I/V drop calculation issues

[Diggsworth]

Hi, I have a small college project due in a few days, in which I need to calculate the voltage/current drops across resistors in the following circuit:
**broken link removed**
Now I've already taken all the readings in a Multisim...simulation, and have all the measured values.
What I need to do now is physically calculate all the drops across the resistors.
Only, all my calculations in the Wheatstone Bridge section of the circuit all come out incorrect.
I tried using Wye-Transform, and while it ultimately was necessary for some calculations, offered no help to specific resistor drop calculations.
If anyone could point me in the right direction towards solving the I/V drops across the resistors, I'd really appreciate it.

 

[MrAl]

Hi,

What is connected between I01 and I02, is that a current source or a voltage source?

 

[Diggsworth]

Oh dang, sorry, a 24-Volt power source is connected, with + at I/O1

 

[MrAl]

Hello again,


Would you like a general method for solving circuits like this, or look for a shortcut that will apply to this circuit?
The general method is called "Nodal Analysis" and is really good to know, but it does require knowing how to deal with simultaneous equations. Sound like something you'd like to learn about? It's not very hard to learn really.

 

[Diggsworth]

I'd love to learn how to solve it properly.
It just kills me to not know how something like this works.

 

[MrAl]

Hello again,


Ok, let me see if i can get you started.

They call it Nodal Analysis because we pay strict attention to each node in the circuit. A node is a junction of two or more components, so this circuit has 6 nodes. We can call one node ground however and declare it to be at 0v, so that reduces this circuit to 5 nodes plus ground. The next thing we would do is assign nodal voltages to each node of the circuit, calling them v1, v2, v3,..., etc., up to the number of nodes. That gives us 5 voltage nodes and the voltage is assumed to be referenced to 0v (the ground node). Note that we will call IO2 the ground node, so it is assumed to be equal to 0v.

What we do next is interesting, as we more or less assume that we already know all of those voltages v1 through v5, and begin calculating the current through each element, 'knowing' those voltages. Of course the current through a resistor is simply I=E/R, we can use that to calculate the currents with each E being the difference in voltage between the two nodes that the component connects between. This means each E comes out to for example v2-v1, which is simply the subtraction of two node voltages.
We also assign currents I1 though however many branches we have. For example, we would call the current though R1 to be I1 and that current would be equal to:
I1=(v1-v2)/R1
Notice that the current arrow would follow conventional current flow and pointing from left to right and that is why we subtract v1-v2 and not v2-v1 above.
We can call the current though R3 to be I3, and that current would be:
I3=(v2-v3)/R3
and following that plan we can calculate the expressions for all the currents through all of the resistors.

Once we have all these expressions, we follow another rule that says that the current into any node is equal to the current leaving that node, or stated differently, the algebraic sum of currents entering a node is zero, so we start to sum currents for each node. For example, the sum of currents for node 2 (junction of R1 and R3) is:
I1-I3-I6=0
Note that we subtracted I3 and I6 because we are assuming that the top of each associated resistor is positive with respect to the bottom.
We continue in this manner until we have done every node, and what we end up with is a set of simultaneous equations in all of these unknown currents, but since we 'know' what each current is we substitute each current with what we calculated previously using the node voltages. What we end up with in the end is a set of simultaneous equations in 5 unknown voltages v1, v2, etc., but we can immediately knock that down to 4 unknowns because we know that v1 is equal to 24 volts. Thus, we end up with 4 equations in 4 unknowns and we then use the mathematics of simultaneous equations to solve for all 4 voltages. This gives us the voltage of every node in the circuit referenced to ground, so now we can go back and really calculate all of the currents by using the sum of currents into the nodes that we used previously, but this time we get the actual value of each current rather than an algebraic expression.

 

[Diggsworth]

Thanks much for the help, the first part of it makes sense, so I'll attempt the calculations.
I'm still a bit fuzzy on how exactly to solve, specifically the part about "knowing" voltages/currents.
But I'll test it out!
Thanks again for the help.

 

[Diggsworth]

Actually, I have no idea what I'm doing.
I tried solving through, and I can't get past putting in a value into the first I1 equation.
I think I might just be looking at it wrong, though.

 

[Diggsworth]

Edited for clarity:
I'm still not clear on why exactly we use I6 instead of I5, wouldn't it be I5 considering it's sum output = sum input, as in to say that; "I1=I3+I5" ?
Or do we use I6 because I5 is in parallel with I3?
The positive/negative aspect of the coupling is a bit confusing to me.
Edit2:
Would I use:
n2: I1 - I3 - I5 = 0
n3: I3 - I2 - I4 = 0
n4: I2 + I5 - I6 = 0
n5: I4 + I6 - I7 = 0

As my formula?

 

[KeepItSimpleStupid]

Just do it with Kirchoff's Current Law and set up three loop equations. You get 3 equations and 3 unknowns.

 

[Diggsworth]

How do you mean?
I originally started with that, and received faulty results.

 

[KeepItSimpleStupid]

Remember to do it, you would need three loop currents. They either have to be clockwise or counterclockwise. What is creating the problems is the signs and the fact that some of the resistors will have 2 currents through them.
e.g. (I1-I2)*R

I'd start by assuming conventional current (pos to neg) for the voltage source and put your loops in that direction.

I don't know which way your voltage source is, but I'll define I1 as left, I2 as right top and I3 as bottom right.

So, for the I1 loop, you get (assuming +24 is at the top) and loops are all clockwise
24 = I1R1 +( I1-I2)*R3 + (I1-I3)*R4 + I1*R7
for the I2 loop, you get
0 = I2*R5 + (I2-I3)*R2 + (I2-I1)*R3
and for the I3 loop
(your turn)

Even I had to do them 2x

For each voltge drop you will have to use either the loop current * the resistor or the difference of two loop currents multipled by the resistor.

It's messy algebra and getitng the polarity right.

 

[MrAl]

Actually, I have no idea what I'm doing.
I tried solving through, and I can't get past putting in a value into the first I1 equation.
I think I might just be looking at it wrong, though.

Hi again,


We should really start with a simpler circuit with say three resistors and then solve that, then move to the more complex circuit. If you like this idea i'll post a complete writeup on how to do this and i think that will clear these things up. In the mean time, you can take a look at Loop Analysis too that the other poster posted and see if you like doing it that way too. It's really a good idea to know both methods anyway.


Here is a better and more complete illustration of the process of analysis using nodal analysis. Note after the last step shown we end up with four equations in five unknowns, but we already know v1 so we dont have to solve for that so that brings us down to four equations in four unknowns. Also note that we could easily substitute 24 in for v1 before we start, and also note that it is simpler if we substitute the values of the resistors too. If we need general formulas we dont do that though
Of course the very last step is to solve the resulting set of simultaneous equations for v2, v3, v4, and v5.

 

[KeepItSimpleStupid]

You need the equation for I3 as well. Then expand the terms. You'll end up with coefficients for I1, I2 and I3 with R's as constants.

Then solve for I1, I2 and I3 either my elimination or substitution. It might be easier to keep the R's as constants until closer to the end of the solution.

For the first therm you'll get, in part, ( R1+R3+R4+R7)*I1-R3(I2)-(R4)(I3)
Now plug in the constants for R.

Do the same for the other two loop eqns.

 

[Diggsworth]

Thanks Al, that's precisely what I was doing.
Where I THINK I went wrong, is during my equation reduction process.
I'll try to get a screen capture of my results quickly.

Edit: fixed...kind of...
**broken link removed**

 

[MrAl]

Hi again,

If you are getting any results for any voltage in the circuit over 24 volts or less than 0 volts then the solution can not be correct. If you did everything the same as my previous post with the circuits (and fully fluent use of arrows ha ha) then the error must be coming in the way you are solving the simuls. Check that over carefully to see if you can get the right results.

 

[Diggsworth]

I know, that's what I'm trying to figure out where the heck I went wrong, but I think it might be for naught.

 

[Diggsworth]

These are the equations I used for the matrices:
**broken link removed**