Understanding Potential Dividers in Circuit Analysis

[suv79]

Homework Statement

how did they get this ?? it is like a potential divider, can anyone explain...

Homework Equations

The Attempt at a Solution

this is what i got[/B]

 

[cnh1995]

Are you familiar with node voltage method?

 

[suv79]

Are you familiar with node voltage method?

yes i know about node voltage, i want the node at V_

 

[cnh1995]

You can write the node voltage equations and rearrange them to get V_. There will be 3 principal nodes.Assume ground node at the bottom wire and proceed.

 

[suv79]

 

[NascentOxygen]

View attachment 95408

this is what i got

Your second term is correct involving V2, and if you had been consistent you'd have a corresponding correct expression involving V1. The method you are using for this is called Superposition.

 

[suv79]

so this

how can i rearrange this to equal this

 

[cnh1995]

so this
View attachment 95449
how can i rearrange this to equal this
View attachment 95450

You are a couple of steps away. Use cross multiplication in the denominator.

 

[suv79]

 

[cnh1995]

View attachment 95451

Using cross multiplication, you will be able to cancel out Z2+Z3 in the denominator with that in the numerator. Then take Z3 common from the numerators and you are done.

 

[suv79]

i can do this ?

 

[suv79]

 

[suv79]

 

[suv79]

when adding fractions why does the denominator for the Z1 Z2 become Z1*Z2 ?

 

[suv79]

i don't really understand the last step here

 

[cnh1995]

View attachment 95455
i don't really understand the last step here

How will you write this term in the denominator
Z₁+Z₂Z₃/(Z₂+Z₃) using cross multiplication? It is used to make the denominators same. It is basic algebra.
e.g 3/4+2/3=[( 3×3)+(4×2)]/(3×4)=17/12.
You can not directly cancel the terms like you did there.

 

[suv79]

 

[cnh1995]

Consider only the first term i.e. the term before + sign. Apply the cross product rule in its denominator. What will you get?
Similarly proceed for the next term. It is a matter of only one step.

 

[suv79]

 

[suv79]

 

[suv79]

now i got this but still need to move Z3

 

[suv79]

 

[cnh1995]

View attachment 95461

I think you are making it unnecessarily complicated. Just apply the cross multiplication rule to the denominator of each term separately. It is just one step and you'll be done.

 

[suv79]

this is better but what happens to Z3^2

 

[cnh1995]

I think you are making it unnecessarily complicated. Just apply the cross multiplication rule to the denominator of each term separately. It is just one step and you'll be done.

Let the two terms be A and B.
So, V_=A+B
(I can't use LaTex on my phone, so this is the best format I can write in..)
Denominator of A is Z1+(Z₂Z₃)/(Z2+Z3)
=[Z1.(Z2+Z3)+Z2.Z3]/(Z2+Z3)
=[z1z2+z2z3+z1z3]/(z2+z3)
This is the denominator of A. Now write A and see what you can cancel out. It is one of the basic algebraic manipulations.

 

[gneill]

Wow. That's a lot of algebra.

If I might make a suggestion? If you begin with a nodal equation at the junction the whole thing becomes much more straightforward.

 

[suv79]

yes please :)

 

[gneill]

yes please :)

Um, that was my suggestion...

Have you been introduced to node voltage analysis ("nodal analysis") method yet?

 

[suv79]

 

[gneill]

Good. Now since the whole thing is set equal to zero you can discard the denominator. With the numerator only now, expand the terms and then gather them by voltage variable. You should see daylight pretty quickly.

 

[suv79]

 

[suv79]

i don't want this Z1Z2

 

[gneill]

No. In post #31 you've performed an illegal operation by multiplying Z₁Z₃Z₂ by zero and ending up with Z₁Z₃Z₂ on the left hand side.

Anything multiplied by zero is zero. So you clear the denominator away by multiplying both sides by it. Poof! it vanishes and you're left with just the numerator to deal with.

 

[suv79]

finished :D
thank you very much