Calculating the Charge of 3 Capacitors

  • Thread starter Thread starter freddy13
  • Start date Start date
  • Tags Tags
    Capacitors Charge
Click For Summary

Homework Help Overview

The discussion revolves around calculating the charge on a capacitor (C3) in a circuit involving three capacitors with given capacitances connected to a voltage source. The subject area includes concepts of capacitance and charge calculations in electrical circuits.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of equivalent capacitance using different formulas and question the clarity of the original poster's equation. There are attempts to clarify the correct order of operations and the proper use of parentheses in the equations.

Discussion Status

Participants are actively discussing the correct formula for equivalent capacitance and recalculating values based on clarifications. Some guidance has been offered regarding the arithmetic involved, and there is a recognition of previous errors in calculations.

Contextual Notes

There is mention of potential unit confusion and arithmetic errors in calculations, which are acknowledged but not resolved. The original poster's attempts to calculate the charge and equivalent capacitance are under scrutiny.

freddy13
Messages
12
Reaction score
0

Homework Statement



Three capacitors having capacitance C1 = 3.00 microF, C2 = 8.50 microF and C3 = 15.0 microF are connected to a 24.0 V battery as shown. Calculate the charge on C3.

Homework Equations



$$ C_{eq}= \left ( \frac{1}{C_1+C_2} + \frac{1}{C_3}\right )^{-1}$$

Q = CΔV

The Attempt at a Solution



First I attempted to find the Capacitance saying that C1+C2= 11.5μC through the parallel addition of capacitors. From here, I plugged everything into the

Ceq=((1/C1+C2(1/C3))-1

equation and got my capacitance which was 2.74 μF. I took the result and plugged all of my values into the Q = CΔV equation, and got a final value of 57.096 μC, or 5.71*10-5 C.

I tried another way and got the same answer, so, any suggestions would be great. I think I messed up somewhere with the units.
 

Attachments

  • prob04a_3capac.gif
    prob04a_3capac.gif
    1.4 KB · Views: 946
Last edited:
Physics news on Phys.org
Can you explain your formula for the equivalent capacitance? Are there some parentheses and an operator missing that might make the order of operations clear (and correct)?
 
Ceq=((1/C1+C2(1/C3))-1
There are more "("'s than ")"'s.

Is that: $$C_{eq} = \left ( \frac{1}{C_1} + C_2 \left ( \frac{1}{C_3} \right ) \right )^{-1}$$
or maybe: $$ C_{eq}= \left ( \frac{1}{C_1+C_2} + \frac{1}{C_3}\right )^{-1}$$

Neither of those equations nets me the same answer you got though.
I think you need to go over your arithmetic carefully.
 
It is the second one! Sorry for the miscommunication!
 
Simon Bridge said:
or maybe: $$ C_{eq}= \left ( \frac{1}{C_1+C_2} + \frac{1}{C_3}\right )^{-1}$$

This one to be more exact!
 
Okay, can you run the calculation again using that formula? What value do you get for the equivalent capacitance? (The value you gave previously is not correct, although that formula is good. So... finger+calculator problems?)
 
I got 6.509 microF this time, does that sound correct?
 
Which, multiplied by the 24 V would give me 156.226 micro C, or 1.56*10^-4 C...is that right?
 
Yes. Much better :approve:
 
  • #10
Thank you very much! Gotta watch those number entering errors!
 

Similar threads

Replies
9
Views
3K
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
14K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K