What am I doing wrong in calculating the charge on each capacitor in a circuit?

In summary, the conversation is about finding the charge on each capacitor in a circuit with 5 microfarad capacitors and a voltage of 30 V. The equations used are C = Q/V, C_{eq} = C_1 + C_2 + ... for capacitors in parallel, and C_{eq} = \frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...} for capacitors in series. The initial attempt at finding the charge for each capacitor was incorrect due to forgetting that current does not flow through a capacitor. The correct method is to consider the equivalent capacitance for the entire circuit and then distribute the charge between the capacitors in series.
  • #1
Baou
6
0

Homework Statement


YF-24-15.jpg
Each capacitor is [tex]5 \mu F[/tex] and [tex]V_{ab} = 30 V[/tex]. I'm trying to find out the charge on each capacitor.


Homework Equations


[tex]C = \frac QV[/tex]
[tex]C_{eq} = C_1 + C_2 +...[/tex] for capacitors in parallel.
[tex]C_{eq} = \frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...}[/tex] for capacitors in series.


The Attempt at a Solution


The first thing I did was calculate the equivalent capacitance for the entire circuit to be 3 microfarads. From here I calculate the total amount of charge to be 9*10^-5 C. Then I consider the circuit with just [tex]C_4[/tex] and the equivalent capacitance of capacitors 1-3. It's just two capacitors in series, and charge should be equally distributed between capacitors in series, right? So I get [tex]C_4 = 4.5 * 10^{-5} C[/tex], but this is incorrect. What am I doing wrong?
 
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  • #2
Baou said:

Homework Statement


YF-24-15.jpg
Each capacitor is [tex]5 \mu F[/tex] and [tex]V_{ab} = 30 V[/tex]. I'm trying to find out the charge on each capacitor.

Homework Equations


[tex]C = \frac QV[/tex]
[tex]C_{eq} = C_1 + C_2 +...[/tex] for capacitors in parallel.
[tex]C_{eq} = \frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...}[/tex] for capacitors in series.

The Attempt at a Solution


The first thing I did was calculate the equivalent capacitance for the entire circuit to be 3 microfarads. From here I calculate the total amount of charge to be 9*10^-5 C. Then I consider the circuit with just [tex]C_4[/tex] and the equivalent capacitance of capacitors 1-3. It's just two capacitors in series, and charge should be equally distributed between capacitors in series, right? So I get [tex]C_4 = 4.5 * 10^{-5} C[/tex], but this is incorrect. What am I doing wrong?
Both capacitors in series would have the 9 and not the 4.5. To understand this realize that a capacitor is a break in the circuit. Current (and therefore charge) do not cross over that break. Instead, one plate has charge moving onto it while the other plate has charge moving off of it. The result is each plate gains a charge equal in magnitude and opposite in sign while current flows in the circuit. It is easy to forget after seeing current that no current actually flows through the capacitor.

In the series connection, the total charge drawn from the battery to the capacitor c(1234) must therefore rest on the first plate it comes in contact with. We have C(123) and C(4). Using this logic -- since again there is a break in the circuit -- all of the charge stores from the applied voltage must be on C(123). Where then, you may ask, does C(4) get its charge? I point finger to the C(123)'s bottom plate whose charge leaves it and therefore heads toward C(4)'s top plate. Then, C(4)'s top plate accepts that charge -- which is equal to the charge on C(123)'s top & bottom plate(read: that is, the total charge of 9). Then, C(4)'s bottom plate has charge leave it, creating current until eventually it too has 9.

Remember, for plates that are parallel, charge is shared by them.
 
Last edited:
  • #3
Okay, that makes sense. Thanks for the explanation; I managed to figure out the rest with your help.
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is made up of two conductive plates separated by an insulating material, called a dielectric.

2. How does a capacitor work in a circuit?

When a capacitor is connected in a circuit, it charges up by accumulating electrons on one of its plates. This creates an electric field between the plates. When the capacitor reaches its maximum charge, it stops accumulating electrons and acts as an open circuit. When a voltage is applied to the circuit, the capacitor will discharge its stored energy, providing a temporary surge of current.

3. What are the different types of capacitors?

The most commonly used types of capacitors are ceramic, electrolytic, film, and tantalum capacitors. Ceramic capacitors are small and inexpensive, but have lower capacitance values. Electrolytic capacitors have higher capacitance values, but are larger and more expensive. Film capacitors have a wide range of capacitance values and are used in high-frequency applications. Tantalum capacitors have high capacitance values and are commonly used in electronic devices.

4. Why are capacitors used in circuits?

Capacitors are used in circuits for a variety of reasons. They can act as filters, smoothing out voltage fluctuations in a circuit. They can also store energy in a circuit, providing a temporary power source. Capacitors are also used in timing circuits, as they can be charged and discharged at a specific rate.

5. What are some common issues with capacitors in circuits?

One common issue with capacitors is their tendency to lose their capacitance over time. This can lead to a decrease in their effectiveness in a circuit. Another issue is the potential for capacitors to fail and become short circuits, which can cause damage to other components in the circuit. Additionally, the size, cost, and capacitance limitations of capacitors can also be limiting factors in circuit design.

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