# Voltage Division and Finding a Current

Discussion in 'Homework Help' started by mikali, Mar 10, 2014.

1. ### mikaliMember

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Problem 1) Picture Name- Circuit Problem 4
I need to find A. Vout/Vs=9 in the circuit find A.
Problem 2) Picture Name- Division
Need to find the voltage of Vo. Use Voltage Division and Current Division.
I get stumped when I look at these two. I know for Problem 2, that current is the same in series and voltage is the same in parallel. But I just don't know how to get started with either. Especially Problem 1 Any help that could just get me started would be great. And once I become fluent in circuit analysis ill contribute to the community. Oh and so far ive only learned about Loop analysis and Voltage/Current division. Next class we learn about Nodal analysis.

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2. ### RatchitWell-Known Member

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

Ratch

3. ### mikaliMember

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The A*I1 has two paths it can take from the looks of it, and this is where I kinda get confused. If A*I1 goes through the 6 ohm, I can figure out the relationship A*I1 has with Vout by Loop analysis. But since it has two paths I kinda get confused. Ive been starring at it for about 3 hours trying various ways to solve. I keep thinking I need a way to establish a relationship to Vs, Vout, and A*I1

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5. ### RatchitWell-Known Member

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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

6. ### mikaliMember

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Ahhhh, that flew over my head lol. Quick side-question-But after finding a Req for the 3 and 6 ohm resistors its best to put the Req inbetween the AI1 and Vout,since were trying to find the Req by Vout. Or it doesn't really matter since either way will have the same result.

But after I find the Req and find that (Vout=2AI1), I then move over to the other side of the circuit and repeat the same process, since it looks like Vs is in parallel with the two 12 ohm resistors, and once I find (Vs=9I1). I then plug it into my Vout equation to find A=40.5?

But a thought occurred to me that since we found the voltage for both pair of resistors in parallel, should I backtrack to apply ohms law to find the current through them?Like divide the 3 ohm and 6 ohm resistor by 2AI1 and then KVL them?

7. ### RatchitWell-Known Member

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You appear to have made an arithmetical error. Your answer is half of what it should be. You can find the error yourself, or we can continue on with my way. It is up to you.

Ratch

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9. ### RatchitWell-Known Member

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OK, now look at the left side of the circuit and obtain I1 in terms of Vs. It is not Vs = 9*I1, because the currents split evenly between the two 12 ohm resistors.

Ratch

10. ### mikaliMember

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if their in parallel and it splits could I just say Vs/12=I,since voltage is the same if in parallel.And I apologize in advance,I know the answer is going to be easy. So far I learned about current division,but we need a current source. I was thinking I could do this if I could convert the Vs to Is...but or since a current divider in this situation is equal to I=(Vs/12)Is

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11. ### RatchitWell-Known Member

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What is the variable "I"? Where did that come from? What is Vs/12?

Ratch

12. ### mikaliMember

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I meant I1. The reason I say Vs/12 is b/c the two 12 ohm resistors in parallel with Vs ,having the same voltage across them, I1 being Vs/12.Though this is wrong

13. ### RatchitWell-Known Member

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You are correct in stating that is wrong. What about the 3 ohm resistor in series with Vs? Won't that 3 ohm resistor drop some voltage? How much voltage? How can you say there be Vs voltage across the two 12 ohm resistors with the 3 ohm resistor in series with Vs?

Ratch

14. ### mikaliMember

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When calculating the Voltage drop across. Its hard to figure out the current going through it since I1 has two ways it can go.Quick Question-Could I perform a Source Transformation on Vs and the 3 ohm resistor.Since the 3 ohm is in series and its difficult to figure out what current is going through it? And if I did transform it,then we would have 3 resistors in parallel with each other,and then try to determine a Req for them and calculate a new value for our source. Would this be a legal move?

So after playing with this idea I got A=3, and after plugging it into the Vout/Vs=(should equal 9) to check if it Vout/Vs does equal 9 and it does. It feels wrong though.....

Last edited: Mar 12, 2014
15. ### RatchitWell-Known Member

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It not only feels wrong, it is wrong. A source transformation is an advanced analysis technique. You have to master the basics first. First, what is the total resistance that Vs sees? Just state the figure without any convoluted reasoning.

Ratch

16. ### MrAlWell-Known MemberMost Helpful Member

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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.

17. ### mikaliMember

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I was thinking about current division awhile ago. But to figure out I1 you need to know (Is). This is why I didn't use current division, b/c I couldn't find what (Is) was ,and got stumped.
So...

I1= (12/12+12)Is
What Im stuck on is the proper labeling of the other 12ohm resistor, would we label this I2 or simplify and find a Req with the 3 and 12 resistor since it seems there in series with one another? And what about (Is),just leave it blank for the time being while writing the current division equation.

18. ### RatchitWell-Known Member

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First, what is the total resistance that Vs sees? Just state the figure without any convoluted reasoning.

Ratch

19. ### mikaliMember

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the first resistor 3 and then the voltage goes through the two 12's. Right...then to calculate the voltage drop ,I thought it was R*I.But to answer your question.3

20. ### RatchitWell-Known Member

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No, it is not 3 ohms. There is more resistance than 3 ohms. What is the resistance after the 3 ohm resistor, that is, what is the resistance of the two 12 ohm resistors in parallel?

Ratch

21. ### mikaliMember

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Since there in parallel a Req of parallel resistors is (1/R1+1/R2)^-1 right?