calculating unknown capacitance

[4electros]

suppose a capcacitor with 10uf capacitance , charged to 12v by connecting it to voltage generator of 12v (DC), then disconnected and substitute the gernerator with another capacitor has unknown capacitance so the voltage across two capacitors became 3v, how could we calculate the unknown capacitance?!

 

[Optikon]

One way would be to discharge it with a constant current source & measure slope of voltage decay.

C(unknown) = I * (V2 - V1) / (T2 - T1)

 

[4electros]

but i don't know current value , this is a homework exercise not practical experiment so i might just have to use formulas relations....

 

[Nigel Goodwin]

Keep it SIMPLE - if you connected an identical 10uF across it the voltage would be half what it was - the relationship is that simple!.

 

[4electros]

oh, i mentioned that the voltage became 3v and the second capcitor has unknown capacitance....it either be you didn't read my first post well or i couldn't understand what you meant!
thanks

 

[Nigel Goodwin]

oh, i mentioned that the voltage became 3v and the second capcitor has unknown capacitance....it either be you didn't read my first post well or i couldn't understand what you meant!

Well I'm not going to do your homework for you!, and if you read my post it gives enough information for you to do it yourself - and yes, I have read your post!.

 

[brian1234]

https://en.wikipedia.org/wiki/Capacitor

This should help you to learn abut capacitors. In particular, the "physics overview" presents the equation relating charge, capacitance and voltage the first relationship you need for this problem,


and "Capacitor Networks (in this case Parellel)" Presents the second relationship you need to solve this problem.

 

[4electros]

Keep it SIMPLE - if you connected an identical 10uF across it the voltage would be half what it was - the relationship is that simple!.

i think that the unknown capacitance here would be 4 times bigger than known one (10uf) then equal to 40uf as the voltage across it decreases 4 times than the previous one (12v), so due to fractional logic we conclude that the unknown capacitor would have more ability to recieve charges from that gernerator capacitor so it will have more capacitance to stand that extra charges...

 

[brian1234]

You're right the capacitance of the old capacitor and the new capacitor in parallel is 4 times bigger than the old capacitor by itself, but keep in mind the voltage measured is measured across both the old capacitor and the newly added capacitor, thus you are measuring the capacitance of both, so the unknown capacitance by itslef is less than 4 times the known capacitance.

 

[lord loh.]

VC=Q

12*10uF=Q1

3*XuF=Q1

Equating...

12*10uF=3*XuF

XuF=12*10uF/3

X=40uF

 

[4electros]

VC=Q

12*10uF=Q1

3*XuF=Q1

Equating...

12*10uF=3*XuF

XuF=12*10uF/3

X=40uF

I thought that i understood when i read nigel's post and quoted it with my previous post here but when i read your post , i found myself some lost in your way..... could you clarify more please?!
thanks

 

[Optikon]

I thought that i understood when i read nigel's post and quoted it with my previous post here but when i read your post , i found myself some lost in your way..... could you clarify more please?!
thanks

The large assumption made for the homework is that charge is conserved between the two capacitors.

For C1, Q1 = V1*C1

for unknown capacitor, C2
Q2 = V2*C2

Since Q1 = Q2 (Charge is conserved)

Then C1*V1 = C2*V2

and go from there.. straight algebra.

 

[lord loh.]

Sorry, the maths was wrong...realised it from Brian1234's post...

I am assuming that the unknown capacitor is not charged...

The charge in the system remains constant. This charge is now distributed across two capacitors.

VC=Q

12*10uF=Q1

now we charge the 2nd capacitor...
The resultant is that 2 capacitors are in parallel... And teh capacitance add up. But the charge is conserved.

3*(10+X)uF=Q1

Equating...
12*10=3*(10+x)
40=10+x
x=30uF

Excuse my blunder...