Complex resistors

[rattus]

Two equations, two unknowns:

Two equations, two unknowns:

The classic way to do this is to write two loop equations,

Define two currents, Ia and Ib, between V1 and ground. Ia and Ib pass through the 5 Ohm resistor in opposite directions. Yes, that is correct!

Then

V1 = 45Ia - 5Ib
V1 = -5Ia + 25Ib

Solve these equations for Ia and Ib.

Or, create Thevenin equivalents connected to each side of the 5 Ohm resistor. Then solve for the unknown voltages. That is, you would have a 66.7V source and a 33.3V source each with a series resistance of 6.67 Ohms.

 

[crossman]

Whew, I finally got this one done.

I did it by creating equations from Kirchoff's voltage loop and current laws, and then solving the simultaneous equations. Luckily, there is some symmetry in this particular problem, which greatly eases the algebra. Since R1 = R4 and R2 = R3, and they are symetrically placed in the circuit, we basically get a system of 3 unknowns for current instead of the 5 unknowns we would get from having each resistor being a different value. Solving simulataneous equations for 5 variables can be pretty tough. With hundreds of manipulations to do, all it takes is one simple mistake like reversing a sign or overlooking a coefficient to bungle the entire matter. Yeah, you can use matrices and determinants but I only vaguely remember that from school 30 years ago.

So here goes:

First, the voltage loop equations (I make an educated guess that current flows from right to left in the R5 resistor)
E1 + E2 = 100v
E3 + E4 = 100v
E1 - E5 + E4 = 100v
E3 + E5 + E2 = 100v

Next, from the current junction law:

I2 = I1 + I5
I3 = I5 + I4

Some identities based on the symmetry of this particular problem:

I1 = I4 ; I3 = I2 ; E1 = E4 ; E3 = E2

Doing some substitution from the identities:

2E1 - E5 = 100v
2E2 + E5 = 100v

Converting these to current formulas using the resistor values and ohm's law for each component:

2(20I1) - (5I5) = 40I1 - 5I5 = 100v
2(10I2) + 5I5 = 20I2 + 5I5 = 100v

Taking the first current law formula, I2 = I1 + I5 and substituting into the second bold equation:

20I1 + 25I5 = 100v

divide both sides by 5 to get an equation which will make it easier to solve the final steps:

4I1 + 5I5 = 20v

Combine this with the first bold equation. Now we are down to two equations with two variables, so we are almost home! What we will do is add the 2 equations together:

4I1 + 5I5 = 20v
40I1 - 5I5 = 100v
add
44I1 = 120v

so I1 = 2.7273 amps

going back through each part, we can solve the rest with Ohm's law and some basic formulas, which I am not going to type.

E1 = 54.5455v
E2 = 45.4545v
E3 = 45.4545v
E4 = 54.5455v
E5 = E1 - E3 = 9.0909v
I1 = 2.7273a
I2 = 4.5455a
I3 = 4.5455a
I4 = 2.7273a

I5 = E5/5 or I3 - I4 so I5 = 1.8182 amps and is moving right to left.

This vindicates that the delta to wye conversion done previously by spsnyder is good.

 

[rattus]

Congrats Crossman,

Now go back and solve the two simultaneous equations in my previous post. It is a lot easier.

The symmetry of this problem shows me that the same currents through the equal resistors. That is, Ia flows through the 10 Ohm resistors, and Ib flows throw the 20 Ohm resistors. That fact allows me to define Ia and Ib as I have.

The Thevenin approach should work for any set of resistors. And a set of node equations could be used as well.

 

[crossman]

I just did it your way, rattus, and I must say it worked right out with no trouble at all. Same answers as spsnyder got with his delta to wye equivalent and that I got with the loop/junction thing. Your way was definitely the easiest for this problem. I didn't try your method earlier because I was focused on doing the loop.junction way, plus I didn't understand where your equations come from. But I just analyzed where your equations come from and I see exactly what you did. Very nice insight into the problem!:smile:

This particular problem allowed some various methods for solving it. But if the arrangement precluded the delta-to-wye conversion and had no symmetry of arrangement or resistor values, then the Kirchoff/simultaneous equation method is the only way to solve this, right? I'm not sure what you mean by Thevinin.

Before I realized the symmetry of the problem, I was going through the whole thing with 5 variables and 5 simultaneous equations, and man, was that a nightmare! I kept making dumb mistakes in the algebra and kept getting wrong answers.

 

[rattus]

Google Thevenin:

Google Thevenin:

Crossman,

Try this:

Call the ends of the 5 Ohm resistor nodes A and B. Now remove the resistor and compute Va and Vb. Now convert the remaining elements to two batteries--one with an emf of 66.7v and the other 33.3V--and compute the parallel combination of 10 and 20 Ohms. This is the internal resistance of each battery. Now connect these two "batteries" to the ends of the 5 Ohm resistor. Now you can compute the current through the 5 Ohm resistor, and then compute Va and Vb.

Another approach is through node equations:

Let Vo = 100V and let the unknown voltages be Va and Vb. Now convert all resistances to conductances.

Now apply Kirchoff's current law to nodes A and B to obtain,

0.1Vo = 0.35Va - 0.2Vb
0.05Vo = -0.2Va +0.35Vb

Now solve for Va and Vb to see if I got it right.

 

[rattus]

Details:

Details:

Crossman, don't read this until you have doped it out yourself.

The attached diagram shows the loop equation and node equation methods for solving this problem.

Please note, the application of the loop equation method is not general in this case. It depends on the fact that the problem is symmetric between Vo and ground. Looks the same right side up and upside down! That being the case, the currents through R1 and R4 are equal; likewise the currents through R2 and R3 are equal. We can then assign the currents I1 and I2 can be assigned as shown. Then the two loop equations may be written and solved for I1 and I2.

The node equation method is general and involves the application of Kirchoff's current law to nodes A and B to obtain the two equations which may be solved for Va and Vb.

 

[rattus]

Wha' hoppen'?

Wha' hoppen'?

Crossman, don't read this until you have doped it out yourself.

The attached diagram shows the loop equation and node equation methods for solving this problem.

Please note, the application of the loop equation method is not general in this case. It depends on the fact that the problem is symmetric between Vo and ground. Looks the same right side up and upside down! That being the case, the currents through R1 and R4 are equal; likewise the currents through R2 and R3 are equal. We can then the currents I1 and I2 can be assigned as shown. Then the two loop equations may be written and solved for I1 and I2.

The node equation method is general and involves the application of Kirchoff's current law to nodes A and B to obtain the two equations which may be solved for Va and Vb.

For some reason I can't edit my own post. The text above has been corrected to make sense.