circuit analysiselectromagnetisminductorresonance
I have a circuit of the following form.
Circuit diagram:
How do I decide the resonant frequency ?
What is the criteria here to select the resonant frequency ?
Background: I am testing the transfer function for the system, and I need to know what is the proper value of "w" (omega) I can use.
Please let me know if anyone has the knowledge on this.
Thanks.
Your calculation of the impedance seen by the source is correct.
Clearly, there is a 'special' (angular) frequency
$$\omega_0 = \frac{1}{\sqrt{LC}}$$
where there is a pole in the impedance - the impedance goes to infinity.
Now, let's look at the dual of the circuit given:
simulate this circuit – Schematic created using CircuitLab
For the dual circuit, the impedance seen by the source is
$$Z = R||(j\omega L + \frac{1}{j \omega C}) = R \frac{1 - \omega^2LC}{1 - \omega^2LC + j\omega RC} $$
and now we have a zero at \$\omega_0\$ - the impedance goes to zero.
In both of these cases, the pole or zero is on the \$j \omega\$ axis. Generally, they are not.
so how do you find the resonance in general?
In this context (RLC), the resonance frequency is the frequency where the impedance of the inductor and capacitor are equal in magnitude and opposite in sign.
Update to address comment and question edit.
From the Wikipedia article "RLC circuit", "Natural frequency" section:
The resonance frequency is defined in terms of the impedance presented to a driving source. It is still possible for the circuit to carry on oscillating (for a time) after the driving source has been removed or it is subjected to a step in voltage (including a step down to zero). This is similar to the way that a tuning fork will carry on ringing after it has been struck, and the effect is often called ringing. This effect is the peak natural resonance frequency of the circuit and in general is not exactly the same as the driven resonance frequency, although the two will usually be quite close to each other. Various terms are used by different authors to distinguish the two, but resonance frequency unqualified usually means the driven resonance frequency. The driven frequency may be called the undamped resonance frequency or undamped natural frequency and the peak frequency may be called the damped resonance frequency or the damped natural frequency. The reason for this terminology is that the driven resonance frequency in a series or parallel resonant circuit has the value1
$$\omega_0 = \frac {1}{\sqrt {LC}}$$
This is exactly the same as the resonance frequency of an LC circuit, that is, one with no resistor present, that is, it is the same as a circuit in which there is no damping, hence undamped resonance frequency. The peak resonance frequency, on the other hand, depends on the value of the resistor and is described as the damped resonance frequency. A highly damped circuit will fail to resonate at all when not driven. A circuit with a value of resistor that causes it to be just on the edge of ringing is called critically damped. Either side of critically damped are described as underdamped (ringing happens) and overdamped (ringing is suppressed).
Circuits with topologies more complex than straightforward series or parallel (some examples described later in the article) have a driven resonance frequency that deviates from \$\omega_0 = \frac {1}{\sqrt {LC}}\$ and for those the undamped resonance frequency, damped resonance frequency and driven resonance frequency can all be different.
See the "Other configurations" section for your 2nd circuit.
In summary, the frequencies at which the impedance is real, at which the impedance is stationary (max or min), and at which the reactances of the L & C are equal can be the same or different and each is some type of resonance frequency.
In which condition the current through the inductor approach INFINITY and how can you explain that physically?
Yes, it is absolutely true - the current in the inductor can become infinite and the current in the capacitor can also become infinite however, because these two currents are anti-phase (one lags the sinewave source by 90deg and one leads the sinewave source by 90deg), the net current taken from an AC voltage source is zero other than what is consumed in the parallel resistor.
You are using a current source and the resultant voltage will be I*R. I'm also presuming you meant R to be a positive resistance value not as shown in your diagram although that won't alter things at all - if you meant R to be negative then the voltage developed across the resistor will be anti-phase to the current through it.
The actual current through the inductor is the voltage across it (as developed by the resistor and current source). Ditto the capacitor.
Best Answer
First find the equivalent impedance of your circuit using, e.g., Thevenin theorem. This impedance will depend on \$\omega\$. The resonant frequency(ies) is(are) the value(s) of \$\omega\$ that makes the reactive part of the impedance zero.