Why Does Impedance Use Positive and Negative J Notation?

[lubo]

Homework Statement

We have a resistor R1 in series with a capacitor C1, then we have a capacitor C2 in parallel with a resistor R2

I don't want to achieve the whole question, just the start as it is confusing me.

Homework Equations

1/Rz = 1/r1+1/r2 (parallel)

The Attempt at a Solution

Total Impedance of the circuit= (R1+Jwc1)+ (R2//C2)

Zt = (R1+jwc1)+(1/(1/R2+1/jwc)

I am presuming 1/zt = 1/r1+1/jwc

zt = (R1+jwc1)+(1/(1/R2+1/jwc)

zt = (R1+jwc1)+(R2/(1+jwc2R2)

This is as far as I want to go. I believe all is right so far. If not please let me know.

My question starts with zt = (R1+jwc1) I thought (R1-jwc1) would be correct. What I mean is -Jwc or +1/jwc. Could anyone enlighten me?

See below to for more information if necessary from a previous question.

Homework Statement

Calculate the J notation Impedance of the network. I only want the initial basic solution.

The problem I have is that sometimes the J notation of C is -ve and sometimes +ve ?

Homework Equations

The Attempt at a Solution

Product/sum of the parallel cct:

jwL x 1/jwC/jwL*1/jwC This is the Inductor and capacitor impedance equation.

The above will be added to:

R -j(1/wC)

My question is therefore, why in the above example at product over sum would it be ok to say jwL x 1/jwC/jwL*1/jwC i.e. * a +ve 1/jwC

When below it I can add it to R and -j(1/wC)

I hope this makes sence, thanks for any help in advance.

hi lubo!


ah, but the first j is on the bottom, while the second is on the top …

and -j = 1/j ​

 

[NascentOxygen]

Total Impedance of the circuit= (R1+Jwc1)+ (R2//C2)

Zt = (R1+jwc1)+(1/(1/R2+1/jwc)

I am presuming 1/zt = 1/r1+1/jwc

Why presume? If you can't confirm it, then to proceed further is likely to be a waste of time and effort.

So start with what you do know. What is X_C for a capacitor? If you haven't yet memorized it, look it up. Is there more than one way to write it?

 

[lubo]

Why presume? If you can't confirm it, then to proceed further is likely to be a waste of time and effort.

Thanks for looking at it, but rather than presume, I was meaning that this is what has been done to get to the next stage.

The start is o.k, the problem is as mentioned. The left hand side does not make sense to me and I am trying to figure out why this has been done. If you have any ideas that would be great. Also see the other questions comments if necessary.

So start with what you do know. What is X_C for a capacitor?

X_C = 1/wC

 

[NascentOxygen]

Shouldn't there be a j in that expression for Xc?

 

[tiny-tim]

hi lubo! happy new year!

Total Impedance of the circuit= (R1+Jwc1)+ (R2//C2)

Zt = (R1+jwc1)+(1/(1/R2+1/jwc)

nooo, that would be the impedance for an inductor circuit …

Z = (R₁+jωL₁)+(1/(1/R₂+1/jωL₂)

for a capacitor, you must always use Z_C = 1/jωC (= -j/ωC)

(as opposed to Z_L = jωL for an inductor)

so, inside that big bracket, you should have 1/(1/jωC), = jωC

(which incidentally is why two or more capacitors "add" when they're in parallel)

try again! ​

(btw, what is the "t" in zt ? )

… What I mean is -Jwc or +1/jwc …

mmm … they're not the same, are they?

 

[lubo]

Shouldn't there be a j in that expression for Xc?

Yes, thanks,

Z_C = 1/jωC (= -j/ωC)

happy new year, to you also "Tiny"-Tim with "big" answers

nooo, that would be the impedance for an inductor circuit …

That is what I thought, the answer must be wrong. It goes like that for a long time, so the autor definitely thinks it is right?

If the full question is: show that the I(current) out for the cct "as above" is given by... Iout =V/Zt would that make any difference?

so, inside that big bracket, you should have 1/(1/jωC), = jωC

(which incidentally is why two or more capacitors "add" when they're in parallel)

try again! ​

I think I have got what you mean for the right hand side (RHS). Thanks.

Ztotal = (R1+Jwc1)+ (R2//C2)

Zt = (R1+jwc1)+(1/(1/R2+1/1/jwc)

Only taking the RHS:

I am presuming 1/zt = (1/R2)+(1/(1/jwc))

zt = (R1+jwc1)+(1/(1/R2+jwc2) I didnt mean to put the 1/jwc2 here

zt = (R1+jwc1)+(R2/(1+jwc2R2)

(btw, what is the "t" in zt ? )

Impedance "Total(t)" I should be more clear, its easy to forget when you are caught up in the question.

mmm … they're not the same, are they? [/QUOTE]

… What I mean is -J(1/wc) or +1/jwc …

Thanks.

 

[tiny-tim]

hi lubo!

(try using the X₂ tag just above the Reply box )

If the full question is: show that the I(current) out for the cct "as above" is given by... Iout =V/Zt would that make any difference?

i can't see what difference that makes … I_total is always equal to V/Z_total

Ztotal = (R1+Jwc1)+ (R2//C2)

Zt = (R1+jwc1)+(1/(1/R2+1/1/jwc)
…
zt = (R1+jwc1)+(R2/(1+jwc2R2)

shouldn't it be 1/jwC₁ ?

 

[lubo]

hi lubo!

(try using the X₂ tag just above the Reply box )


i can't see what difference that makes … I_total is always equal to V/Z_total


shouldn't it be 1/jwC₁ ?

I have found the X₂ tag, thanks.

Sorry about that, yes it should be on the Left hand side (LHS) "(R1+(1/jwc₁))", I hope this is what you meant. All I wanted was confirmation that the answer I had in my book was not correct. It is frustrating looking at something that should be correct and it is not. It is another to guarantee that without some help! No need go any further at this time.
Thanks again.

 

[NascentOxygen]

No need go any further at this time.

Glad you are sorted. But I can't refrain from pointing out where you have been typing 1/jwC there are many who contend you should be in the good habit of typing 1/(jwc) since that is what you mean here.

We have all found there comes a time when learned sloppiness has been our undoing. :grumpy:

 

[tiny-tim]

… you have been typing 1/jwC there are many who contend you should be in the good habit of typing 1/(jwc) …

aren't they the same?

 

[NascentOxygen]

aren't they the same?

1/2mv² is not the same as 1/(2mv²)

 

[tiny-tim]

1/2mv² is not the same as 1/(2mv²)

erm … yes it is!

if you want to write the characters in that order, you should put a bracket round the fraction, as (1/2)mv²

(alternatively you can type the fraction as a single character, as ½mv²,

or you can leave a gap, as 1/2 mv²)

 

[NascentOxygen]

if you want to write the characters in that order, you should put a bracket round the fraction, as (1/2)mv²

No need for brackets here, because multiplication does not have precedence over division.

or you can leave a gap, as 1/2 mv²)

Gaps are not recognized mathematical operators. They mean nothing, fortunately.