Circuit Problem

Discussion in 'Homework Help' started by Digit0001, Mar 5, 2011.

  1. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    Hi

    The following circuit i have tried to find the current and and voltage but i get wrong answer. Can someone suggest how would they approach this question. Also where have a gone wrong?

    [​IMG]

    This is what i have done:
    I have choose to use mesh analysis to determine the currents in the circuit.

    KVL Mesh 1

    470i_{1} + 4700(i_{1} - i_{2}) + 2200i_{1} = 10

    KVL Mesh 2

    10000i_{2} + 10000(i_{2} - i_{3}) + 4700(i_{2} - i_{1}) = 0

    KVL Mesh 3

    4700(i_{3} - i_{4}) + 10000(i_{3} - i_{2}) = 0

    KVL Mesh 4

    2200i_{4} + 4700(i_{4} - i_{3}) = 0


    simplied as follows

    7370i_{1} - 4700i_{2} = 10
    23700i_{2} - 10000i_{3} - 4700i_{1} = 0
    14700i_{3} - 4700i_{4} - 10000i_{2} = 0
    6900i_{4} - 4700i_{3} = 0

    KCL i_{1} = i_{3} + i_{4}

    replace i_{1} in the KVL
    7370(i_{3} + i_{4}) - 4700i_{2} = 10

    24700i_{2} - 10000i_{3} - 4700(i_{3} + i_{4}) = 0

    14700i_{3} - 4700i_{4} - 10000i_{2} = 0

    now i use matrix to find the values, however i realized that my answer was incorrect when i got 0 for determinate i_{2}.

    P.S
     
    Last edited: Mar 5, 2011
  2. Vahe

    Member

    Mar 3, 2011
    75
    9
    I think you might be mixing the mesh currents with the currents that are marked as i_1, i2, i3 and i4. Maybe you should write the mesh currents as I_{m1}, I_{m2}, I_{m3} and I_{m4}. Based on this, note that i_1=i_2=I_{m1}, i_3=I_{m1}-I_{m2} and i_4=I_{m2}.

    Your mesh equations seem correct to me. I suggest that you write the currents in those equations as I_{m1}, I_{m2}, etc. Also in your simplified expressions under "simplified as follows" in the 2nd equation, your coefficient for i_2 should be 24700 and not 23700 - it might be a typo. When I solve it I get the following mesh currents,

    <br />
I_{m1}= 1.67 \text{mA}\\<br />
I_{m2}= 490 \mu \text{A}\\<br />
I_{m3}= 426 \mu \text{A}\\<br />
I_{m4}= 290 \mu \text{A}<br />

    Try it yourself to see if you get these values. I did not double-check these, so use these with caution.

    Cheers,
    Vahe
     
  3. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    ok if i replace the i's with I's then when i how would i simplify my equation?
     
  4. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    for the equation 1), 2) and 3) originally i wanted to use KCL i1 = i2 + i3 to simplify in order to calculate using matrix form, however now that i have changed the symbol how would i solve the equation.
     
  5. mjhilger

    Member

    Feb 28, 2011
    119
    16
    You are confusing your mesh currents with the labeled currents. The labeled current i1 = mesh 1 current, but it ends there. Your KCL: KCL [​IMG] = [​IMG] + [​IMG]
    is wrong because of this. From your simplified equations, you have 4 equations and 4 unknowns. Use the matrix to solve from there for your mesh currents. You should be able to determine the identified currents from there. By inspection you should note some of these are equal and other easy solutions for all desired labeled currents once you have mesh current 1 & 2.
    BTW I'm sure you are to analyze the circuit using node and loop techniques. However, so you might check your mesh currents you should note that you can brute force the value of R5,R6&R7 in parallel, then that in series with R4, and further that value in parallel with R3. Then you just have R1 + R2 + (calculated vaule of the collapsed right string). You should be able to eaisly calculate the currents in this brute force manner to check your work of the mesh currents.
     
    Last edited: Mar 6, 2011
  6. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    ok i have tried using just nodes and loops but i get stuck becuase i cannot solve the value of current with the equations found:

    These are the equation i got from the circuit:

    When using KVL:

    V1 + V3 + V2 = 10
    V4 + V5 -V3 = 0
    V6 - V5 = 0
    V7 - V6 = 0

    When using KCL:

    i1 = i3 + i4
    i2 = i3 + i5
    i4 = i5 + i6 +i7

    the two equations i end up with are:

    1) 470(i3 + i4) + 4700i3 + 2200(i3 + i5) =10
    7370i3 + 470i4 + 2200i5 = 10

    2)-4700i3 + 10000i4 + 10000i5
     
  7. mjhilger

    Member

    Feb 28, 2011
    119
    16
    Your mesh equations are the right ones (save the correction Vahe pointed out).
    7370 i1 - 4700 i2 = 10
    24700 i2 - 10000 i3 - 4700 i1 = 0
    14700 i3 - 4700 i4 - 10000 i2 = 0
    6900 i4 - 4700 i3 = 0

    These are your 4 equations and 4 unknowns. Solve this by whatever means you have at your disposal. A matrix solving calculator should do this 4x4 in a few seconds. Or you can use your algebra to solve for i1,i2,i3 & i4. These values are your mesh loop currents - not the ones you are looking for. But, once you have these loop currents you can determine the values you are seeking. I1 and I2 as labeled on the schematic are = to your i1 mesh current. I3 (schematic labeled) is mesh current i1 - i2. And I4 should be easy to relate also by inspection of your schematic.
     
  8. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    well i did get the same answers as Vahn did when doing the Mesh method i just wanted to see if i done it using nodes and loops, if it will give me the same results as the mesh method
     
  9. mjhilger

    Member

    Feb 28, 2011
    119
    16
    KCL, KVL, Norton, & Thevenin are tools for you to use. Like saw, drill, screwdriver, file, etc are in a wood working workshop. You are asked to preform the homework problems to become familiar with those tools. You will soon see, as perhaps you tried, that some tools work way better for a particular task than others. For instance you can cut a board with a drill by drilling several holes until the piece is cut, but it takes a whole lot of work, leaves lots of room for error and usually the result is not pretty. KVL KCL etc. are your tools, so for this situation loop current equations provide a fairly direct approach, the others - not so much.
     
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