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Kirchoff's Rules problem.

  1. Mar 11, 2005 #1
    Using Kirchhoff's rules, calculate the current in R1 with the directions indicated in the figure below. Assume that R1 = 1.00 kΩ, R2 = 3.00 kΩ, R3 = 5.00 kΩ, E1 = 60.0 V, E2 = 50.0 V and E3 = 65.0 V (Diargram in attachment). This problem has kind of stumped me. I tried setting up three equations:

    V1 = I1 * R1
    V2 = I2 * R2
    V3 = I3 * R3
    But Im getting stuck after this on how to solve for my unknowns. Anybody have any suggestions. Thanks.
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2005 #2
    Using Kirchoff's Law write two equations for the loops (say bcfa and cdef). Your third equation is from point rule. I2 = I1 + I3.

    Three equations, three unknowns.
     
  4. Mar 11, 2005 #3
    What you have there is two loops. Using Kirchoff's laws you know the sum of the voltages in both loops will be zero. To start you can look at each loop individually, assume there is a current flowing clockwise in both loops. Call the current in loop one I1 and the current in loop 2 I2.

    Now go around each loop and find the sum of all the voltages you come across. I'll show you what i mean by doing the loop on the left...

    E1 - E2 - R2(I1-I2) - R1*I1 = 0

    Use a negative when the voltage drops (due to resistance or an opposing voltage) and positive when you come across a voltage reinforcing the current.

    E1 is positive because it is reinforcing a current in the clockwise direction, while E2 is negative because it is opposing clockwise currents in loop 1. The voltage drop across the resistor R2 is given by V=RI but since it is touching the other loop there are two currents influencing its voltage. The I2 current is negative since we assumed all currents travel in a clockwise direction, and it is opposing the current flow of I1.

    After you do the same analysis on the second loop you will have two equations and two unknowns, then you can solve for I1 and I2. If it turns out get get a negative current you know the current is flowing in the opposite direction, but the magnitude is correct.

    Trying to explain how to do this in a post is much harder than doing it once you figure it out. After just a few of these problems you'll be an expert.
     
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