Complex resistors

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Gary Shumaker

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Hi guys,
I had so much fun from the last challenge; let’s see what you can do with this. I used to know the exact formula some 30 years ago but I am having a little trouble remembering.

Since I can't seem to copy and paste from power point. I will describe the circuit:

100VDC source. The first parallel leg has a 20 ohm resistor on top in series with a 10 ohm resistor. Parallel leg 2 has a 10 ohm resistor on top and a series20 ohm resistor on the bottom. Running between the two parallel legs and between the two series resistors on each parallel leg is a 5 ohm resistor.

I’m looking for the total resistance, voltage drops and current flow. Note that the current flowing in the 5 ohm resistor is from right to left.
 
That's not "complex," since there are no capacitive or inductive loads. :wink: :D The description is clear enough. But of course you realize that as soon as you connect the 5 ohm resistor, none of the resistive elements is in series or in parallel with any other. I'd go about it with "loop currents," if I had the energy. But I would need more coffee first.
 
I'll have to read it more carefully and draw it out, but here are the formulas we need.

Resisters in series add.... Req=R1 + R2 + ... Rn

Resisters in Parrallel Req = 1/[(1/R1)+(1/R2) + ...(1/Rn)]

Find one Req and use Ohms Law (V=IR or E=IR) Lets compare answers...
 
Well, it is a little harder than just using the series and parallel resistor equations because of the 5 ohm resistor. We need to use some simultaneous equations based on Kirchoff's law for current and some voltage drop equations.

Hey Gary Schumaker, what you need to do is hypothesize an incorrect formula and claim that it works. Now THAT will get this thread rolling!:grin:
 
If I understood you right, this is another wheatstone bridge. Again, we could take one side which is a delta, and covert it to a wye. However, the formulas are kind of confusing on what resistor you use where.

However, I think I have found an even easier way to solve for the voltage and current across the 5 ohm resistor.

Do a search for Thevenin. You covert the other 4 resistors to a thevein resistance (I get 13.33 ohms), and the voltage to a thevein voltage (I get 33.3 volts). You then are left with a simple series circuit - 33.3 volts in series with 13.33 ohms and 5 ohms.

From that, I get that there is 12.5 volts across the resistor (+ on the left) and 2.5 amps through the resistor.

Edit: But I still cant add. :) Ignore my numbers above. The current should be 33.3 /(13.33+5) = 1.81 amps. That gives 9.08 volts across the resistor.

Steve
 
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The way I drew it from the description does not require any convertions or complex (I should say complicated:) ) equations. A 10 and 20 in parrellel equal a 6.6667 ohm resister, which add to the 5 for a 11.66667 in parrellel with a 20 yeilds a 7.37 in series with a 10. Final Req is 17.3684 ohms if I drew it correctly. (Big IF...)

Total current then 5.7576 amps. Work backwards to calc currents in each resister calculating voltage add nodes etc.
 
Let me give it a shot

Let me give it a shot

Again, it's been 30 years. But if memory serves, let's take the circuit and form into something that looks like a bridge rectifier. Replace the diodes with resistors and add a horizontal resistor in the bridge circuit.

Now let?s label the resistors from left to right as R1 20Ω, R2 10Ω, R3 5Ω, R4 10Ω and R5 20Ω.

My first step would be to calculate the total currents through R2 as a single series resistor and R3 in series with R4, both paralleled with R5.

Step 2 would be to perform a similar process but drop out R5.

Step 3 would be to average the results from step 1 and 2.

Not really sure.
Gary
 
Gary,

Is this the circuit you described?

Gary_Schumaker_Circuit_web.jpg
 
Ah,

From your last post I see my assumed convention for R3, R4 and R5 are skewed from yours.
 
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Thats the circuit I was working with. (Not to be picky, but if memory serves me correct, I think you battery is backwards and the
+ sign should go on the long bar.)

I get 1.81 amps flowing left to right through the 5 ohm resistor. (+ Conventional current flow.)

Steve
 
spsnyder said:
A 10 and 20 in parallel equal a 6.6667 ohm resister . . .
Click to expand...
The 10 and the 20 are neither in parallel nor in series. No two resistors are in series, and no two are in parallel. Now that a sketch has been posted, perhaps that will be more obvious.
 
kspifldorf said:
Thought the 5ohm was right to left.
Click to expand...
Doesn't matter. If the current proves to actually flow right to left, then the value of I5, as shown in the sketch,will be negative.
 
Charlie b - That's not the circuit I had sketched so disregard my answer as I was working a different problem.

The convention for a battery is the long line is positive - FYI
 
Gary, I'll try to post the pdf showing how to use delta - wye transform to solve this problem. After that is done the problem becomes fairly simple to solve. Hope this works.

If it works you see that I get Is (current through the battery) is 7.2727 Amps with a Req = 13.75 Ohms. Is that what you get?
 
OK, using the sketch I made and attached last post, here's what I got.

I1 = 2.7273 A
I2 = 4.5455 A
I3 = 4.5454 A
I4 = 2.7273 A
I5 = -1.8182 A

Va = 100V
Vb = 45.455V
Vc = 54.546V
Vd = 0 V

regards.
 
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