little boy Posted August 23, 2011 Share Posted August 23, 2011 My link This is a the page of my exercise. After the switch is moved to position B, it says a parallel circuit is established, but circuit seems not completed because one terminal of the cell is not connected. Link to comment Share on other sites More sharing options...
TonyMcC Posted August 23, 2011 Share Posted August 23, 2011 (edited) The question concerns the following actions. First you charge C1 to 12V using the battery. By moving the switch to position B you disconnect the battery and use the charged capacitor to put charge into capacitors C2 and C3. At this stage you can ignore the 12V battery. Just considering the capacitors, you have a network that can be described as C1 in parallel with C2 and C3 which are in series. So how much charge does C1 accept when the switch is in position A? How does that charge get shared by the 3 capacitors when the switch is moved to position B? Edited August 23, 2011 by TonyMcC Link to comment Share on other sites More sharing options...
little boy Posted August 24, 2011 Author Share Posted August 24, 2011 now I understand, thankyou Link to comment Share on other sites More sharing options...
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