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Finding the currents of four resistors in a circuit.

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data


    [tex]\epsilon[/tex]batt = 12V
    Find the following currents.
    The current I1 through the resistor of resistance R1 = 15.0 ohms.
    The current I2 through the resistor of resistance R2 = 45.0 ohms.
    The current I3 through the resistor of resistance R3 = 25.0 ohms.
    The current I4 through the resistor of resistance R4 = 25.0 ohms.

    2. Relevant equations
    I = V/R

    3. The attempt at a solution
    First I reduced the circuit to find the equivalent resistance of all resistors R = 56.5 ohms. From this I was able to determine that the current from the equivalent resistor is 12/56.5 = .212 A.

    I am really confused as to how to solve for the currents through each element since I don't know the ∆V of each and some are in series while others are in parallel.

    Any help is appreciated!
    Last edited: Mar 31, 2008
  2. jcsd
  3. Mar 31, 2008 #2
    You don't need it. Take a look at Kirchhoff's circuit laws. Maybe it will help.
  4. Mar 31, 2008 #3
    Do you know Kirchhoff's laws?

    If we consider I2 flowing upwards while I3 and I4 flow downwards, we have the following system

    I3 = I4 (R3 and R4 are in series)

    I2 = I1 + I3 (The current I2 distributes between the two branches)

    and then the voltage drop between the upper node and the lower one is the same along the three branches

    e - I2 R2 = I1 R1 = I3(R3+R4)

    you have a linear system of four equations that gives you the four currents.
  5. Mar 31, 2008 #4
    I completely follow you, until I get here:

    Why does the voltage drop on the right have to the the same as the voltage drop on the right?

    Is it because the wires are at a junction at the top (and therefore must be at the same voltage) and there is a junction at the bottom (so these must also be at the same voltage)?

    That would make sense since the ∆V would have to be the same...
  6. Mar 31, 2008 #5
    Also, what is the best way to solve the group of equations? I tried using a matrix, but I did not get the correct answer. It has been a looong time since I have attempted one, so it is possible I made an error.
  7. Mar 31, 2008 #6
    First off, give a direction to all of your currents in that circuit, and use Kirchhoff's Loop and Junction Laws:

    At a junction: [tex]\Sigma I_{in}=\Sigma I_{out}[/tex]

    Around any closed loop: [tex]\Sigma \Delta V = 0[/tex]

    You'll have three equations, and three unknowns. It may be easier to use a matrix, but if you don't know how to use one, do it algebraically.
  8. Mar 31, 2008 #7
    I am still stuck on this problem. I came up with the equations:

    I3 = I4

    I2 = I1 + I3

    (e-I2R2) - (I1R1) = 0

    (e-I2R2) - (I3R3 + I4R4) = 0

    I'm not sure why you say I have three equations and three unknowns.... isn't it four? Or maybe you eliminated one since I3 = I4. Honestly, my algebra skills must be terrible, because I cannot seem to solve this.

    I have spent way too much time on this, and I am pretty close to conceding... giving it a few more go's. Someone please let me know if you find something wrong with my equations or method.

  9. Mar 31, 2008 #8
    That's right, it's only three because I3=I4, so you don't need to worry about an I4 term. Replace all I4's with I3's to make your problem easier.

    Other than that, everything else looks good. Start by solving for one variable in one equation, and sticking it into another.
  10. Apr 1, 2008 #9
    Because it's a voltage drop. The voltage difference depends only on the initial and end points and not on the path followed to go from one point to the other.

    If we call Va to voltage of the top node and Vb the voltage of the bottom one, we have for the left branch

    Va - Vb = I1 R1

    for the right one

    Va - Vb = I3 R3 + I4 R4 = I3(R3 + R4)

    and for the central one, taking into account the dc source

    Va - Vb = e - I2 R2
  11. Apr 1, 2008 #10
    It is also possible to simplify by replacing R3 and R4 by R5=50 ohm, to make the equations look nicer.
  12. Apr 1, 2008 #11
    Thanks for your help everyone... I was able to solve this using my three equations. It was just a matter of plugging them into each other and avoiding stupid mistakes (which I am prone to).

    I1 = .163
    I2 = .212
    I3 = I4 = .0490
  13. Apr 1, 2008 #12
    Here's my method to do the same problem:

    Combine all resistors into one equivalent resistor across the battery.
    When that can be done, I wouldn't write a set of simultaneous equations.

    Let R34 = R3+R4 = 25+25 = 50 ohms
    Let R123 = R1 parallel R34 = (15)(50)/(15+50) = 11.54 ohms
    Let R equivalent = R134+R2 = 11.43+45 = 56.54 ohms

    battery current = E / R equivalent :
    IE = 12 V / 56.54 ohm = 0.212A
    (That's what you called I2)

    voltage across R2 given by ohm's law:
    VR2 = IE R2 = (0.212A)(45 ohm) = 9.55 V

    voltage across R1 equals battery minus voltage across R2 :
    VR1 = E - VR2 = 12 V - 9.55 V = 2.44 V

    current through R1 given by ohm's law:
    IR1 = VR1 / R1 = 2.44 V / 15 ohm = 0.163 A
    (That's what you called I1)

    Current through R3 and R4 = battery current minus current through R1 :
    IE - IR1 = 0.212 A - 0.163 A = 0.049 A
    (That's what you called I3)
    Last edited: Apr 1, 2008
  14. Apr 2, 2008 #13
    You can easily solve this with circuit reduction, if you think of it like this:


    It's pretty straight forward, 3+4 series, 1+34 parallel, 134+2 series

    (Sorry the image is crappy but you still get the general idea.)
  15. Apr 2, 2008 #14
    I have heard it called a principle of circuit topology -- you can visualize a circuit as being bent, twisted, flipped over, mirror imaged, etc., if it helps a person to see it better in their imagination -- as long as you don't add or remove any connections. After doing about a dozen problems a person will no longer find it necessary.
  16. Apr 2, 2008 #15
    I've never heard it called anything, but yeah after a while you just start to automatically re-arrange the circuit in your mind so that it makes more sense.
  17. Apr 6, 2008 #16
    There is another, systematic way of analyzing this kind of problems:

    We consider each loop separately and label a current that flows around the loop (let's say, counterclockwise). In this case we have two loops: let I1 and I2 be the currents. We construct a vector with these currents


    Now, we build a resistance matrix n x n, (being n the number of loops, in this case 2 x 2). The diagonal elements R_ii are the sum of resistances along each loop. The non-diagonal term R_ij is the resistance common to loops i and j, positive if both currents flow in the same direction, negative if they go in opposite directions. In our case

    [tex]\mathsf{R}=\left(\begin{array}{cc}R_1+R_2 & -R_2\\-R_2 & R_2 + R_3 + R_4\end{array}\right)[/tex]​

    Finally, we construct another vector with the emfs along each loop. Positive if we enter at them through the cathode, negative if we enter through the anode. In our case


    We obtain the currents solving the matricial equation

    [tex]\mathsf{R}\cdot\mathbf{I}= \mathbf{V}[/tex]

    [tex]\left(\begin{array}{cc}R_1+R_2 & -R_2\\-R_2 & R_2 + R_3 + R_4\end{array}\right)\cdot \left(\begin{array}{c}I_1\\I_2\end{array}\right)=\left(\begin{array}{c}\mathcal{E}\\-\mathcal{E}\end{array}\right)[/tex]​

    The solution of this equation (inverting R) gives I1 and I2. To obtain the current flowing through each resistance you add the currents of the adjacent loops (with their sign), i.e.

    [tex]I(R_1)=I_1\qquad I(R_2)=I_1-I_2\qquad I(R_3)=I(R_4)=-I_2[/tex]​
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