Calculating the Charge of 3 Capacitors

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Three capacitors with capacitances of 3.00 μF, 8.50 μF, and 15.0 μF are connected to a 24.0 V battery, and the charge on C3 needs to be calculated. The equivalent capacitance was initially miscalculated, but after clarifying the formula, it was determined to be 6.509 μF. Using this corrected value in the equation Q = CΔV yielded a charge of approximately 156.226 μC on C3. The discussion emphasized the importance of careful arithmetic and unit management in calculations. Ultimately, the correct approach led to a satisfactory solution.
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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.
 

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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!
 
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