Solving Series Capacitors: C1 & C2 Charge & Voltage

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In summary, a battery with an emf of 60 V is connected to two capacitors in series: C1 = 12 microFarads and C2 = ?. After the connection, the charge on C2 is 540 uC. Using the fact that series capacitors have the same charge, the charge on C1 was found to be 720 uC. Using the equation C = Q/V, C2 was then solved to be 12 microFarads. However, this was incorrect since the voltage was divided between the two capacitors. Knowing that the charges are equal, and with the value of C1 given, the voltage drop across C1 was determined to be 45 V. Therefore, the voltage drop across
  • #1
1. A battery with an emf of 60 V is connected to the two capacitors shown in the figure . Afterward, the charge on capacitor 2 is 540 uC.

The capacitors are in series, with C1 = 12microFarads and C2 = ?

Since series capacitors have the same charge Q, I used C1 and solved for Q in C = Q/V, obtaining Q = 720uC. I then solved C2 = 720/60 = 12microFarads, which is incorrect.
 
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  • #2
The charge across each capacitor isn't 60 V because the 60V is divided across the two capacitors. However, you know what Q is across the first capacitor. It's 540 uC, because if you consider the two capacitors as one unit, it would have the same charge as the two individual capacitors.
 
  • #3
So.. what is C2 then? I got 18 and that was incorrec
 
  • #4
So.. what is C2 then? I got 18 and that was incorrect
 
  • #5
Old post

Hey! The Original Post is over 11 months old!
 
  • #6
I didn't realize that but this is a question I am currently working on
 
  • #7
jadenjaden said:
So.. what is C2 then? I got 18 and that was incorrect
(What units?)

Show how you got that.
 
  • #8
The capacitors are in series. If the charge on C2 = 540 μC, how much charge is on C1 ?

What's the voltage drop across C1 ?
 
  • #9
I am also working on this question, and I don't know where to begin. i know that the charge on both capacitors is equal
 
  • #10
danielcheung1 said:
I am also working on this question, and I don't know where to begin. i know that the charge on both capacitors is equal

If the charges are equal, and you know the value of C1, what does that tell you about the voltage across C1?
 
  • #11
i am not sure,
but is it that Q = CV
and so i can say C1V1 = C2V2
and i think the voltage drop has to be different for the two capacitors..
 
  • #12
danielcheung1 said:
i am not sure,
but is it that Q = CV
and so i can say C1V1 = C2V2
and i think the voltage drop has to be different for the two capacitors..

Yes, but the problem statement gives you the charge on one of the capacitors, right? You've also determined that the charges must be equal for both capacitors. The problem statement also gives you the value of one of the capacitors. Apply your Q = CV to find the voltage on that capacitor.
 

1. How do you calculate the total capacitance of a series circuit with two capacitors?

In a series circuit, the total capacitance is equal to the sum of the individual capacitances. So, the formula for calculating the total capacitance is: Ctotal = C1 + C2.

2. What happens to the voltage across each capacitor in a series circuit?

In a series circuit, the voltage across each capacitor is the same as the voltage across the entire circuit. This means that the voltage is divided between the two capacitors, with each capacitor receiving a portion of the total voltage.

3. How do you calculate the charge on each capacitor in a series circuit?

The charge on each capacitor in a series circuit is equal to the total charge of the circuit, which can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

4. Can the capacitors in a series circuit have different capacitance values?

Yes, the capacitors in a series circuit can have different capacitance values. However, the total capacitance will always be less than the smallest individual capacitance in the circuit.

5. How does the total capacitance change if more capacitors are added in series?

If more capacitors are added in series, the total capacitance will decrease. This is because the capacitors are connected in series, meaning that the charge is divided among them, resulting in a lower overall capacitance.

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