Charge on a capacitor in a series

In summary, the conversation is about a circuit with five capacitors and a battery. The values for the capacitors are given, as well as the battery voltage. The question is asking for the charge on capacitor C5. The equations used are C=Q/V, and the attempt at solving the problem involves treating the parallel system as one capacitor of equivalent capacitance. However, the answer obtained is incorrect, and the confusion lies in the assumption that the voltage across series capacitors is the same. The conversation ends with a question about the resultant of C2, C3, and C4 and whether it is equal to 3.9 μF.
  • #1
acdurbin953
42
0

Homework Statement



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A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 3.9 μF, C2 = 3.9 μF, C3 = 5.7 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

What is Q5, the charge on capacitor C5?

Homework Equations


C=Q/V

The Attempt at a Solution


Okay, I thought that voltage across series capacitors was all the same across each one. Treating the parallel system as one capacitor of equivalent capacitance, I thought you could just divide the voltage by 3, and then multiply by C5 to get the charge on C5. I tried this and the answer is wrong. I'm really confused.
 
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  • #2
acdurbin953 said:

Homework Statement



103szyq.png

[/B]
A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 3.9 μF, C2 = 3.9 μF, C3 = 5.7 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

What is Q5, the charge on capacitor C5?

Homework Equations


C=Q/V

The Attempt at a Solution


Okay, I thought that voltage across series capacitors was all the same across each one. Treating the parallel system as one capacitor of equivalent capacitance, I thought you could just divide the voltage by 3, and then multiply by C5 to get the charge on C5. I tried this and the answer is wrong. I'm really confused.

The voltage across capacitors in series is not the same, unless the capacitances are equal. What is the resultant of C2, C3 and C4? Is it equal to 3.9 μF?
I think you mix charge with voltage. The charge is same on series capacitors .
 

1. What is the formula for calculating the charge on a capacitor in a series?

The formula for calculating the charge on a capacitor in a series is Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

2. How does the charge on a capacitor in a series change when the voltage increases?

As the voltage across a capacitor in a series increases, the charge on the capacitor also increases. This is because the charge on a capacitor is directly proportional to the voltage across it, according to the formula Q = CV.

3. What is the effect of adding more capacitors in series on the total charge?

When capacitors are connected in series, the total charge is divided among them. This means that adding more capacitors in series will decrease the charge on each individual capacitor. The total charge, however, remains the same.

4. How does the capacitance of capacitors in series affect the charge on each capacitor?

The capacitance of capacitors in series has an inverse relationship with the charge on each capacitor. This means that as the capacitance of the capacitors increases, the charge on each individual capacitor decreases.

5. Can a series of capacitors have a negative charge?

No, a series of capacitors cannot have a negative charge. Capacitors can only store positive charges on their plates, and when connected in series, the total charge will always be positive.

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