Series RLC Impedance Calculation

Hello! Welcome to your fourth lesson on AC circuits.

In our last lesson, we used phasor diagrams as a graphical tool to analyze the phase relationships in series RLC circuits. We saw how the individual voltage phasors for the resistor, inductor, and capacitor combine to determine the overall phase angle between the source voltage and current. This led us to the "voltage triangle" and the formula $\varphi =\arctan \left(\frac{X_L-X_C}{R}\right)$.

While graphical methods are intuitive, they can be cumbersome for calculation. Today, we will achieve the learning outcome: Calculate impedance for series RLC circuits using phasors or complex numbers. We will formalize the phasor concept using complex numbers, a powerful mathematical tool that combines magnitude and phase into a single entity. Given your background and current studies in pure mathematics, this lesson will provide a direct and powerful application of complex number theory to a physical system.

From Graphical Phasors to Complex Numbers

Representing sinusoidal AC quantities as rotating vectors (phasors) is effective, but adding them graphically is imprecise. The complex plane provides a perfect algebraic framework for this. A complex number $z=x+jy$ can represent a vector with a real component $x$ and an imaginary component $y$. We can use this to represent our phasors.

In electrical engineering, the imaginary unit is denoted by $j$ instead of $i$ to avoid confusion with current, $i(t)$.

Let's start with a short video that introduces this idea and defines the central concept of this lesson: impedance.

As the video explained, we can represent a sinusoidal voltage $V(t)=V_{peak}\cos (\omega t+\varphi )$ by a complex number whose magnitude is the RMS value ($V_{rms}=V_{peak}/\sqrt{2}$) and whose angle is the phase $\varphi$. The central idea is a generalization of Ohm's Law:

• In DC circuits: $V=IR$
• In AC circuits: $V=IZ$, where $V$, $I$, and $Z$ are all complex numbers.

Impedance ($Z$) is the measure of the total opposition to current flow in an AC circuit. It's a complex quantity that accounts for both resistance and reactance.
$Z=R+jX$
where:

• $R$ is the resistance, the real part of impedance.
• $X$ is the reactance, the imaginary part of impedance.

The Complex Impedance of R, L, and C

The real power of this method comes from defining a unique complex impedance for each component. We can derive these by assuming a sinusoidal current $I(t)$ and finding the resulting voltage $V(t)$ across each component. Using Euler's formula, $e^{j\omega t}=\cos (\omega t)+j\sin (\omega t)$, we can represent the current as the real part of a complex exponential: $I(t)=\text{Re}(I_0{e}^{j\omega t})$. For simplicity in calculations, we work with the full complex forms $I=I_0{e}^{j\omega t}$ and $V=V_0{e}^{j\omega t}$.

1. Resistor (R): The voltage is simply $V_R=I\cdot R$.
   The impedance is $Z_R=\frac{V_R}{I}=R$. It's purely real, indicating no phase shift.

2. Inductor (L): The voltage-current relationship is $V_L=L\frac{dI}{dt}$.
   If $I=I_0{e}^{j\omega t}$, then $\frac{dI}{dt}=j\omega (I_0{e}^{j\omega t})=j\omega I$.
   Substituting this in gives $V_L=L(j\omega I)=(j\omega L)I$.
   The impedance is $Z_L=\frac{V_L}{I}=j\omega L$. This is purely imaginary and positive. The $j$ factor represents the +90° phase shift (voltage leads current).

3. Capacitor (C): The relationship is $I=C\frac{d{V}_C}{dt}$, or $V_C=\frac{1}{C}\int Idt$.
   Integrating $I=I_0{e}^{j\omega t}$ gives $\int Idt=\frac{1}{j\omega }(I_0{e}^{j\omega t})=\frac{1}{j\omega }I$.
   So, $V_C=\frac{1}{C}\left(\frac{1}{j\omega }I\right)=\left(\frac{1}{j\omega C}\right)I$.
   The impedance is $Z_C=\frac{V_C}{I}=\frac{1}{j\omega C}$.
   Since $\frac{1}{j}=\frac{-j}{j(-j)}=-j$, we can write $Z_C=-j\frac{1}{\omega C}$. This is purely imaginary and negative, with the $-j$ factor representing the -90° phase shift (voltage lags current).

This derivation shows how the calculus of derivatives and integrals elegantly transforms into simple multiplication and division in the complex frequency domain.

Analyzing a Series RLC Circuit

One of the great advantages of complex impedance is that components in series are handled by simple addition, just like resistors in a DC circuit.

For an RLC circuit, the total impedance $Z_{total}$ is:
$Z_{total}=Z_R+Z_L+Z_C$
$Z_{total}=R+j\omega L+\frac{1}{j\omega C}$
$Z_{total}=R+j\left(\omega L-\frac{1}{\omega C}\right)$

This is a central result. The total impedance is a complex number where:

• The real part is the resistance $R$.
• The imaginary part is the total reactance $X_{total}=X_L-X_C=\omega L-\frac{1}{\omega C}$.

Rectangular and Polar Forms

We can express $Z_{total}$ in two ways:

1. Rectangular Form: $Z=R+j{X}_{total}$
2. Polar Form: $Z=|Z|\angle \varphi$

Where:

• The magnitude $|Z|$ is calculated using the Pythagorean theorem:
  $|Z|=\sqrt{R^2+X_{total}^2}=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$
  This magnitude gives the ratio of the RMS voltage to the RMS current: $|Z|=V_{rms}/I_{rms}$.
• The phase angle $\varphi$ is the argument of the complex number:
  $\varphi =\arctan \left(\frac{X_{total}}{R}\right)=\arctan \left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)$
  This is the same phase angle we found using the voltage triangle, but now it emerges directly from the algebra of complex numbers.

Worked Example

Let's apply this to a concrete problem. The following resource provides a clear, step-by-step example.

Let's summarize the steps from that example:

1. Given: A series RLC circuit with $R=400\Omega$, $L=400\text{mH}$, $C=200\mu \text{F}$, and a source voltage with $\omega =1000\text{rad/s}$.
2. Calculate Individual Impedances:
   • $Z_R=400\Omega$
   • $Z_L=j\omega L=j(1000)(400\times 10^{-3})=j400\Omega$
   • $Z_C=-j\frac{1}{\omega C}=-j\frac{1}{(1000)(200\times 10^{-6})}=-j5\Omega$
3. Calculate Total Impedance (Rectangular Form):
   • $Z=Z_R+Z_L+Z_C=400+j400-j5=400+j395\Omega$
4. Convert to Polar Form:
   • Magnitude: $|Z|=\sqrt{400^2+395^2}\approx 562\Omega$
   • Phase Angle: $\varphi =\arctan (395/400)\approx 44.6^{\circ }$
   • So, $Z\approx 562\Omega \angle 44.6^{\circ }$

The final result tells us everything we need to know: the total opposition to current is $562\Omega$, and the voltage will lead the current by $44.6^{\circ }$ (an inductive circuit).

Test your understanding!

Consider a series RLC circuit with $R=100\Omega$, $L=20\text{mH}$, and $C=50\mu \text{F}$.

At what angular frequency $\omega$ will the total impedance $Z$ be purely resistive (i.e., its imaginary part is zero)? What is the value of the impedance at this frequency?

Show answer

The total impedance is $Z=R+j\left(\omega L-\frac{1}{\omega C}\right)$.

For the impedance to be purely resistive, the imaginary part must be zero:
$\omega L-\frac{1}{\omega C}=0$
$\omega L=\frac{1}{\omega C}$
${\omega }^2=\frac{1}{LC}$
$\omega =\frac{1}{\sqrt{LC}}$

This specific frequency is known as the resonant frequency. Plugging in the values:
$\omega =\frac{1}{\sqrt{(20\times 10^{-3}\text{H})(50\times 10^{-6}\text{F})}}=\frac{1}{\sqrt{1\times 10^{-6}}}=1000\text{rad/s}$

At this frequency, the impedance is simply the resistance, since the reactances cancel out:
$Z=R=100\Omega$

Conclusion

Today we have transitioned from a graphical understanding of AC circuits to a more powerful and efficient algebraic one using complex numbers.

Key Takeaways:

• Complex impedance ($Z$) generalizes resistance to AC circuits, combining resistance (real part) and reactance (imaginary part) into a single complex number: $Z=R+jX$.
• The impedances for individual components are:
  • Resistor: $Z_R=R$
  • Inductor: $Z_L=j\omega L$
  • Capacitor: $Z_C=1/(j\omega C)=-j/(\omega C)$
• For a series RLC circuit, the total impedance is the simple sum: $Z_{total}=R+j(\omega L-1/\omega C)$.
• The magnitude of the total impedance, $|Z|$, determines the ratio of voltage to current amplitudes ($V/I$).
• The argument (angle) of the total impedance, $\varphi$, is the phase angle between the voltage and the current.

Preview of the Next Lesson:

Now that we have a solid grasp of impedance, we are ready to tackle the crucial topic of power in AC circuits. You've likely seen power ratings on appliances given in Watts (W), but in AC systems, you'll also encounter Volt-Amps (VA) and VAR. In our next lesson, we will explore real, reactive, and apparent power, and see how the impedance triangle we've implicitly worked with today has a direct analogue in the "power triangle". This will be essential for understanding power factor and efficiency in real-world systems.