jlee said:If initially (before "s1 open, and s2 connected to ground" ) both capacitors have no charges,
then the final result is Vo = 7.5V
ptmsl said:Are you talking about ideal devices or real world capacitors? In the case of real world capacitors the output voltage will depend more on the leakage resistance of the two capacitors more than their values. For ideal devices the charge will devide the voltage if the capacitors have equal values.
jlee said:I assumed S1 and S2 are perfect switch... no leakage.
The final answer should be 7.5V because:
1. Initially, there was no charge on both capacitors. So voltage across both capacitors were zero.
2. After " s1 close, after enought time ", the right-side capacitor was charged to 5; accumulated charge on the top place is positive charge Q = 5*C. The left-side capacitor wass floating... no effect, no charge accumulated.-------WHY?
3. Now net charge on the top plates of two capacitors = 5*C. Consider conversation of charges.
4. After " open s1 and switch s2 to 10v", there was -2.5*C negative charge on the top plate of the left capacitor, and there was +7.5*C positive charge on the top plate of right capacitor. Total net charge was conserved. Therefore, the final Vo= +7.5V.
What is the answer your interviewer told you? If you think 7.5V is correct, please give me a help. Thanks.
so i think that we are finding the charge consewrvation at the top node (Vo-10) and not (10-Vo).eecs4ever said:"
When S1 is closed, 2*5C = 10 C amount of charge is stored on the top node.
eecs4ever said:"
if S2 were connected to 20v then , we can again solve
Qinitial = Qafter
(vo-20v) * C + vo * C = 10 C
=> vo = 15v
now is a voltage drop across the left capacitor, meaning that there is charge on it.
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