Re: resonant circuit
mujee,
In a series circuit, the currents in all elements are equal (forgive me for stating the obvious). The fundamental relationship among voltage, inductance and time is: V(L) = Ldi/dt. This means that the voltage across the inductor leads the current by 90deg. (The derivative of the sin is the cosine). The fundamental relationship among voltage, capacitance and time is: I = C dv/dt. Solving for V, and ignoring the constant of integration, v=(1/C)∫Idt. This means that the voltage lags the current by 90deg. (The integral of the sine is the negative of the cosine).
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We now have two voltages that are exactly out of phase with each other. At some frequency (f), the inductive reactance Xl= 2pif will exactly equal the capacitive reactance Xc=1/(2piC). At this frequency, not only will the two voltages be exactly out of phase (This is true at all frequencies), but they will be equal in magnitude (The reactance are equal, and the currents are equal), so the current goes to infinity, the individual voltage magntudes go to infinity, and the net impedance across the capacitor-inductor combination is zero. This, of course is an impossible situation in the real world, However, if we insert a resistance into the series circuit, the the net impedance will simply be the resistance of the resistor, but the circuit will still be in resonance. The current will equal the applied voltage divided by the resistance of the resistor.
Regards,
Kral