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Tips on solving resistor ladders

  1. Apr 2, 2009 #1
    I'm trying to find the equivalent resistance for this circuit
    IIRC, it's a resistor ladder.

    I think I've seen this solved by working backwards, but I can't for the life of me remember how to do it now. I've tried rewriting the network in other ways to expose any series/parallel/voltage division sections, but I can't seem to figure it out.

    Any tips on how to solve these circuits would be appreciated.
  2. jcsd
  3. Apr 2, 2009 #2
    Welcome to PF, stimpyholder!

    You could describe solving this one backwards. You just need work your way through the circuit resistor pair at a time. Star with the connection of the two [tex]20k\Omega[/tex] resistors.
  4. Apr 2, 2009 #3
    Sorry, having one of those days. How do I determine if the pair are series or in parallel?

    For some reason, I have a brain fart whenever I hit resistor networks like these.
  5. Apr 2, 2009 #4
    Are those two terminals connected to anything else? If so, you cannot add the two 20 ohm resistors. If not, you can ignore the two terminals and redraw the circuit without them. Are you trying to find the Thevenin equivalent with respect to the two open terminals on the right?

    Resistors are in series if they share only one node. They are in parallel if they share two nodes. There are four nodes in your original circuit. If you cannot see that, reply so we can help you understand nodes (this can be a hard topic for some students).

    For thevenin resistance: start by shorting out the voltage source. After that, do you see the parallel combination?
    if those nodes are not connected to anything else and you don't want the thevenin equivalent: start by removing those wires that connect to nothing. They're unimportant and there to throw you off. After that, can you see the possible series combination?
    Last edited: Apr 2, 2009
  6. Apr 2, 2009 #5
    The question does go on to add a [tex]1k\Omega[/tex] load, but asks to first find the voltage across the 2 teminals.

    I think suggesting counting nodes has helped. As I understand it;

    The 2 [tex]20k\Omega[/tex] resistors share a single node so they are in series, these are then connected in parallel with the next [tex]10k\Omega[/tex] resistor, which is then in turn connected in parallel with the last [tex]10k\Omega[/tex] resistor, giving,

    [tex]R_{eq} = 10k\Omega || (10k\Omega || 20k\Omega + 20k\Omega)[/tex]
  7. Apr 2, 2009 #6
    An ideal voltage source has an internal resistance which is zero. Therefore you could replace the voltage source by its internal resistance. This gives a network with all resistors. You will notice that the two 10 kOhm resistors are in parallel, giving an equivalent of 5 kOhms. This equivalent is in series with the 20 kOhm resistor which gives 25 kOhms, which in turn is in parallel with 1the 20 kOhm resistor across terminala a-b. The parallel combination of 25 kOhms and 20 kOhms is (100/9) kOhms.
  8. Apr 2, 2009 #7
    Yeah, those terminals on the right prevent you from adding the two 20 ohm resistors. This is a thevenin resistance problem, so you should start, as stated above, by shorting the voltage out. Start combining from the left until you get one resistor in parallel with that last 20 ohm resistor.
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