Capacitor charging and discharging

In summary, the conversation discusses a circuit problem involving charging and discharging a capacitor. The main points discussed include the use of total capacitance and resistance, the role of the switch in connecting to different junctions, and how to simplify the circuit. The conversation also reveals a lack of understanding on basic circuit concepts and suggests going back to study the fundamentals.
  • #1
Prince1281
26
1

Homework Statement


1. In the circuit shown below calculate how long itwill take to charge the capacitor C2 to 4 V and how long it will take then to discharge it to 1V.
upload_2014-11-5_17-51-34.png

Homework Equations


upload_2014-11-5_17-51-53.png


The Attempt at a Solution


I tried to solve it by taking t=0 when v=0 to get the k which is then ln(VBat). But I am confused if the resistance are connected in series or parallel? Can someone confirm. And for the C in RC do I take the total capacitance or just the C2? Please someone confirm. Many thanks
 
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  • #2
I'm guessing for charging, the switch is connected to a and for discharging the switch is connected to b? Can you specify is that is the case?
 
  • #3
I have no idea sir. This is exactly the information I have. I think its supposed to be some general case. I have weak understanding of this whole circuit concept.

Is the total resistance of the circuit R3+R2 then parallel with R1 or is it R1+R2+R3?
 
  • #4
Well, I think it would be fair to assume the switch connects to the "a" junction only during charging and to the "b" junction only during discharging. But there could be a possibility for both "a" and "b" to be connected at the same time. Without knowing for sure which case we are talking about, we can't do this problem.

If indeed "a" is connected during charging and "b" is connected for discharging, you have two different circuits, can you draw the two different circuits?
 
  • #5
So if its connected to junction a, we can ignore R3 right? can we ignore anything else for the charging part
 
  • #6
Yes, for "a" you can ignore R3, and for "b" you can ignore the voltage source. That's the point of the switch really.

But the capacitors are connected in parallel, what can you do about that?
 
  • #7
But it asks only for C2 so shouldn't C1 be ignored somehow?
 
  • #8
You can't just ignore it...it's part of the circuit. But it is in parallel with C2, so that should tell you something! What do parallel circuit elements have in common?
 
  • #9
Well capacitors got charge common when in parallel...Not sure how that helps. Explain please
 
  • #10
Charge on a capacitor is not equal when in parallel... any circuit elements in parallel has equal voltage across them.

Do you know how to simplify circuits?
 
  • #11
Matterwave said:
Charge on a capacitor is not equal when in parallel... any circuit elements in parallel has equal voltage across them.

Do you know how to simplify circuits?
Oh yeah man sorry you're right
I suck at simplifying...if I ignore r3 I will have capacitors in parallel. But what about resistors will they be in parallel too?

And also how does having same voltage help...
 
  • #12
Prince1281 said:
Oh yeah man sorry you're right
I suck at simplifying...if I ignore r3 I will have capacitors in parallel. But what about resistors will they be in parallel too?

And also how does having same voltage help...

The capacitors are in parallel, then the rest of the circuit is a series circuit. It seems that you are missing some very basic knowledge on circuits. I would suggest you go back and study the basics before trying to solve this more complicated problem.
 
  • #13
Matterwave said:
The capacitors are in parallel, then the rest of the circuit is a series circuit. It seems that you are missing some very basic knowledge on circuits. I would suggest you go back and study the basics before trying to solve this more complicated problem.
No its fine. I just don't understand how can I use only c2 when both the capacitors are in parallel. Just because they share same voltage does it mean I can ignore c1?
 
  • #14
Prince1281 said:
No its fine. I just don't understand how can I use only c2 when both the capacitors are in parallel. Just because they share same voltage does it mean I can ignore c1?

You can't ignore c1, but you can turn C1 and C2 into an equivalent capacitor.
 
  • #15
Matterwave said:
You can't ignore c1, but you can turn C1 and C2 into an equivalent capacitor.
 
  • #16

Yes we can combine the capacitors in parallel and get a total capacitance. So is the total capacitance to be used with the total resistance in the relevant equation that I uploaded for this problem in order to get the charging time. Please confirm. Thanks
 
  • #17
Do you know what is the value of a capacitor equivalent to C1 and C2 in parallel?
 
  • #18
Yeah its 156 nf.
 

Related to Capacitor charging and discharging

1. What is a capacitor and how does it charge and discharge?

A capacitor is an electronic component that can store and release electrical energy. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to the capacitor, one plate accumulates positive charge and the other accumulates negative charge. This process is known as charging. When the voltage is removed, the capacitor discharges, releasing the stored energy.

2. What factors affect the charging and discharging of a capacitor?

The charging and discharging of a capacitor is affected by several factors, including the capacitance value, voltage applied, and the resistance of the circuit. The capacitance value determines how much charge the capacitor can store. A higher voltage will result in a faster charging and discharging process. The resistance of the circuit also plays a role, as a higher resistance will slow down the charging and discharging process.

3. How does a capacitor behave in a DC circuit?

In a DC circuit, a capacitor initially acts as an open circuit, blocking the flow of current. As the capacitor charges, it behaves as a short circuit, allowing current to flow. Once fully charged, the capacitor acts as an open circuit again, preventing further current flow. When the voltage is removed, the capacitor discharges, releasing the stored energy.

4. What is the time constant of a capacitor charging or discharging?

The time constant of a capacitor charging or discharging is a measure of how quickly the capacitor reaches its maximum charge or discharge. It is calculated by multiplying the resistance of the circuit (in ohms) by the capacitance value (in farads). A larger time constant indicates a slower charging or discharging process, while a smaller time constant indicates a faster process.

5. How is the energy stored in a capacitor calculated?

The energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage applied. This formula shows that the energy stored in a capacitor is directly proportional to both the capacitance value and the square of the applied voltage.

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