Why Are My Capacitance Calculations Incorrect?

In summary, the conversation discusses solving a problem involving four capacitors connected in a circuit. The equivalent capacitance for part a is found to be 5.96uF, which is the correct answer. For part b, the correct charges for Q1 and Q2 are found to be 26.3uC and for Q3 and Q4, the correct charges are 63.2uC and 89.5uC respectively. The conversation further explains how to find the voltage and charge across each capacitor in the circuit.
  • #1
jhess12
11
0

Homework Statement



http://books.google.com/books?id=1D...are+connected&sig=iwUiJRqZDNHDGOfQfHHvh_9aQ3Y

problem 41 on this link. I got the equivalent capacitance for part a. for the two capacitors with 15 and 3uF- [tex]C_{eq}[/tex]=2.5uF
for capacitor with 6uf- i got the [tex]C_{eq}[/tex]=8.5uF. final answer for a.=[tex]C_{eq}[/tex]=5.96uF. this final anwer is correct according to the answers given to me.

Homework Equations





The Attempt at a Solution


capacitor with 15uF-[tex]Q_{1}[/tex]
capacitor with 3uF-[tex]Q_{2}[/tex]
capacitor with 6uF-[tex]Q_{3}[/tex]
capacitor with 20uF-[tex]Q_{4}[/tex]
for part b. i know that the Q for [tex]Q_{1}[/tex] and [tex]Q_{2}[/tex] are the same because they are in series. the Q i get for them is Q=[V/(1/C1+1/C2)]. If V=15V and
C1=15 and C2=3, then the answer i get is 37.5uC. but the answer in the book says [tex]Q_{1}[/tex] and [tex]Q_{2}[/tex] =26.3uC

for [tex]Q_{3}[/tex]- Q=[tex]C_{3}[/tex]V If [tex]C_{3}[/tex]=6uF and V=15V, then Q should be 90uC. OR if Q=[tex]C_{eq}[/tex]V, then Q=8.5uF(15)=127.5uC. but the book says it should be 63.2uC

for [tex]Q_{4}[/tex]- Q=[V/(1/[tex]C_{eq}[/tex])] If [tex]C_{eq}[/tex]=5.96uF and V=15V, then Q should be 89.5uC- this one i got correct.

I don't understand how i can get certain parts right, but others wrong, unless i am using the wrong equation. Please help, i need to know this for a test.
 
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  • #2
Start with C4... you did that right... the charge on C4 is just the charge on the Ceq...

What is the voltage across C4? Use V = Q/C...

So what is the voltage across C1 and C2? this is where you made your mistake... it's not 15V...
 
  • #3
ok well, i figured that if Q4 is 89.5uC, then Q4=Q, then V3=Q/C=89.5/8.5=10.53V. if
V3C=Q, then 10.5(6uF)=63.2uC=Q3. this appears to be correct, even though i don't understand why. as for Q1 and Q2, i am still confused, i asked my professor to explain it, but he says i should know how to solve this so to figure it out on my own. I keep thinking that it should be Q=[V/(1/C1+1/C2)]. i can see that V=V1+V2, but beyond that...i just don't know
 
  • #4
jhess12 said:
ok well, i figured that if Q4 is 89.5uC, then Q4=Q, then V3=Q/C=89.5/8.5=10.53V. if
V3C=Q, then 10.5(6uF)=63.2uC=Q3. this appears to be correct, even though i don't understand why. as for Q1 and Q2, i am still confused, i asked my professor to explain it, but he says i should know how to solve this so to figure it out on my own. I keep thinking that it should be Q=[V/(1/C1+1/C2)]. i can see that V=V1+V2, but beyond that...i just don't know

No. what I meant what you know the charge on Q4 is 89.5uC. You know the capacitance of Q4 is 20uF. So the voltage across Q4 is 89.5uC/20uF = 4.475V.

15V - 4.475 = 10.525V

So the voltage across Q3 is 10.525V... so the charge across Q3 is 10.525*6uF = 63.15uC. etc...
 
  • #5
i got it earlier from your earlier post, thanks so much
 

FAQ: Why Are My Capacitance Calculations Incorrect?

What is capacitance charge?

Capacitance charge is the amount of electric charge that can be stored in a capacitor at a given voltage. It is measured in units of Coulombs (C).

How is capacitance charge calculated?

Capacitance charge can be calculated by multiplying the capacitance of the capacitor (in Farads) by the voltage across the capacitor (in Volts).

What factors affect capacitance charge?

The capacitance charge of a capacitor is affected by the size and shape of the capacitor, as well as the material used for the capacitor's plates and the distance between the plates. It is also affected by the voltage applied to the capacitor.

How is capacitance charge related to electric potential energy?

Capacitance charge is directly related to electric potential energy, as the electric potential energy stored in a capacitor is equal to one half of the product of the capacitance and the square of the voltage. This can be expressed as U = 1/2 * CV^2.

How is capacitance charge used in practical applications?

Capacitance charge is used in various practical applications, such as in electronic circuits for storing electric charge, in power systems for power factor correction, and in measuring instruments for capacitance measurements. It is also used in energy storage devices, such as batteries and supercapacitors.

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