Capacitor networks HELP!

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Yatayyat
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#1
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For this question that I am stuck on, I need to somehow find the voltage of both capacitors when current comes to a stop.

Q: A 470𝜇F capacitor is charged using a 10V battery. It is then disconnected, and connected to an uncharged 220𝜇F capacitor. Calculate the voltage across the capacitors once the current has stopped flowing. (Hint: capacitors are effectively in parallel, and total charge has not changed.)

What I have tried already is that I found that the total capacitance must be the sum of capacitances of the two capacitors because it is connected in parallel to each other. Hence '470𝜇F + 220𝜇F = 690𝜇F'

I really don't know what to do next now. I know that when it is disconnected from the battery and onto the next capacitor somehow voltage is being shared and the current is coming down, but I don't know what equation I can use to relate to that.

Any help would be really great. Thanks!
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username3249896
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#2
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The initial charge can be found.

When the 2 capacitors are connected in parallel, there will be flow of charge and a new voltage will be establised.

Since charge is conserved, the total charge stored by the capacitors must be equal to the inital charge
then C1V+C2V=Qo
can be used to find the new voltage
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Yatayyat
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#3
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(Original post by BobbJo)
The initial charge can be found.

When the 2 capacitors are connected in parallel, there will be flow of charge and a new voltage will be establised.

Since charge is conserved, the total charge stored by the capacitors must be equal to the inital charge
then C1V+C2V=Qo
can be used to find the new voltage
Thanks a lot! Makes sense
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mjg583
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#4
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(Original post by Yatayyat)
For this question that I am stuck on, I need to somehow find the voltage of both capacitors when current comes to a stop.

Q: A 470𝜇F capacitor is charged using a 10V battery. It is then disconnected, and connected to an uncharged 220𝜇F capacitor. Calculate the voltage across the capacitors once the current has stopped flowing. (Hint: capacitors are effectively in parallel, and total charge has not changed.)

What I have tried already is that I found that the total capacitance must be the sum of capacitances of the two capacitors because it is connected in parallel to each other. Hence '470𝜇F + 220𝜇F = 690𝜇F'

I really don't know what to do next now. I know that when it is disconnected from the battery and onto the next capacitor somehow voltage is being shared and the current is coming down, but I don't know what equation I can use to relate to that.

Any help would be really great. Thanks!
Does anyone have an explanation for the fact that the "capacitors are effectively in parallel"?

It appears to be implying that they are not in fact in parallel but act as if they are.

Also in the video, the instructor explains this procedure but draws two capacitors which are clearly in series and treats them as if they are in parallel.
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