why do capacitors look like open circuit at low frequency in mid band model?
If you think about it, a capacitor will charge to the DC value across it thereby acting as a barrier to DC. Therefore when choosing a capacitance you make it so it blocks low frequency and yet pass the frequency of interest. When coupling amplifiers a coupling capacitor is used to separated there respective DC bias and still pass the frequency of interest.
capacitor has reactance of XC=1/2*PI*f*C where f is equal to da frequency . now if u consider a signal of low frequency than what happens is XC tends to infinity as 1/0 is infinity.. so infinity resistance is present and so it becomes an open circuit... if wrong plz correct
True, I tried to explain without the mathematical expression so as to get a general feel for how capacitors work. When considering the band of frequency that a RC network will pass you must consider it according to the equivalent time constants and find what is known as the 3 dB points that are used to set a margins for the pass band.
what sort of materials is the capacitor made of...does it store electricity, photons. or some other emf frequency...
The basic capacitor is two metallic plates separated by a insulating material so apparent current passing through the capacitor actually builds up charge on the capacitor. Basic Equation CV = Q C capacitance V voltage across the capacitor Q the charge differential across the capacitor Differential C (dV/dt) = dQ/dt = i
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