2 goes counter-clockwise.

    Equation from mesh 1:
    (4+10+1)(i1) - (10)(i2) = 77
    Equation from mesh 2:
    -(10)(i1) + (2+10+3)(i2)=16

    I got the following:
    i1 = 10.52
    i2 = 8.08

    looking at the direction of i1 and i2, I can see that i1 = i_a and that i2= i_c.
    I can also see that i_a + i_c = i_b

    Therefore,
    i_a = 10.52
    i_c = 8.08
    i_b = 18.6

    (all of these are incorrect, the correct answer is i_a = 7.96, i_b = 3.72, and i_c = -4.24)

    I am just trying to understand how the problem using mesh analysis. I assume that it would be the same procedure when doing it for the second time with the polarities of the 77V source reversed.
     
  2. jcsd
  3. Mar 9, 2015 #2
    There's a problem with the signs in both your equations. If the direction of the mesh currents is the same through the middle resistor, then the sign of the voltage drops should be the same as you traverse the loop in KVL.

    As a sidenote: It's usually better to stick to one direction for the mesh currents as a convention. It's a matter of preference, of course, but it'll probably help you avoid errors later.
     
  4. Mar 9, 2015 #3
    So what you're saying is that in both equations, i2 should have the same sign? so it should be:
    equation from mesh 1:
    (15)(i1) - (10)(i2) = 77
    equation from mesh 2:
    (-10)(i1) -(15)(i2) = 16


    also, the reason I chose to pick two different directions for the mesh currents is because of the polarities of each of the voltage sources. Had I chosen both mesh currents to go clockwise, would the second equation be (-10)(i1)+(2+10+3) = -16? In other words, does the direction of the current and the polarity of the voltage source matter? Having mesh 2 go clockwise would mean that the current i2 is going from the positive to negative in through the voltage source so I was unsure if it would equal 16 or -16.
     
  5. Mar 9, 2015 #4
    I meant that, for mesh 1, if you begin at v1 and traverse the mesh clockwise, you have:
    -v1 + 4*i1 + 10*(i1 + i2) + 1*i1 = 0

    This is because i1 and i2 both have a direction downwards through the middle resistor. If the reference directions for i1 and i2 were instead both clockwise, you'd have:
    -v1 + 4*i1 + 10*(i1 - i2) + 1*i1 = 0

    Nothing in the schematic has any impact on what reference directions you should choose for your mesh currents, and whatever your choice may be, it has no impact on anything else in the schematic.

    All you have to worry about is applying Ohm's law and KVL correctly.
     
  6. Mar 9, 2015 #5
    I understand now. Making both meshes go clockwise gives me the following equations

    (15)(i1)-(10)(i2)= 77
    (-10)(i1)+(15)i2) = -16

    solving for this gives me i1 = 7.96 and i2 = 4.24
    i1 = i_a since the direction of i1 is the same as the direction if i_a. i2 = -i_c since the direction of i2 is the opposite direction of i_c. And i_b is equal to i_a + i_c by KCL. Doing all of this gives me the correct answer

    Thank you!
     
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