DC. They are then disconnected from the source and from each other and are reconnected to each other in parallel but with the leads from the 4 microfarad capacitor interchanged. Find the final voltage across the parallel combination?
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The charge on the 6 µF cap is Q = CV = 6x120 = 720 µC
on the 4 µF, that is 480 µC
When they are cross-connected, there is a big spark, and 480 µC on the second cancels –480 on the first, leaving 240 µC on both.
Total C is now 10 µF, Q = CV, or V = Q/C = 240/10 = 24 volts
on the 4 µF, that is 480 µC
When they are cross-connected, there is a big spark, and 480 µC on the second cancels –480 on the first, leaving 240 µC on both.
Total C is now 10 µF, Q = CV, or V = Q/C = 240/10 = 24 volts
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Almost certainly 0V . capacitors of that kind of size capacitance will almost certainly be electrolytic and so 99% of the time polarised. The voltage that will initially be applied on the switch over is gonna be the charging voltage in reverse and reverse voltages of even a 1V reverse voltage will destroy the oxide layers on a plain electrolytic capacitor's plates- leading to capacitor failure. And that's not even considering the massive current flow- that's another failure mechanism.
Also no clear definition of final time has been given so I think it would have to be when the voltage stops changing. All real capacitors have leakage so the only time the capacitor voltages will stop changing is after an infinite time, when it theoretically reaches 0V
Also no clear definition of final time has been given so I think it would have to be when the voltage stops changing. All real capacitors have leakage so the only time the capacitor voltages will stop changing is after an infinite time, when it theoretically reaches 0V