I have a question that I need a little help with.
A 5.0 uf capacitor is connected to a 3.5 uf capacitor, and both of them are connected to a 6 volt battery.
2) With the battery still connected, the 3.5 uf cap is accidentally shorted out. As a result, what are the charge and voltage of the 5.0 uf cap?
3) The shorted out cap is replaced by a new 3.5 uf cap; the system is charged up again and then the battery is disconnected. Again, the 3.5 uf cap is accidentally shorted out. What now is the charge and voltage of the 5.0 uf capacitor?
I think I may have 1) and 2) figured out, but I am not too sure about 3). Any help would be appreciated.
For 3), think about the charge before we disconnect the battery and short out the 3.5 uF cap.
On one side of the cap is a charge +Q and on the other side is a charge -Q. When we disconnect the battery, nothing should change. Why? Because the circuit was in DC before, and in DC capacitors are open circuits. So there were no currents. When we disconnect the battery there are still no currents, so no change in the ...
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