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January StudentZone Quiz Solution


To find the voltage across the current source, we can find the equivalent resistance of the entire circuit.  Taking this a step at a time...

First, we can combine the 3k and 7k series resistors (3k + 7k = 10k), and combine the resulting series equvalent resistor value with the 10k resistor in the original circuit (10k || 10k = 5k).  You have now combined three resistors (3k, 7k, 10k) into a single equivalent resistance (5k):

At this point, there are no more parallel/series reductions we can make.  Now, we can make a difficult circuit more workable by applying a Y-delta transform to the three resistors at the edge of the circuit:

In this example, we are transforming R1, R2, and R3 in the "Y" circuit to an equivalent "delta" circuit of Ra, Rb, and Rc values.  Since R1, R2, and R3 are all equal, Ra, Rb, and Rc will also all be equal, and we only have to calculate one of the values:

Ra = [5k*5k + 5k*5k + 5k*5k] / 5k = 75k/5k = 15k

Now, we have two pairs of resistors that we can combine in parallel: 5k || 15k = 3.75k.  Redrawing the circuit, again, we've run out of series/parallel resistors to simplify.  So, again, let's apply a Y-delta transform.  This time,  there's a bit more math, b/c the resistor values aren't equal:

So, using the the node numbers above, Ra sits between nodes 2 and 3, Rb site between nodes 1 and 3, and Rc sits between nodes 1 and 2.  Ra = 3.75k, Rb = 15k, and Rc = 3.75k.

Using equations in Y-Δ transform - Wikipedia:

R1 = (Rb*Rc) / (Ra + Rb + Rc) = (15k * 3.75k) / (3.75k + 15k + 3.75k) = 56.25k / 22.5k = 2.5k

R2 = (Ra*Rc) / (Ra + Rb + Rc) = (3.75k * 3.75k) / (3.75k + 15k + 3.75k) = 14.0625k / 22.5k = 0.625k

R3 = (Ra*Rb) / (Ra + Rb + Rc) = (3.75k * 15k) / (3.75k + 15k + 3.75k) = 56.25k / 22.5k = 2.5k

Once we've transformed the "delta" circuit to a "Y" circuit, we now have several series/parallel simplifications that can be made.  First, combine 5k and .625k into an equivalent 5.625k resistor value.  Then, similarly, 5k and 2.5k into 7.25k value.  These two series equivalent values are now in parallel, and can be combined: 5.625k || 7.25k = (5.625k * 7.5k) / (5.625k + 7.25k) = 42.1875k / 13.125k = 3.2143k.

At last, all we have left is series resistors: 1k + 2.5k + 3.2143k = 6.7143k.

So, to find the voltage across the current source, we use the current value and the resistor value to find the voltage across the equivalent resistor: 1mA * 6.7143k = 6.7143 V.

  • I converted 3 delta networks to y networks.  On the surface this seems a more complicated solution, however because all resistors are the same value, only one calculation is needed for the 3 delta conversions to y.    (5x5/(5+5+5)).

  • engr57,

    Now that you mention it, you're correct - there is a much simpler approach to this problem.  Except, I wasn't able to find a way to solve the problem with three Delta-Y transformations - I only used two.  But you're absolutely correct - it looks like I took a more difficult route to solve this problem initially! 


  • Engr57, 

    You're assuming that ALL of the resistors are equal, but they are not. There's a 3k, 10k, 7k, and 1k resistor thrown in there for fun.  

    That said, while you *could* use three delta-wye transforms, the third one is a little more difficult b/c the values aren't equal.  And that third "delta" can also be easily reduced using series/parallel techniques.

    So, if I had to do it again, I would use your simpler approach of identifying the two simple delta-wye transforms, and the remaining series/parallel combinations.

    Thanks for the lesson, Engr57.  When I have time, maybe I'll update the solution with your approach.  


  • I just love this dialog.  The problem started with 4 delta

    networks.  By combining the 3k and 7k with the 10k, the problem

    reduces to 3 delta networks all of which have equal 5k legs.  The 1k

    resistor is sitting outside of the 3 delta networks and can be dealt with last.

    So you see that the 3 delta networks can be transformed to 3 wye

    networks with one calculation.


  • Haha, I'm enjoying this discussion as well.

    We might be taking the same approach, but let me show you what I mean when I say I'm only using two delta-y transformations.

    As you state above, first, combine the 3k, 7k, and 10k into a single 5k resistor.

    Then, as you can see in the second circuit drawing below, I've identified *two* delta networks to transform.

    Once you've done that, the rest is just series/parallel reductions.  

  • You are absolutely correct. ( See Attached).

    Lets go onto a new problem.

    Some people like crossword puzzles, but solving circuit problems is

    more interesting.