Some theory help please

[Herr Masel]

I started learning electronics from these books by the open university, it is going well and I am enjoying it but I am really stuck on one thing here. The topic is analyzing circuits. In the following diagram I copied (quite poorly, I should add ) from the book I am supposed to find the current going through R3. This is done by dividing the diagram into two and working out the current of each section seperately, then subtracting the smaller from larger one which leaves you with the current going through R3.
The way I reckon it works this way (also going by their earlier examples):
Ua=Ia(R1+R3)+IbR3
and for the other section:
Ub=Ib(R2+R3)+IaR3


     R1=15ohm               R2=70ohm
  ---\/\/\/\----- |---------\/\/\/\---
|                   |                        |
|                   \                        |
__ Ua=6v       /                      ___ Ub=6v
-                   \ R3=10ohm       -
|                     |                      |
|                     |                      |
|-----------------|-----------------|


I will not include my math here because it would make the post needlessly long, but I can if you want me to. The thing is I keep getting the wrong results. The answer is I3=O.27A according to the book. Can you help understand what is wrong? I had my father look at the math and he said it was correct, but he doesn't know electronics so I guess something is wrong with my equations? I hope I am not getting too hung up on this, but I don't want to keep going with the book until I've solved it. Thanks alot.

[R.G.]

Well, I can tell you that the answer they've given you is correct, but I got the answer the way you're not supposed to.

What they want you to do is to write the two loop equations, then solve them. I did it more simply. I noted that the batteries are both 6V, so the same voltage appears at the Ua end of the 15 ohm resistor as appears at the Ub end of the 70 ohm resistor. If the voltages at two points are the same, then I can connect them, and no currents will flow - effectively I have not changed the network. So I just connected the 70 ohm to Ua, then realized that R1 and R2 are in parallel. I calculated R1||R2 at 12.352 ohms, added R3=10 ohms for 22.352 ohms, then divided that into 6V to get 0.26842 amperes.

What they want you to do is to set up

Lessee:
Ua = IaR1 +IaR3 +IbR3 so Ua = Ia(R1 + R3) + IbR3 and then
Ib = Ua/R3 - Ia*(R1+R3)/R3

From Ub = Ib*(R2+R3) + IaR3
substituting for Ib gives
Ub = (R2+R3)(Ua/R3 - Ia*(R1+R3)/R3 ) +IaR3
multiplying by R3 to get rid of the "/R3"
gives UbR3 = Ua(R2+R3)-Ia(R1R2+R1R3+R2R3)-IaR3^2 + IaR3^2

canceling the R3^2 terms and rearranging,
Ia = (Va(R2+R3)-UbR3)/(R1R2+R1R3+R2R3)
or Ia = 0.221

subbing back in
Ib = 6/10 - 0.221((15+10/10))
  = 0.6 - 0.221(2.5) = 0.0475

so the current in R3 = 0.221+0.0475 = 0.2685

I learned a trick about keeping things straight in these messes of equations. Equal things always have equal units (i.e. no number of apples ever equals any number of oranges) so at each equation, if what's on one side is a voltage, all the other terms must come down to being a voltage, so  if you have a V on one side, the other side terms must all individually be equal to a voltage, as I*R or voltage squared divided by voltage, or resistor*voltage divided by resistor. The values you don't know, but you do know that a Rx and Ry are both resistors, Ix and Iy are both currents, and Vx and Vy are both voltages. So you can quickly find mistakes by seeing that you somehow got a resistance squared on one side being equal to a current on the other side.

Hope that helps.

[The Tone God]

R.G. beat me to it. I look at it the same way. Treated Ua and Ub as the same and ran R1 and R2 parallel. If they really wanted you to go through the full excerise they could have made Ua and Ub different.

Andrew

[davebungo]

Have the books introduced Thevenin's theorem yet?  This basically states (this isn't the exact theorem as I'm going from a distant memory here) that any 2 terminals of a network of voltage sources, current sources and impedances can be simplified to be a single ideal voltage source (with no in-built source impedance) in series with an impedance.  These are the so called Thevenin voltage and Thevenin Impedance although in this case we are talking about simple resistance.  So this is how it looks:

                                 RTh
                      ------/\/\/\/\-----o
                      |
                    ( VTh )
                      |
                      -------------------o

The advantage of this approach is that you can then connect your load (in this case R3) to this circuit and perform a very simple analysis to obtain the voltage and current.

So how do you determine VTh and RTh?  Well VTh is simply the open circuit voltage which appears at the terminals (i.e. with no load connected).  RTh is the impedance/resistance seen looking into the circuit with all voltage sources replaced by short circuits.  (An ideal voltage source has no impedance so the mpedance seen looking into it is zero).

So if we take your circuit and re-draw it slightly we have:

                                 R1
                      ------/\/\/\/\------------o
                      |                  R2        |
                      |         ------/\/\/\/\---
                    ( Ua )    |
                      |       ( Ub )
                      |         |
                      ------------------------o

VTh = Ub + R2.(Ua-Ub)/(R1+R2)
RTh = R1//R2

Now we can connect R3 to work out the current I3 which is simply VTh / (RTh + R3).

In your example Ua = Ub so VTh is simply Ub (or Ua) (as R.G. pointed out).

Therefore, I3 = Ub / ( R1//R2 + R3 ) = 6 / ( 12.35 + 10 ) = 0.268.

You may be thinking that this is all a bit long-winded, but this approach comes into its own with more complex circuits and it also has its uses when analyzing transistor circuits.

[Herr Masel]

Just sat down to this now. Thanks alot guys. The book hasn't introduced Thevenin's theorem yet, it's coming up in three pages, at the beginning of the next chapter. They wanted me to do it RG's way I guess, which in fact I tried several times. It is very strange because I always got the wrong answer, even when my father helped me and he is very good at maths (but like I said not electronics). What I did was get the first equations - Ua = IaR1 +IaR3 +IbR3 so Ua = Ia(R1 + R3) + IbR3. After that I placed the numbers and went from there. After coming back here, I tried it by going with the equations as far as I could (again, like RG did) and only putting the numbers in at the end. For some reason this made a difference and I got the right answer! I guess I must of missed a step and confused one of the values, though I went over it so many times I thought there must have been a different problem. I'll take it as a lesson to use the equations, not the numbers.

I learned a trick about keeping things straight in these messes of equations. Equal things always have equal units (i.e. no number of apples ever equals any number of oranges) so at each equation, if what's on one side is a voltage, all the other terms must come down to being a voltage, so  if you have a V on one side, the other side terms must all individually be equal to a voltage, as I*R or voltage squared divided by voltage, or resistor*voltage divided by resistor. The values you don't know, but you do know that a Rx and Ry are both resistors, Ix and Iy are both currents, and Vx and Vy are both voltages. So you can quickly find mistakes by seeing that you somehow got a resistance squared on one side being equal to a current on the other side.

Hope that helps.

I'm not sure what you mean by that advice... obviously if V = a whole mess then that whole mess = V but I will try to use it next time I get stuck on something and see if it makes sense then.

Thanks again, good night!

[R.G.]

I'm not sure what you mean by that advice... obviously if V = a whole mess then that whole mess = V but I will try to use it next time I get stuck on something and see if it makes sense then.

It's not deep philosophy. It's just if V = whole bunch of terms then *each of the terms all by itself has to have units of volts*. It's not possible that you could have volts come out to be equal to amps or ohms, so if one side of the equation is in volts, then the other side has to be calculable as volts, not something else. If any of them evaluate to something that's not a volt, then you made a mistake.

So each term could be a V, or an I*R, or Vx*Ry/Rz, or I*Rn^2/(Ro+Rp), as long as each term could be reduced to voltage if you actually knew what the values were. You don't know the values, but you do know a resistance from a current from a voltage, so it's an independent check. If you multiplied or divided wrong, chances are the units of one or more terms will be incorrect. Volts add up to volts, ohms add up to ohms, and amps add up to amps. Six volts can't be equal to four amps, any more than it can equal four oranges.

Note that raw numbers have to be taken as the correct units;  (V+1) is take as meaning (an unknown number of volts plus one volt).

If one side is in I^2, then all the other terms have to equivalent to current squared; ditto if one side is a resistance, all the other terms have to evaluate out to resistances.

Does that make more sense?

[mac]

Just sat down to this now. Thanks alot guys. The book hasn't introduced Thevenin's theorem yet, it's coming up in three pages, at the beginning of the next chapter. They wanted me to do it RG's way I guess, which in fact I tried several times. It is very strange because I always got the wrong answer, even when my father helped me and he is very good at maths (but like I said not electronics).

You say you got the wrong answer many times. It sound more like an algebraic misunderstanding. How good are you at algebra? Hope better than my english.
I ask you this because basic dc circuits with no power dissipation calculations are mostly a set of linear equations which can be easily solved if you have some basic algebraic rules. Fuzz face bias is an example. But solving ac circuits requires some understanding of differential equations that can be a problem for people with basic math formation.
Fortunately, there are plenty of info about math & electronics in the net to guide you.
Or simply, throw away your book and let RG do the math   

mac

[Herr Masel]

Well  I left high-school at tenth grade, but I sort of have my maths "diploma" (what is it in english? GED? SAT? I keep forgetting). I have a pretty sharp mind but I just don't have much experience with it so alot of basic rules I forget, or I overlook trivial mistakes when I try to solve something complicated. It's a matter of practice really, which through this book I am getting.

R.G. yes it is clear thank you. I understood the math/logic behind it the first time, just not how it could be helpful, but I got it.