Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kirchoff’s Rules

  1. Feb 25, 2008 #1

    A triangular array of resistors is shown in the figure. What current will this array draw from a 35.0 V battery having negligible internal resistance if we connect it across (a) ab, (b) bc, (c) ac?

    Kinda confused... just anything to start me off on the right track would be very helpful.
  2. jcsd
  3. Feb 25, 2008 #2
    Have a good look at the circuit the triangular arrangement is probably throwing you off.
    If you put a voltage source between b (positive terminal) and c (negative terminal) the c node becomes your circuit ground. So you can stretch this point and make the circuit look more rectangular. This would result in the voltage source Vbc in parallel with the 10 ohm resistor which is in series with the 15 Ohm resistor which is then in series witht he 20 Ohm resistor. Redraw the circuit that way and you should have no problem figuring out the rest of it.

    hope this helps.
  4. Feb 26, 2008 #3
    Thanks a ton
  5. Mar 4, 2011 #4
    This is actually a homework problem that I am also stuck on. Following the above method equals the wrong answer. I can honestly say it doesn't make sense either. That doesn't make a rectangle.
  6. Mar 5, 2011 #5
    There are two resistors in series, which are parallel with the 3rd resistor and voltage source in each case. You should easily be able to find the equivalent resistance of the circuit and thus the current knowing this.

    For a), I get 3.5A as the answer.
  7. Mar 5, 2011 #6
    Still doesn't help. Trying to teach myself via text this due to lack of interaction from professor. Ended up buying the solution online. Would have never seen the setup, still don't follow how it becomes that.
  8. Mar 5, 2011 #7
    Is 3.5A the correct answer?

    For a), by adding a voltage source across the terminals AB, you have a voltage source in parallel with the resistor.

    The other two resistors are in series with another, i.e. all current entering BC must exit through CA (and vice versa).

    Because the two resistors are in series, you can add up their resistances to get the equivalent resistance (that is, you can replace the 10 and 20 ohm resistors by a single resistor of 30 ohms).

    You're then left with the voltage source in parallel with the 15 ohm resistor, in parallel with the 30 ohm resistor. To find the equivalent resistance of two parallels resistors, you use:
    [tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]
    which simplifies to:
    [tex]R_{eq} = \frac{R_{1}*R_{2}}{R_{1}+R_{2}}[/tex]

    So, Req = 15*30 / (30+15) = 10 ohms.

    Then, to find the current, you simply apply Ohm's Law: i = V/R = 35/10 = 3.5A.

    Does that clear it up for you? If not, which of the above steps don't you follow? It might be easier for you to visualise it as a rectangle (i.e. the AB resistor on the left, BC resistor on the top, AC resistor on the bottom, with nothing on the right line).
  9. Mar 6, 2011 #8


    User Avatar

    Staff: Mentor

    Here's a picture showing rearrangements of the circuit for case (a). You should be able to convince yourself that the underlying topology of each version is the same -- moving components and wires around the page doesn't affect the circuit so long as all connections remain the same.

    Attached Files:

  10. Jul 24, 2012 #9
    Thank you Gneil, your picture helped a ton.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook