4.5 Phasors: sinusoidal steady state and the series rlc circuit  (Page 2/2)

The voltage dropped across the capacitor is

The integral may be moved through the $Re\left[\right]$ operator to produce the result

Thus the phasor representation of $V_C\left(t\right)$ is

We call $\frac{1}{j\omega C}$ the impedance of the capacitor because $\frac{1}{j\omega C}$ is the complex scale constant that relates “phasor voltage $V_C$ " to “phasor current $I$ .”

Kirchhoff's Voltage Law. Kirchhoff's voltage law says that the voltage dropped in the series combination of $R$ , $L$ , and $C$ illustrated in [link] equals the voltage generated by the source (this is one of two fundamental conservation laws in circuit theory, the other being a conservation law for current):

If we replace all of these voltages by their complex representations, we have

An obvious solution is

where I is the phasor representation for the current that flows in the circuit. This solution is illustrated in [link] , where the phasor voltages $RI,j\omega LI$ , and $\frac{1}{j\omega C}I$ are forced to add up to the phasor voltage $V$ .

Impedance. We call the complex number $R+j\omega L+\frac{1}{j\omega C}$ the complex impedance for the series RLC network because it is the complex number that relates the phasor voltage $V$ to the phasor current $I$ :

The complex number $Z$ depends on the numerical values of resistance $\left(R\right)$ , inductance $\left(L\right)$ , and capacitance $\left(C\right)$ , but it also depends on the angular frequency $\left(\omega \right)$ used for the sinusoidal source. This impedance may be manipulated as follows to put it into an illuminating form:

The parameter ${\omega }_0=\frac{1}{\sqrt{LC}}$ is a parameter that you will learn to call an "undamped natural frequency" in your more advanced circuits courses. With it, we may write the impedance as

The frequency $\frac{\omega }{{\omega }_0}$ is a normalized frequency that we denote by $\nu$ . Then the impedence, as a function of normalized frequency, is

When the normalized frequency equals one $(\nu =1)$ , then the impedance is entirely real and $Z=R$ . The circuit looks like it is a single resistor.

The impedance obeys the following symmetries around $\nu =1$ :

In the next paragraph we show how this impedance function influences the current that flows in the circuit.

Resonance. The phasor representation for the current that flows the current that flows in the series RLC circuit is

The function $H\left(\nu \right)=\frac{1}{Z\left(\nu \right)}$ displays a "resonance phenomenon." that is, $\left|H\right(\nu \left)\right|$ peaks at $\nu =1$ and decreases to zero and $\nu =0$ and $\nu =\infty$ :

When $\left|H\right(\nu \left)\right|=0$ , no current flows.

The function $\left|H\right(\nu \left)\right|$ is plotted against the normalized frequency $\nu =$ $\frac{\omega }{{\omega }_0}$ in Figure 3.14. The resonance peak occurs at $\nu =1$ , where $\left|H\right(\nu \left)\right|=$ $\frac{1}{R}$ meaning that the circuit looks purely resistive. Resonance phenomena underlie the frequency selectivity of all electrical and mechanical networks.

Circle Criterion and Power Factor. Our study of the impedance $Z\left(\nu \right)$ and the function $H\left(\nu \right)=\frac{1}{Z\left(\nu \right)}$ brings insight into the resonance of an RLC circuit and illustrates the ffequency selectivity of the circuit. But there is more that we can do to illuminate the behavior of the circuit.

This equation shows how voltage is divided between resistor voltage RI and inductor-capacitor voltage $j(\omega L-\frac{1}{\omega C})I.$

or

In order to simplify our notation, we can write this equation as

where $V_R$ is the phasor voltage $RI$ and $k\left(\nu \right)$ is the real variable

[link] brings very important geometrical insights. First, even though the phasor voltage $V_R$ in the RLC circuit is complex, the terms $V_R$ and $jk\left(\nu \right)V_R$ are out of phase by $\frac{\pi }{2}$ radians. This means that, for every allowable value of $V_R$ , the corresponding $jk\left(\nu \right)V_R$ must add in a right triangle to produce the source voltage $V$ . This is illustrated in [link] . As the frequency $\nu$ changes, then $k\left(\nu \right)$ changes, producing other values of $V_R$ and $jk\left(\nu \right)V_R$ that sum to $V$ . Several such solutions for $V_R$ and $jk\left(\nu \right)V_R$ are illustrated in Figure 3.15(b). From the figure we gain the clear impression that the phasor voltage $V_{>R}$ lies on a circle of radius $\frac{V}{2}$ centered at $\frac{V}{2}$ Let's try this solution,

and explore its consequences. When this solution is substituted into [link] , the result is

or

If we multiply the left-hand side by its complex conjugate and the right-hand side by its complex conjugate, we obtain the identity

This equation tells us how the angle $\psi$ depends on $k\left(\nu \right)$ and, conversely, how $k\left(\nu \right)$ depends on $\psi$ :

The number $cos\psi$ lies between $-1$ and $+1$ , so a circular solution does indeed work.

The equation $V_R=\frac{V}{2}(1+e^{j\psi })$ is illustrated in [link] . The angle that $V_R$ makes with $V$ is determined from the equation

In the study of power systems, $cos\phi$ is a "power factor" that determines how much power is delivered to the resistor. We may denote the power factor as

But $cos\psi$ may be written as

But $cos\psi$ may be written as

Therefore the square of the power factor η is

The power factor is a maximum of 1 for $k\left(\nu \right)=0$ , corresponding to $\nu =1$ $(\omega ={\omega }_0)$ . It is a minimum of 0 for $k\left(\nu \right)=\pm \infty$ , corresponding to $\nu =0,\infty$ $(\omega =0,\infty )$ .