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Capacitor when connected in parallel act as fitler why?
Harshithshah
A capacitor impedance is effected by frequency; the capacitor can be modelled as: 1/(j*w*C), j = i = imaginary number, w = omega (frequency in radians), C = capacitance in Farads.
To understand filtering, you must look at the output vs. the input relationship - Vo/Vi.
If you have a capacitor in SERIES with a resistor, and your output is across the capacitor, this boils down to a simple voltage divider:
Vo = Vi*1/(j*w*C) / (R + 1/(j*w*C)) = Vi * 1 / ((j*w*C) + 1)
Therefore, Vo / Vi = 1 / ((j*w*C) + 1)
So you can see as the frequency increases (w increases), Vo will become smaller and smaller - thus this is a lowpass filter. Similarly, if the output is taken across the resistor you will find this is a high pass filter.
I'm not sure what you're meaning by connected in parallel. If I have misunderstood your question, please post an update to the answer or comment in the discussion area.
Wiki User
∙ 14y ago
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