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HCHopeful

Just doing one last run through of my notes. I was wondering if anyone could give me an answer to my question.

Since a different current flows through circuits in parallel, if we had a larger capacitor and a smaller capacitor, which gets more current?

Obviously in resistors the smaller magnitude gets more current, but I am unsure if that applies to capacitors as well.
 

Cawolf

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Well they will each have the same voltage drop across them.

Q = CV

If we look at the time derivative:

dQ/dt = d(CV)/dt = (C)dV/dt

i(t) = C(dV/dt) at each capacitor.

So the instantaneous current is time dependent on the change in voltage.

If you plotted voltage across the plates on the x-axis and i(t) (instantaneous current) on the y-axis you would see that the slope is C.

Therefore at some time t, the capacitor with the greatest capacitance, has the largest current.

These are my thoughts, so they are long because I am just applying some things I know - it is possibly incorrect.
 

Jumb0

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Well they will each have the same voltage drop across them.

Q = CV

If we look at the time derivative:

dQ/dt = d(CV)/dt = (C)dV/dt

i(t) = C(dV/dt) at each capacitor.

So the instantaneous current is time dependent on the change in voltage.

If you plotted voltage across the plates on the x-axis and i(t) (instantaneous current) on the y-axis you would see that the slope is C.

Therefore at some time t, the capacitor with the greatest capacitance, has the largest current.

These are my thoughts, so they are long because I am just applying some things I know - it is possibly incorrect.
That makes sense.
What about capacitors of unequal capacitance in series?
Circuit elements in series are supposed to have the same current. Does it follow that the series capacitor with the larger capacitance has a smaller voltage?
What about steady state condition? i.e. After each capacitor has reached its fully charged state? Which one will have a greater voltage?
 
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Cawolf

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Capacitors in series will always have the same charge. The voltage across them will vary.

V = Q/C so you are correct in saying that capacitance is inversely related to voltage drop.

When the capacitors are fully charged there is no flow of charge, current is 0.

There you once against just look at V = Q/C and see that if charge is constant, then the potential difference is smallest in the capacitors with the largest capacitance.
 

Cawolf

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I think that you seem to know it pretty well. Just remember your simple formulas and best of luck tomorrow - get some rest! :)
 
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