- #1
Master J
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An AC circuit consists of 3 segments: a capacitor, then another capacitor which is in parallel with a resistor, and then finally another capacitor, these 3 segments being in series with each other.
The formula which describes the charge Q(t) on the left hand side plate of the first capacitor is A.sin(wt). I need to find the formula which describes the charge on the right hand plate of the last capacitor (no.3).
Attempt:
I tried working in currents I. So the equation above becomes wAcos(wt). The total impedance Z of the 2nd segment is Z_1.Z_2 / Z_1 + Z_2, which is just R / ( 1 + RwC ).
I figured that the current at the last capacitor would be V / Z = I, where V is the potential. Since the current and voltage are out of phase by 90 degrees in the first capacitor, I got V = A.cos(wt).
So From this I get I, and just integrate to get Q(t). I don't think I am correct tho. I always have trouble with electrical circuits. Could someone perhaps point me in the right way?
The formula which describes the charge Q(t) on the left hand side plate of the first capacitor is A.sin(wt). I need to find the formula which describes the charge on the right hand plate of the last capacitor (no.3).
Attempt:
I tried working in currents I. So the equation above becomes wAcos(wt). The total impedance Z of the 2nd segment is Z_1.Z_2 / Z_1 + Z_2, which is just R / ( 1 + RwC ).
I figured that the current at the last capacitor would be V / Z = I, where V is the potential. Since the current and voltage are out of phase by 90 degrees in the first capacitor, I got V = A.cos(wt).
So From this I get I, and just integrate to get Q(t). I don't think I am correct tho. I always have trouble with electrical circuits. Could someone perhaps point me in the right way?