Voltage Division and Finding a Current

ohhh,I see what I did wrong. In my notes I have the 6 ohm in the middle but forgot to draw it in the picture. Sorry.

But as for combining, Starting at the left side, it looks like the 3.6, 9, 18 are in series. The middle- the 1.2 and 6 ohm are in series. The right- the 3.6 and 18 are in series.

After combining I get three resistors in parallel with values start from left-middle-right (30.6, 7.2, 21.6) and the 1 ohm in series with the set-up. After combining again I get 4.59 ohm resistor in series with 1 ohm and finally a Req=5.59

Although I love the idea of putting a 1V source at the two inputs,i'll go actually build this on my breadboard

And I found a weird problem, that involves Req in a kinda wheatstonebridge set-up. I would just like a conformation on some things. Should I post it here or on a new thread?

Hi,

What three resistors in parallel? There are no set of three that can be connected in parallel.
Try again, and i've redrawn the new configuration in the attachment. You should also see now why we like to have a second way to calculate, so that we can check our answers.

Yes post the new problem here, that should be fine.

going from left to right -cant we combine the 9 and 3.6, the 6 and 1.2, and the 18 and 3.6. I posted picture. I'll post the other problem once I can get this one

Hi,

Yeah but you are missing the OTHER terminal. I told you a few posts back that you have to DRAW the other terminal too in order to see this clearly. I even took the time to redraw it with the other terminal so you could see it better
Look at that drawing then see what you can muster up. You'll see a difference then.

Sure no problem, once you get this we can move on to the next problem and hopefully i can help with that too.

hrmm,I added the extra terminal,it seems like the top two on the right are in parallel,dont mind what it says on the picture. I meant the top two right resistors are in parallel

Hi again,

Yes, that's right, and you can see that after you combine them in parallel then you have a simpler circuit, which is mostly just series and then parallel.
It helps to draw each stage out. You drew the first stage after transforming Delta to Wye, now draw it again after combining the parallel resistances. See what you get then. You'll see how simple this comes out.

In cases where you cant transform you can always resort to the voltage excitation method where you apply 1v DC and calculate (or measure) the current then use Ohm's Law to calculate the total resistance. We got lucky with this problem that one transformation did the job.

After combining the two 18 ohm in parallel, which is 9. The 9 and 1 are in series so...10 Req Im surprised it turned out as clean as it did lol I posted the steps in a picture.But as for the other method voltage excitation. I replace the Req with a 1V and then use a KVL to find the currents?Is this what your saying?

Hi,

Ok that's better

With the other analysis method, you apply 1v at the two input terminals then calculate the current going into or out of an input terminal.
For this you might use nodal analysis and that gives you the current, then use Ohm's Law.
Nodal analysis is a set technique for doing just about any circuit. Once we calculate the voltage at the node just to the left of the 1 ohm resistor, we can calculate the current with Ohm's Law, and that gives us the required current. We then use Ohm's Law one more time to get the total resistance.
Remember we do it this way assuming we dont know how to combine resistances like with the topology transformation. This is an entirely different way of doing it.

hrmm interesting. I'll have a go at it later today and get back to you. But as for that other problem. I posted a picture. It requires to find a variety of Req with respect to different terminals, i'll just start with the first problem of the picture and that's finding a Req of A and B. Now I know that if we have other open circuits such as C,D,E and F.Dont we leave out or re-write the circuit w/o those open circuits and whatever is connected to them, if im not mistaken?

I posted an example of re-writing and leaving the components out.called Example picture

Hi again,

Well, for these you do the same thing...try to combine resistances first in order to simplify the network and then go from there.
In the first circuit "wheatstone" this should be easy. But you really have to get used to looking for configurations that make the circuit simpler yet dont change the electrical properties. Do you see the simplification in the wheatstone one? Sometimes you just have to redraw it a little to see the simplification.

Ive tried to redrawn it a bunch of times and this is what I came up with,im kinda lost with this one.

Hello,

Well why didnt you combine the resistances in series next? That simplifies the network and then you can look for other series or parallel combinations.

After combining the inseries restors on top and bottom, it seems like both 6 ohm resistors are in parallel since it seems that both endpoints of both 6 ohm resistors are connected. But im not 100% sure since E and F are connected as well. What do you think.

Hi,

Well, to get from A to B you do not need E and F because they are open, so just forget about E and F.

It's always the same with these networks. First you try to combine series or parallel, and you might have to do that a couple times or more. If you cant do that then you have to resort to something else like applying a 1v DC test signal (or 1 amp DC test signal) and doing some nodal (or other) analysis. Then you can calculate the unknown (current or voltage) and then use Ohm's Law to get the total resistance.

You really do have to examine the network and think sometimes. "What resistors can be combined in series, what ones in parallel, then after that, can it be done again, or even a third or more times?"

Hrmm interesting. But after removing E and F,we have two 6 ohm resistors in parallel,after combining. A 3 ohm in series inbetween A and B, so a 9=Req

Thanks for helping me with everything so far, btw.

Ratchit said:

Let's start with prob. 1. What is A*I1 is terms of Vout?

Ratch

Click to expand...

For Problem 1 the answer for Vout is 0 (zero) volts. CAREFULLY EXAMINE the circuit. the branch circuit where you are calculating Vout is incomplete. As there is no V+ there is no Vout.... the circuit is incomplete

for your voltage divider problem a small amount of math savvy is required but not much. Each branch of the divider is done the same way so it is relatively easy. The resistor relationship is always 2/1 in this circuit. so the voltage being tapped is 1/3 of the total branch voltage. Since you have the final output voltage of .2 volts across the 1 ohm resistor in a divider of 3 ohms total resistance the divider relationship is 2/1. To figure the total voltage you have .2 volts across 1 ohm which gives you .4 volts across the 2 ohm resistor for a total of .6 volts across the entire divider branch. Stepping back to the next higher level you have the same 2/1 relationship, giving a total branch voltage of 1.8 volts... And once again this brings us up one level, this time to the primary voltage divider, which also uses the same 2/1 relationship. Again you would simply multiply the output of this branch by three giving you your Vin. 1.8 x 3= 5.4 volts, Vin = 5.4volts. But there is MORE to this lesson. CONSISTENCY. Sticking with the same resistor values would have simplified this exercise substantially, and in a real world application would have reduced the number of different values of resistors required. Further, with a bit of practiced math the entire circuit could have been reduced to just 2 resistors of the proper values, cutting the parts count by four resistors (or to on third of the original parts count). Now the challenge for you with a Vs of 5.4 volts calculate a two resistor voltage divider that gives you a Vout of .6 volts. This is the most cost effective method of manufacturing an electronic device : reduced parts count. It also increases product reliability as there is less to go wrong.

Ratchit said:

A*I1 can split into two paths, the 3 ohm resistor and the 6 ohm resistor. The voltage Vout is the same across both resistors. What is Vout? Hint: Find out parallel resistance of 3 ohms and 6 ohms.

Ratch

Click to expand...

there is no Vout , the circuit is incomplete and there is no voltage in this branch.

MrAl said:

Hello mik,

It sounds like you need to look up current division in resistor circuits. It is similar to voltage division.

For two resistors R1 and R2 in series forming a voltage divider with R1 the upper resistor fed by a voltage Vs we have:
vR2=Vs*R2/(R1+R2)

For two resistors in parallel R1 and R2 forming a current division circuit the current divides, and the current through R2 is:
iR2=Is*R1/(R1+R2)

where Is is the total current that splits through the two resistors.

Note that for both of these equations the sum of the two resistances is in the denominator. For voltage division the resistor that is in the numerator is the one we want to find the voltage across, and for current division the resistor in the numerator is the opposite resistor to the one we want to find the current through.
For current division if the two resistors are the same then half the current goes through one and half goes through the other so we get the total current divided by 2 in either resistor.

You should look this over carefully as you will run into this many times. Just remember that current division has the 'opposite' resistor in the numerator.

Also, always first check to see if you can combine resistors to simplify the analysis. To do this you would look at how to combine two resistors in parallel or in series.

The second circuit looks interesting too. I cant wait to see how you approach that one.

Click to expand...

Current is not divided in resistor circuits. OHMS LAW... you may have branch current but total current is ALWAYS based on the total resistance of the circuit. As the first circuit ONLY has a total of 9 ohms the current is Vo/9 (parallel resistance law of note in this instance is resistors of equal value in parallel are calculated by the value of one resistor divided by the number of parallel resistors hence 12ohms divided by 2 = 6 ohms, plus the single 3 ohm resistor =total value of 9 ohms) and remember the rest of the resistors are just for show since they are not part of the circuit as there is no current path available to them.

mikali said:

Hrmm interesting. But after removing E and F,we have two 6 ohm resistors in parallel,after combining. A 3 ohm in series inbetween A and B, so a 9=Req

Thanks for helping me with everything so far, btw.

Click to expand...

No you have 2 12 ohm resistors in parallel which give a total resistance of 6 ohms. You must be VERY CAREFUL how you word things when describing circuits. for example 2 6ohm resistors in parallel will give you a resistance of three ohms a substantial error. This isn't nitpicking about verbage, since it substantially changes the value of the given answer. This is something you need to be careful about. Making it a habit early on can save you many issues down the road