Sensor - Calculate the Capacitance

[2013]

1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)?

 

[gneill]

1. Vc = 12V * C1/(C1+C2+Cc)

2. Vc = -12V * C2/(C1+C2+Cc)


?

Yes. Now each of those expressions for Vc contribute to the actual Vc (you need to sum them; that's how superposition works). Solve for Cc.

 

[2013]

You mean:
1. + 2.

2Vc = (C1+C2) / (2C1+2C2+2Cc)

Is this right? Have I understand you right?

 

[gneill]

You mean:
1. + 2.

2Vc = (C1+C2) / (2C1+2C2+2Cc)

Is this right? Have I understand you right?

No. First, each of the expressions contribute to the total of Vc. Think of them as Vc_a and Vc_b, and so the total is Vc = Vc_a + Vc_b. That's how superposition works. You sum up the contributions due to each source to yield the result of both sources acting together. (Surely you must have covered superposition in your circuit theory classes?)

Second, you have to add the fractions properly. You don't add the common denominators; Is 1/3 + 1/3 equal to 2/6?

 

[2013]

We have done superposition, but only in a short launch. We have set different priorities in the first semester.

I`m so sorry that I have no idea what I exactly should do.

Can you show me the beginning?

 

[gneill]

We have done superposition, but only in a short launch. We have set different priorities in the first semester.

I`m so sorry that I have no idea what I exactly should do.

Can you show me the beginning?

Sum the two expressions (their right hand sides). That will yield the final expression for Vc. This is just algebra; you should be able to sum fractions, especially when they have the same denominators.

 

[2013]

12V * C1/(C1+C2+Cc) + (-12V * C2/(C1+C2+Cc)) = Vc
Vc = (C1+C2) / (C1+C2+Cc)

?

 

[gneill]

12V * C1/(C1+C2+Cc) + (-12V * C2/(C1+C2+Cc)) = Vc
Vc = (C1+C2) / (C1+C2+Cc)

?

What happened to the +12V and -12V terms

 

[2013]

12V * C1/(C1+C2+Cc) - 12V * C2/(C1+C2+Cc)) = Vc
12V * (C1/(C1+C2+Cc) - (C2/(C1+C2+Cc)) = Vc

Vc = 12V * ((C1-C2)/(C1+C2+Cc))

?

 

[gneill]

12V * C1/(C1+C2+Cc) - 12V * C2/(C1+C2+Cc)) = Vc
12V * (C1/(C1+C2+Cc) - (C2/(C1+C2+Cc)) = Vc

Vc = 12V * ((C1-C2)/(C1+C2+Cc))

?

Yes.

 

[2013]

Vc = 12V * ((C1-C2)/(C1+C2+Cc))
Vc*Cc=12V * ((C1-C2)/(C1+C2)
Vc*(Qc/Vc)=12V * ((C1-C2)/(C1+C2)
Qc=12V * ((C1-C2)/(C1+C2)=12V*((2,46*10^-11As/V - 2,01*10^-11As/V) / (2,46*10^-11As/V+2,01*10^11As/V)
= 1,21As

Cc=(Qc/Vc)=1,21As/0,15V=8,05As/V

Is this right? Why is it not the same capacity as in the other method?

 

[gneill]

Vc = 12V * ((C1-C2)/(C1+C2+Cc))
Vc*Cc=12V * ((C1-C2)/(C1+C2)
Vc*(Qc/Vc)=12V * ((C1-C2)/(C1+C2)
Qc=12V * ((C1-C2)/(C1+C2)=12V*((2,46*10^-11As/V - 2,01*10^-11As/V) / (2,46*10^-11As/V+2,01*10^11As/V)
= 1,21As

Cc=(Qc/Vc)=1,21As/0,15V=8,05As/V

Is this right? Why is it not the same capacity as in the other method?

The answers should be identical. Your algebra has errors.

 

[2013]

Can you say me at which step, I have done a mistake.

?

 

[gneill]

Can you say me at which step, I have done a mistake.

?

Your second line is incorrect. You cannot eliminate a single additive term from the denominator by multiplying the fraction by that term. Also, why are you introducing Qc at this point? You have Cc alone as a variable in the expression, and all other quantities have known values. Don't make things harder! Just do the algebra to isolate Cc.

 

[2013]

Vc=12V*(C1-C2)/(C1+C2+Cc)
Vc*(C1+C2+Cc)=12V*(C1-C2)
how can I get C1 and C2 to the other site?

 

[gneill]

Vc=12V*(C1-C2)/(C1+C2+Cc)
Vc*(C1+C2+Cc)=12V*(C1-C2)
how can I get C1 and C2 to the other site?

Elementary operations: Divide both sides by Vc. This will leave the LHS (Left Hand Side) as a simple sum of terms. Move everything that isn't Cc to the RHS.

 

[2013]

I have to say thank you to you, I have already the solution and it is the same as in the other method.
Thank you, Thank you for helping me :) :)

 

[gneill]

I have to say thank you to you, I have already the solution and it is the same as in the other method.
Thank you, Thank you for helping me :) :)

*Whew*! I'm glad we got there in the end

You're welcome. Good luck in your studies.