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In the given figure, initially switch is closed. On opening the switch the charge that will flow through the battery will be ![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FVLEBT_RootRepository%2FResources%2Fe25b51ec-072b-4585-a22c-aaccd2faea34.jpg&hash=de47c2bcb55f6a3bd3a9eb72d1f0c882)
![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FBTStudent%2FAssessmentBuilder%2FAMSResources%2FStylesheets%2FTemplates%2FXSLTemplates%2FSkins%2FImages%2FImages%2Fhorizontal_line.gif&hash=6b21877ed7daecfc3b92b1a2bfe7849a)
![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FVLEBT_RootRepository%2FResources%2F17274075-ed96-4afc-9890-f252237211d0.jpg&hash=74de69bdbce53c280054cf758e07f0e7)
![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FVLEBT_RootRepository%2FResources%2Fa0a57bcc-6043-4f97-882e-43ce06f8659e.jpg&hash=85a89b022ddc160bb3075c0e47259eaa)
![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FVLEBT_RootRepository%2FResources%2F59b829a4-b7ca-4884-a7e4-2d671261eae7.jpg&hash=8578f6f30f52754032992ca5e2a0c5f3)
![[IMG]](proxy.php?image=http%3A%2F%2F203.196.176.41%2FVLEBT_RootRepository%2FResources%2F097ca71d-4c37-4e9f-b1f5-5e9ca619d90f.jpg&hash=20d324d0dbae85a3360c24ddf5a9c341)
Replies
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reachrkataZero (assuming ideal caps)!!!
Anything you do with only capacitors on DC sources makes no difference.
-Karthik
😁 -
piyushhhmm it will be zero ..because..just after or before opening the switch the capacitor(connected close to voltage source in series) will behave like an open circuit..finally no charge that will flow through the battery on opening the switch just after it was been closed for a long time(ideally infinte)
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