Easy Capacitor question

  • #1
The related question says:http://[ATTACH=full]199761[/ATTACH]

[ATTACH=full]199762[/ATTACH]

The capacitor 1 is initially charged by having the flying lead attached to point C. Then, the flying lead is moved to point D. Capacitor 1 discharges.

C2 has no charge intially.

When the [B]flying lead is attached to D[/B], which one of these statements is true?

[LIST]
[*]Current flows from Y to C2
[*]Current flows from C2 to Y
[*]There is no current flowing between C2 and Y
[/LIST]
Explain why you have chosen this option.

I think it is option C. Because the path R2 and C2 the current could take has higher resistance than the current just passing through R2. But, my question is wouldn't the C2 get charged and hence making it the second option true.? Please help.
 

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  • #2
phinds
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First off, do you not understand the concept of a voltage divider? Second off, do you think the cap C2 is an open circuit when you first move the switch to D ?

Also, your image doesn't show up for me.
 
  • #3
First off, do you not understand the concept of a voltage divider? Second off, do you think the cap C2 is an open circuit when you first move the switch to D ?

Also, your image doesn't show up for me.
Yes. I know what a voltage divider is. But, how can you apply that on capacitors?
Plus, I think it is a closed circuit. But, wouldn't some current flow through capacitor 2 as that path has less resistance?
 
  • #4
phinds
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Yes. I know what a voltage divider is. But, how can you apply that on capacitors?
Plus, I think it is a closed circuit. But, wouldn't some current flow through capacitor 2 as that path has less resistance?
If it is a closed circuit, how can current NOT flow through it?
 
  • #5
If it is a closed circuit, how can current NOT flow through it?
Is this theory correct, Sir? current always passes through a path that low resistance. Since resistor has a resistance. Would it pass through Y to Capacitor 2? Is this correct?
 
  • #6
CWatters
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I think it is option C. Because the path R2 and C2 the current could take has higher resistance than the current just passing through R2.
Option C is not correct.

If you have two parallel paths electricity doesn't just take one with the least resistance - it's shared in proportion. So even if the C2 path has higher resistance than the R2 path some will still flow into C2.

But, my question is wouldn't the C2 get charged and hence making it the second option true.? Please help.
C2 would be charged (at least for awhile) but option 2 says "Current flows from C2 to Y". That's not correct.

Typically when a capacitor charges through a resistor you get a voltage curve that is asymptotic towards the supply voltage. That's not true in this case. What voltage is C2 charging towards?
 
  • #7
Option C is not correct.

If you have two parallel paths electricity doesn't just take one with the least resistance - it's shared in proportion. So even if the C2 path has higher resistance than the R2 path some will still flow into C2.



C2 would be charged (at least for awhile) but option 2 says "Current flows from C2 to Y". That's not correct.

Typically when a capacitor charges through a resistor you get a voltage curve that is asymptotic towards the supply voltage. That's not true in this case. What voltage is C2 charging towards?
I don't know the voltage. But, as you said, that's only one that makes sense. Thank you!
 
  • #8
Option C is not correct.

If you have two parallel paths electricity doesn't just take one with the least resistance - it's shared in proportion. So even if the C2 path has higher resistance than the R2 path some will still flow into C2.



C2 would be charged (at least for awhile) but option 2 says "Current flows from C2 to Y". That's not correct.

Typically when a capacitor charges through a resistor you get a voltage curve that is asymptotic towards the supply voltage. That's not true in this case. What voltage is C2 charging towards?
Can you also summarise a reason why?
 
  • #9
NascentOxygen
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To see what the final voltage that a particular capacitor is aiming for, as a general rule you an imagine: if that capacitor were not in the circuit what would be the voltage across the two points where we see it connected. Then that will be the voltage it is charging towards.

If you thoughtfuly apply that concept here, you'll get the answer.
 

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