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Physics: Capacitors and Potential difference?
Hi! Below is an attached image of the problem.
Before the switches are closed, I know that the potential difference between two plates (Correct me if I'm wrong).
According to one of my peers, charge redistribution takes time for capacitors (and immediately for normal circuits).
Since I dont understand 2a, there is no way I can understand 2b. Can someone explain 2a? I don't really get what it is asking. What is charged and what is uncharged in the diagram?
Sorry for this mess. Thanks in advance to anyone who can help!
- WhoLv 72 months agoFavorite Answer
the capacitor on the left is charged
("Ideal wire" has zero resistance - "real wire" has resistance)
1) immediately after closing there will be no PD between the bottom of the capacitors
2) current has to flow between the top of the LD capacitor and the top of the RH capacitor
with ideal wire this will take zero time (cos there is nothing to limit the current flow)
But with real wire that current will be limited by the resistance of the wire - so it will take time to fully charge
i.e with "real wire" it will take some time for the RH capacitor to "charge up" - whereas with an ideal wire it will take zero time
- DixonLv 72 months ago
Well, it is using idealised component symbols in the diagram and then asking about a non ideal circuit. Despite the diagram given, the question actually wants you consider that when the switches have both just closed, you get this, where r is the lumped resistance of the wires;