Charge and Voltage on Capacitors.

In summary, the problem is to determine the charge on each capacitor and the voltage across each, given C = 3 μF and V = 4.0 V. The circuit consists of three capacitors, initially labeled as C1, C2, and C3, connected in parallel. The solution presented in the conversation is incorrect as it assumes that the capacitors in series have different charges. The correct solution involves adding the two capacitors in parallel (C2 and C3) to obtain an equivalent capacitance (Ceq). The charge on each capacitor in the equivalent circuit would be equal, and the voltage across each capacitor can be calculated using the equation V = Q/C.
  • #1
Yut
7
1

Homework Statement


Determine the charge on each capacitor and the voltage across ech, assuming C=3yF(micro) and the battery voltage is V=4.0 Vhttps://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11006374_10152682348496806_273021698509210726_n.jpg?oh=128f2d03e7e0ad275cb982e11d827c2c&oe=54F5E264&__gda__=1425403951_252b026b8105e21db6642c5d28a9f753

Homework Equations


so,
C23 = 2C/3 = 2microF
Qeq = Q1 + Q23 = Ceq V <--- why is Qeq is the sum of the series capacitors, i thought they had the same charge
Q eq = 8. micorC
Q1 = Q23 = 4.00 microC
V1= Q1/C = 1.33 V
V23 = V - V1 = 2.67 V
Q2 = Q3 = CV23 = 2.0 micro C <--- not sure how did they get this value, when I plug in the found valued I don't get 2.

The Attempt at a Solution

 
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  • #2
The problem statement is unclear to me. You state that C = 3 μF, does that mean the circuit is fundamentally made up of a several 3 μF capacitors? If so, how do you arrive at C23 = 2 μF?

Capacitors in series, unless there are circuit changes when capacitors are already charged, should bear the same charge as you indicate.

Is the solution that you present as Relevant equations your solution or someone else's?
 
  • #3
This is an example problem from one of my lectures, and I was not sure how they arrive to the solution.
Originally the C23 capacitor, was C2 and C3 capacitors in parrallel. They combined them into Ceq by additing them two.
 
  • #4
Can you list all the initial capacitor values and the initial circuit conditions?
 
  • #5
https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11042003_10152682449531806_1029796180_n.jpg?oh=fcaa90dacb0f79caf862a01ccfe11094&oe=54F5C0C6&__gda__=1425387735_717d66c414ad18a91e2dbebad58aa77e
 
  • #6
Okay, it's much more clear now that I can see the whole circuit. Their derivation of the equivalent capacitance looks fine.

In your first post you showed and questioned the step:

Qeq = Q1 + Q23 = Ceq V

which I agree is incorrect. Series connected capacitors should have the same charge. So what would be your version of a solution?
 

FAQ: Charge and Voltage on Capacitors.

What is the relationship between charge and voltage on capacitors?

The charge and voltage on capacitors are directly proportional to each other. This means that as the charge increases, the voltage also increases, and vice versa. This relationship is described by the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

How does the charge on a capacitor affect its ability to store energy?

The charge on a capacitor is directly related to its ability to store energy. The more charge a capacitor has, the more energy it can store. This is because capacitance, which is a measure of a capacitor's ability to store charge, is directly proportional to the amount of charge on the capacitor. Therefore, the higher the charge, the higher the capacitance, and the more energy the capacitor can store.

What happens to the voltage on a capacitor when it is connected to a voltage source?

When a capacitor is connected to a voltage source, the voltage on the capacitor initially increases, but then levels off as the capacitor becomes fully charged. This is because the capacitor acts as an open circuit at first, and the voltage source charges it until the voltage across the capacitor is equal to the voltage of the source. Once the capacitor is fully charged, the voltage across it remains constant.

How does the capacitance of a capacitor affect its ability to hold charge?

The capacitance of a capacitor is directly proportional to its ability to hold charge. This means that the higher the capacitance, the more charge a capacitor can hold. This is because capacitance is a measure of a capacitor's ability to store charge, and a higher capacitance means that the capacitor can hold more charge for a given voltage.

Is the charge on a capacitor constant?

No, the charge on a capacitor is not constant. In fact, the charge on a capacitor can change over time, depending on the voltage across it. When a capacitor is connected to a voltage source, the charge on the capacitor increases until it reaches a maximum value. Similarly, when a capacitor is discharged, the charge on it decreases until it reaches zero. However, the total charge on a capacitor always remains the same, as it cannot be created or destroyed.

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