home
Mission
////////
home
////////

X_Complex Number

Tags

The usage of complex numbers in analyzing Alternating Current (AC) Resistor-Inductor-Capacitor (RLC) circuits to find impedance and phase angle

Session: May 2023
Page count: 20

Introduction and Aim

The exploration aim is to explore mathematical methodologies used to analyze AC circuits, and to demonstrate that mathematical methods using complex numbers may be more  efficient than those using only real numbers, even for real-life applications of mathematics.  Findings shall be verified experimentally to determine whether the theoretical methods used are accurate. The focus is on impedance and phase angle for resistors, inductors, and capacitors  connected to an AC power supply. This analysis explores a multitude of topics from the IB AA  HL curriculum, including Euler’s form of complex numbers, Maclaurin series, Argand  diagrams, vectors and phasors, trigonometric functions because of AC’s periodic motion,  differentiation, integration, as well as differential equations to highlight the need for complex numbers in AC circuit analysis. This investigation will first introduce the mathematical physics  required, then why complex numbers should be used, the impedances of the three individual  circuit components, their combined impedance represented visually, a worked example problem using our findings, and finally, an experimental verification.
I chose this exploration because I have a passion for Physics, especially for the electromagnetism topic. I have a keen interest in circuit building and have always wanted to  know more about the electronic devices we use in our daily lives. Personally, I find it fascinating how imaginary numbers may make real life applications of mathematics easier, especially in circuits which is arguably more focused on the experimental as opposed to the theoretical part of physics. To see whether this seemingly contradictory idea of using imaginary numbers in real-life applications is valid, I will build my own circuit to compare the experimental value to my manually-calculated value.

Contextualizing Physical Aspects

To thoroughly understand the mathematics applied in circuits, the physics behind it must briefly be established in the context of mathematics. Alternating current (AC) is current, or moving charges, that changes direction periodically (Tsokos). AC is sinusoidal (Figure 1) as electrons oscillate with a frequency, which means instantaneous voltage VV and current ii can be defined with trigonometric equations using angular frequency. Angular frequency is defined  by ω=2πf\omega=2\pi f, where ω\omega has the unit rads⎺¹, and ff is the frequency in Hertz (Tsokos). Voltage,  also known as potential difference, is defined as the electrical potential energy transferred from an electron during its movement from one point in the circuit to another, and is measured in Volts (Tsokos). Current is defined as the rate of change of charge q, is measured in Amperes, and can be algebraically represented through Equation 1: i=ΔqΔt=dqdti=\frac{\Delta q}{\Delta t}=\frac{dq}{dt} where tt is the time interval in which charge flows (Tsokos). Chosen values of ii and ff in Figure 1 are examples. We have established that voltage and current are sinusoidal as AC is sinusoidal. In the standard sine function form f(x)=asin⁡(bx−d)+Kf(x)=a\sin(bx-d)+K, aa is the amplitude, hence we can say the following where V0V_0 and i0i_0 are the maximum/peak voltage and current, respectively:
V(t)=V0sin⁡(ωt)V(t)=V_0\sin(\omega t) and i(t)=i0sin⁡(ωt)i(t)=i_0\sin(\omega t)
Figure 1: Sinusoidal current in AC-powered circuits. Graphed on Desmos
Resistors are circuit components which resist, or oppose, the flow of charges, and resistance is measured in ohms and denoted by RR (Tsokos). Ohm’s Law relates resistance, voltage, and current: R=ViR=\frac Vi. Inductors are circuit components which oppose change in charge through storing inductance, which is measured in Henry and denoted by LL (Electronics  Tutorials Editors. “The”). Inductance is defined by the following equation with VLV_L being the induced voltage of the inductor, Equation 2: VL=LΔiΔt=LdidtV_L=L\frac{\Delta i}{\Delta t}=L\frac{di}{dt} . A negative sign sometimes accompanies this equation to represent the opposition of motion. This explains why induced current lags voltage by a phase difference of π2\frac \pi 2, because the derivative of a negative sinusoidal ii is a negative cosine graph, meaning a cosine graph reflected over the xx-axis (Figure 2). Capacitors are circuit components which can store charge across parallel plates, where capacitance is measured in Farad and denoted by CC (Tsokos). Ohm’s Law for a capacitor states the following, with VCV_C being the induced voltage of the capacitor, i=CdVcdti=C\frac{dV_c}{dt}(All About Circuits Editors), leading to Equation 3: VC=1C∫i dtV_C=\frac1C\int i~dt, as capacitance is constant in a given circuit. This relation explains that the induced current leads voltage by π2\frac \pi 2 because the derivative of the sinusoidal ii is cosine (Figure 3). The chosen values of current and voltage are examples. These components’ circuit symbols are depicted below in a series RLC circuit diagram (Figure 4).
Figure 2: current (green) lags VLV_L( (red) by π2\frac \pi 2. Graphed on Desmos
Figure 3: current (blue) leads VCV_C (red) by π2\frac \pi 2. Graphed on Desmos
Figure 4: RLC circuit with an AC power supply (bottom)
Impedance, denoted by ZZ and measured in ohms, is similar to resistance as it is also a measure of opposition of flow of charge and can hence be defined as Z=V0i0Z=\frac{V_0}{i_0}. However, it also has complex components from the capacitor and inductor (Analog Devices Editors). The complex characteristic of capacitors and inductors shall be proven later, however, in short, they are complex because voltage and current are defined by ω\omega, making them frequency-dependent. Impedance is made up of resistance and reactance, the latter of which is denoted by XX and is essentially the resistance arising in capacitors and inductors when supplied by AC power rather than Direct Current (DC) (TechTarget Contributor). To find the value of impedance in circuit analysis, we must find the magnitude of impedance, ∣Z∣|Z|.

AC Circuit Analysis without Complex Numbers: The Need for Complex Numbers

AC circuits may be analyzed without complex numbers as well; however, this becomes unnecessarily elaborate as it involves complicated calculus, which shall be demonstrated. By Kirchhoff’s Second Law, V(t)=VR+VL+VCV(t)=V_R+V_L+V_C (Tsokos), where VRV_R is the resistor’s voltage. By Ohm’s Law and the established sinusoidal characteristic of voltage,
V(t)=i(t)×R=V0sin⁡(ωt)V(t)=i(t)\times R=V_0\sin(\omega t)
VR=V0=iRV_R=V_0=iR as the voltage of the resistor is frequency-independent.
From Equation 2: VL=LdidtV_L=L\frac{di}{dt}
From Equation 3: VC=1c∫0τi dtV_C=\frac1c\int_0^\tau i~dt, where τ\tau represents the time interval being considered. Combining these to find combined voltage V(t)V(t), which is required to find impedance:
V0sin⁡(ωt)=iR+Ldidt+1C∫0τi dtV_0\sin(\omega t)=iR+L\frac{di}{dt}+\frac1C\int_0^\tau i ~dt
We want to find combined voltage in terms of current, so we attempt to get rid of the integral:
(V0sin⁡(ωt))′=(iR)′+(Ldidt)′+(1C∫0τi dt)′(V_0\sin(\omega t))'=(iR)'+(L\frac{di}{dt})'+(\frac1C\int_0^\tau i~dt)'
From the chain rule and standard derivatives, differentiating with respect to tt gives us:
ωV0cos⁡(ωt)=Rdidt+Ldi2d2t+1Ci\omega V_0\cos(\omega t)=R\frac{di}{dt}+L\frac{di^2}{d^2t}+\frac1Ci
Clearly, with second-order differential equations, circuit analysis becomes a bit tangled for our syllabus. Hence, the usage of complex numbers is necessary to simplify these operations.

Defining Euler’s Form and Complex Numbers

Now that the need for complex numbers has been established, we can detail how they shall be utilized within this exploration. As seen above, the variable ii has been used to represent current. Hence, j=−1j=\sqrt{-1} shall be used to represent complex numbers. It is extremely valuable that this is considered a constant as it is numerical, rather than another variable. Expressing physical quantities such as complex impedance which requires multiplication and division of variables becomes much easier with Euler’s form. Euler’s form can be found through the Maclaurin Expansion of exe^x, where ee is Euler’s number e≈2.71828e ≈ 2.71828, which is detailed below:
The Maclaurin series states f(x)=ex=∑k=0∞xkk!=1+x+x22!+x33!+⋅⋅⋅f(x)=e^x=\sum_{k=0}^\infty\frac{x^k}{k!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdot\cdot\cdot
If x=jθx=j\theta, θ\theta being the angle between the positive real axis and the imaginary axis for the complex number in an Argand diagram, as shall be discussed later, then:
f(jθ)=ejθ=∑k=0∞(jθ)kk!=1+jθ+(jθ)22!+(jθ)33!+⋅⋅⋅f(j\theta)=e^{j\theta}=\sum_{k=0}^\infty\frac{(j\theta)^k}{k!}=1+j\theta+\frac{(j\theta)^2}{2!}+\frac{(j\theta)^3}{3!}+\cdot\cdot\cdot
As j2=−1j^2=-1 , f(jθ)=f(j\theta)=
=1−θ22!+θ44!−θ66!+⋅⋅⋅=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdot\cdot\cdot This is the Maclaurin series for cos⁡θ\cos \theta
+j(θ−θ33!+θ55!−θ77!+⋅⋅⋅+j(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+\cdot\cdot\cdot Maclaurin series for sin⁡θ\sin \theta multiplied by jj
Hence, ejθ=cos⁡θ+jsin⁡θe^{j\theta}=\cos\theta+j\sin\theta and rejθ=r(cos⁡θ+jsin⁡θ)=rcisθre^{j\theta}=r(\cos\theta+j\sin\theta)=rcis\theta, where rejθre^{j\theta} is  Euler’s form, rr is the modulus of xx, and θ\theta is the argument of xx. This reinforces that Re(j)=rcos⁡θRe(j)=r\cos\theta. The real component of voltage is V(t)=V0cos⁡(ωt)V(t)=V_0\cos(\omega t) if we use V0=rV_0=r and ωt=θ\omega t=\theta.
Originally, I was confused by why instantaneous voltage and current, previously defined to be sinusoidal, may be defined using cosine when calculating impedance. However, by exploring this derivation of Euler’s form, I was able to understand why cosine is used for calculations of impedance – to find the real component of voltage. The imaginary component of voltage cannot be compared to experimental values, so this finding was critical.
In Euler’s form, we express voltage as Equation 4: V(t)=V0ejωtV(t)=V_0e^{j\omega t}. This shall be used for most calculations further on. There are also two other forms of complex numbers; if xx is a complex number, in Cartesian form it is expressed as x=a+jbx=a+jb, and in polar form it is expressed as x=rcis(θ)=rcos⁡(θ)+jrsin⁡(θ)x=rcis(\theta)=r\cos(\theta)+jr\sin(\theta). rr is calculated through r=a2+b2r=\sqrt{a^2+b^2}when converting from Cartesian into polar form or Euler’s form. θ\theta can be calculated through θ=tan⁡−1(ba)\theta=\tan^{-1}(\frac ba) when converting from Cartesian form.
As it shall be proven, Z=R+jXZ=R+jX, where Re(Z)=RRe(Z)=R, and Im(Z)=XIm(Z)=X where XX is comprised of XLX_L and XCX_C found in inductors and capacitors, respectively.

Impedance of a Resistor

From Equation 4 we know as VR=V0ejωtV_R=V_0e^{j\omega t}.
From Ohm’s Law and Equation 4, iR=VRR=V0ejωtRi_R=\frac{V_R}R=\frac{V_0e^{j\omega t}}{R}.
From Ohm’s Law for impedance, ZR=VRiR=V0ejωtV0ejωtR=11R=RZ_R=\frac{V_R}{i_R}=\frac{V_0e^{j\omega t}}{\frac{V_0e^{j\omega t}}{R}}=\frac1{\frac 1R}=R.
Hence, ZR=RZ_R=R. This reinforces that resistance is an only real component.

Impedance of an Inductor

We assume no internal resistance of the inductor itself, as the focus of this exploration is on the imaginary component, reactance.
We know VL=V0cos⁡(ωt)V_L=V_0\cos(\omega t) when we take the real component of voltage in polar form. We now have the voltage, and must find the current. From Equation 2, we also know:
didt=VLL\frac{di}{dt}=\frac{V_L}L and hence didt=V0cos⁡(ωt)L\frac{di}{dt}=\frac{V_0\cos(\omega t)}L
To find the current iLi_L from this formula, we must integrate both sides of this equation with indefinite integrals with respect to tt:
∫didt=∫(V0Lcos⁡(ωt)) dt\int \frac{di}{dt}=\int(\frac{V_0}{L}\cos(\omega t))~dt
The integral of cos⁡(x)\cos (x) is positive sin⁡(x)\sin(x), and from the ‘reverse chain rule’, the equation must be divided by ω\omega when we take V0V_0, LL, and ω\omega as constants:
iL=V0Lωsin⁡(ωt)+Di_L=\frac{V_0}{L\omega}\sin(\omega t)+D
The integrating factor DD can be calculated by substituting time t=0t=0, where ii is also 00 because charge cannot instantaneously flow when no time has passed. This means:
When t=0t=0 and i=0i=0, 0=V0Lωsin⁡(ω×0)+D0=\frac{V_0}{L\omega}\sin(\omega\times0)+D
As sin⁡0=0\sin 0 = 0, we can say D=0D=0, so iL=V0Lωsin⁡(ωt)i_L=\frac{V_0}{L\omega}\sin(\omega t)
Sine and cosine are cofunctions, so trigonometric equations using sine may also be expressed through cosine as cos⁡(π2−x)=sin⁡x\cos(\frac \pi 2-x)=\sin x. From Figure 2 we also know that current lags behind voltage with a phase difference of π2\frac \pi 2. As cosine is an even function, cos⁡(∅)=cos⁡(−∅)\cos(∅) = \cos(−∅). We can hence also express iLi_L( as the following, where ∅=π2−x∅ = \frac \pi2-x, and x=ωtx=\omega t:
iL=V0Lωcos⁡(−(π2−ωt))=V0Lωcos⁡(ωt−π2)i_L=\frac{V_0}{L\omega}\cos(-(\frac \pi2-\omega t))=\frac {V_0}{L\omega}\cos(\omega t-\frac \pi2)
By the Fourier transform, a phase difference of π2\frac \pi2 and the fact that current is frequency dependent implies that the reactance of an inductor must be an imaginary quantity (Schleider). At first, I was a little hesitant on how to continue to find ZLZ_L as voltage VLV _L was in Euler’s form, while current iLi_L was in polar form. However, I took some time to reflect and realized that rather than converting from polar form to Euler’s form by calculating the modulus and argument, it would be easier to simply redo the integration using Equation 4, which gave the voltage expressed in Euler’s form as seen below.
∫didt=∫(V0Lejωt) dt\int\frac{di}{dt}=\int(\frac{V_0}{L}e^{j\omega t})~dt
From the reverse chain rule, where jj is also a constant,
i=V0jLωejωti=\frac{V_0}{jL\omega}e^{j\omega t}
As ZL=VLiLZ_L=\frac{V_L}{i_L}, we can use the derived values for voltage and current:
ZL=V0ejωtV0jLωejωt=11jLω=jLωZ_L=\frac{V_0e^{j\omega t}}{\frac{V_0}{jL\omega}e^{j\omega t}}=\frac1{\frac1{jL\omega}}=jL\omega
Hence, ZL=jLωZ_L=jL\omega. This reinforces that inductive reactance is an imaginary component.

Impedance of a Capacitor

Once again, we assume no internal resistance of the capacitor itself.
From Equation 4 we know as VC=V0ejωtV_C=V_0e^{j\omega t}.
Rather than integration, which was used to calculate the impedance of an inductor, we must use differentiation to find the impedance of a capacitor.
From Equations 1 and 3 we know, as q=CVq=CV and i=ΔqΔti=\frac{\Delta q}{\Delta t},
iC=dqdt=ddt(CV0ejωt)i_C=\frac{dq}{dt}=\frac{d}{dt}(CV_0e^{j\omega t})
From the chain rule and derivative of ee, iC=jωCV0ejωti_C=j\omega CV_0e^{j\omega t}.
Expressed in polar form, the real component is iC=ωCV0cos⁡(ωt+π2)i_C=\omega CV_0\cos(\omega t+\frac \pi 2). The presence of ωt\omega t proves that current depends on angular frequency. The addition of π2\frac \pi 2 is justified by Figure 3, as current is depicted to lead the voltage by a phase difference of π2\frac \pi 2. By the Fourier transform, a phase difference of π2\frac \pi 2 implies that this must be an imaginary quantity (Schleider).
As ZC=VCiCZ_C=\frac {V_C}{i_C}, we can combine our findings of VCV_C and iCi_C to give:
ZC=V0ejωtjωCV0ejωt=1jωCZ_C=\frac{V_0e^{j\omega t}}{j\omega CV_0e^{j\omega t}}=\frac 1{j\omega C}
Here, the denominator is complex because of jj. This makes it difficult to work with as we are unable to separate the real and imaginary components of ZCZ_C. In the Cartesian form of complex numbers, if a+jba+jb is a complex number, the conjugate will be a−jba-jb. However, in this case, there is no purely real component as aa is zero, hence the conjugate will simply be  −jb-jb, where bb is ωC\omega C. This means we multiply both the numerator and denominator on the right hand side of the equation by −jωC-j\omega C, giving −jωC−jωC\frac{-j\omega C}{-j\omega C} which can simplify to 11 on the left-hand side:
ZC=1jωC×−jωC−jωC=−jωC−j2ω2C2Z_C=\frac 1{j\omega C}\times\frac{-j\omega C}{-j\omega C}=\frac{-j\omega C}{-j^2\omega ^2C^2}
Hence, ZC=−jωCZ_C=-\frac j{\omega C} by j2=−1j^2=-1. This reinforces that capacitive reactance is an imaginary component.

Combining Impedance, Visual Representations, and Phase Angle

We can say that total impedance of the circuit is equal to the sum of impedances of individual components, or Ztot=ZR+ZL+ZCZ_{tot}=Z_R+Z_L+Z_C.
To reiterate, the individual impedances are:
ZR=RZ_R=R, ZL=jLωZ_L=jL\omega and ZC=−jωCZ_C=-\frac j{\omega C} where ZL+ZC=XZ_L+Z_C=X, ZL=jXLZ_L=jX_L and ZC=jXCZ_C=jX_C
Expressing this in a single equation,
Ztot=R+jLω+(−jωC)Z_{tot}=R+jL\omega+(-\frac j{\omega C})
Equation 5: Ztot=R+j(Lω−1ωC)Z_{tot}=R+j(L\omega -\frac1{\omega C})
This proves that, as aforementioned, impedance is a complex quantity whereby Re(Z)=RRe(Z)=R and Im(Z)=XL−XCIm(Z)=X_L-X_C.
We may also visually represent this on an Argand diagram with phasors to simplify problem-solving for equations. An Argand diagram allows plotting of complex numbers. The xx-axis from a Cartesian plane is the real axis, and the yy-axis from a Cartesian plane represents the imaginary axis on an Argand diagram. For example, the complex number x=4+5jx=4+5j may be represented as shown in Figure 5 below.
Figure 5: Example of plotting on an Argand diagram. Graphed on Desmos
Phasors are essentially vectors made specifically for AC quantities and have similar characteristics to vectors. However, it is important to note that the modulus, or magnitude, of  phasors refers to the root-mean-squared (rms) values of voltage and current, whereas vectors use the peak values of voltage and current (Electronics Tutorials Editors. “Phasor”). rms values are the equivalent DC values when regarding power-dissipation in AC and DC (Tsokos).
To represent impedance, we can plot RR on the real axis, and XL−XCX_L-X_C on the imaginary axis. As XLX_L is positive and XCX_C is negative, we can say XCX_C is the negative part of the imaginary axis and plot the value of XCX_C on the negative imaginary axis (Figure 6). Let us use an example to highlight how one may calculate impedance from an Argand diagram. Say XL=7X_L=7 ohms, and XC=2X_C=2 ohms. By phasor addition, we can conclude that the resultant phasor is 5 units upwards in the positive imaginary axis direction (green on Figure 6). Say R=4R=4 ohms (blue on Figure 6). We may plot these two phasors on an Argand diagram and use the tip-to-tail method  of phasors to find the resultant phasor (red on Figure 6). Impedance is the modulus, or magnitude of the resultant phasor. This is represented algebraically by the following:
Equation 6: ∣Z∣=R2+(XL−XC)2=R2+(Lω−1ωC)2|Z| = \sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+(L\omega-\frac 1{\omega C})^2}
In this example, we can conclude ∣Z∣=42+52≈6.403|Z| = \sqrt{4^2+5^2}≈ 6.403 ohms.
We can also use an Argand diagram to find the phase angle of our circuit. This is solved through θ=tan⁡−1(ba)\theta=\tan^{-1}(\frac ba). In this example, we can conclude θ=tan⁡−1(54)≈0.896\theta = \tan^{-1}(\frac 54) ≈ 0.896 rad.
Figure 6: Impedance plotted on an Argand diagram. Graphed on Desmos
Before exploring impedances of individual components and then combining them algebraically, I had briefly looked into visual representations of finding impedance. However, investigation into these visual representations had left me confused as to why XCX_C was in the  negative part of the imaginary axis of the Argand diagram. Hence, algebraically deriving these impedances was essential in clarifying to me why it was negative on diagrams. We can now use this method to easily solve problems involving impedance.

Applying Knowledge: Example Problem & Parallel Circuits

After deriving the impedance and phase angle for a series circuit, I started wondering how one may analyze a parallel circuit using these concepts with real values. As it is difficult to experimentally determine phase angle, for the following sections there will be an emphasis on impedance only. In a series circuit, impedance is rather easy to find: substitute the values in  the equation we have derived, Equation 6, using the fact that current is equal for all components  in a series circuit. I remembered that voltage is constant through a parallel circuit, but that current is divided into ‘branches’, and that the total current must be equal to the sum of the  currents of each branch (Tsokos). For parallel circuits, 1Rtotal=∑n=1k1Rn\frac 1{R_{total}}=\sum_{n=1}^k\frac1{R_n}, so the reciprocal of the total resistance equals the sum of the reciprocals of the resistances of individual components (Tsokos). As impedance is complex resistance, we can say:
1Ztot=1ZR+1ZL+1ZC\frac1{Z_{tot}}=\frac1{Z_R}+\frac1{Z_L}+\frac1{Z_C} and so Ztot=11ZR+1ZL+1ZCZ_{tot}=\frac1{\frac1{Z_R}+\frac1{Z_L}+\frac1{Z_C}}
Replacing this with the individual component impedances, we can say: Ztot=11R+j(1ωL−ωC)Z_{tot}=\frac1{\frac 1R+j(\frac1{\omega L}-\omega C)}
We now multiply by the conjugate to find a real denominator:
Ztot=11R+j(1ωL−ωC)×1R−j(1ωL−ωC)1R−j(1ωL−ωC)Z_{tot}=\frac1{\frac1R+j(\frac1{\omega L}-\omega C)}\times\frac{\frac 1R-j(\frac1{\omega L}-\omega C)}{\frac1R-j(\frac1{\omega L}-\omega C)}
Ztot=1R−j(1ωL−ωC)(1R)2−(j(1ωL−ωC))2Z_{tot}=\frac{\frac1R-j(\frac1{\omega L}-\omega C)}{(\frac1R)^2-(j(\frac1{\omega L}-\omega C))^2}
As j2=−1j^2=-1,
Ztot=1R−j(1ωL−ωC)1R2+(1ωL)−ωC)2Z_{tot}=\frac{\frac 1R-j(\frac1{\omega L}-\omega C)}{\frac1{R^2}+(\frac 1{\omega L})-\omega C)^2}
Ztot=1R1R2+(1ωL−ωC)2−j(1ωL−ωC)1R2+(1ωL−ωC)2Z_{tot}=\frac{\frac1R}{\frac1{R^2}+(\frac1{\omega L}-\omega C)^2}-j\frac{(\frac1{\omega L}-\omega C)}{\frac1{R^2}+(\frac1{\omega L}-\omega C)^2}
To find the impedance, we may find the modulus of this expression:
Equation 7: ∣Zparallel∣=(1R1R2+(1ωL−ωC)2)2+(−(1ωL−ωC)1R2+(1ωL−ωC)2)2|Z_{parallel}|=\sqrt{(\frac{\frac1R}{\frac1{R^2}+(\frac1{\omega L}-\omega C)^2})^2+(-\frac{(\frac1{\omega L}-\omega C)}{\frac1{R^2}+(\frac1{\omega L}-\omega C)^2})^2}
To avoid further complicating this expression, we will not expand this and hence leave it in this form for further calculations using real values. In our example problem giving a numerical answer, I will be using the following values which were chosen based on feasibility in experimentally verifying findings:
•
R=20R=20 ohms
•
C=010C=010 F
•
L=5L=5 H
•
f=1f=1 Hz
Before substituting in values to find the impedance, the calculation of angular frequency  ω=2πf\omega=2\pi f is required:
ω=2π(1)=2π\omega=2\pi(1)=2\pi in exact value
Now we can substitute these values to find the impedance:
∣Zparallel∣|Z_{parallel}|
=(1201202+(1(2π)(5)−(2π)(0.10))2)2+(−(1(2π)(5)−(2π)(0.10))1202+(1(2π)(5)−(2π)(0.10))2)2=\sqrt{(\frac{\frac1{20}}{\frac1{20^2}+(\frac1{(2\pi) (5)}-(2\pi)(0.10))^2})^2+(-\frac{(\frac1{(2\pi)(5)}-(2\pi)(0.10))}{\frac1{20^2}+(\frac1{(2\pi)(5)}-(2\pi)(0.10))^2})^2}
∣Ztot∣=1.67062|Z_{tot}| = 1.67062 ohms
I also decided to explore a different method of deriving the parallel impedance, using rules of operations with complex numbers and algebraic manipulation.
For parallel circuits, 1Ztot=1Z1+1Z2+1Z3+⋅⋅⋅+1Zn\frac1{Z_{tot}}=\frac1{Z_1}+\frac1{Z_2}+\frac1{Z_3}+\cdot\cdot\cdot+\frac1{Z_n}
For our purposes, 1Ztot=1ZR+1ZL+1ZC=ZLZC+ZRZC+ZRZLZRZLZC\frac1{Z_tot}=\frac1{Z_R}+\frac1{Z_L}+\frac1{Z_C}=\frac{Z_LZ_C+Z_RZ_C+Z_RZ_L}{Z_RZ_LZ_C}
Which gives us Equation 8: Ztot=ZRZLZCZLZC+ZRZC+ZRZLZ_{tot}=\frac{Z_RZ_LZ_C}{Z_LZ_C+Z_RZ_C+Z_RZ_L}
Replacing these variables with previously-derived expressions,
Ztot=(R)(jωL)(−j1ωC)(jωL×−j1ωC)+(R×−j1ωC)+(R×jωL)=−j2ωLRωCj2ωLωC−jRωC+jωLRZ_{tot}=\frac{(R)(j\omega L)(-j\frac 1{\omega C})}{(j\omega L \times -j\frac 1{\omega C})+(R\times -j\frac1{\omega C})+(R\times j\omega L)}=\frac{-j^2\frac{\omega LR}{\omega C}}{j^2\frac{\omega L}{\omega C}-j\frac R {\omega C}+j\omega LR}
We know j2=−1j^2=-1, so:
Ztot=LRCLC−jR(1ωC−ωL)Z_{tot}=\frac{\frac{LR}C}{\frac LC -jR(\frac 1 {\omega C}-\omega L)}
At first, I was confused as to why we now have XC−XLX_C-X_L( which differed from Equation  6 stating XL−XCX_L-X_C, so I took some time to explore how reciprocals of complex numbers are  represented on an Argand diagram, as highlighted below in Figure 7. While the magnitude is  the same, the direction of the reciprocal is as though reflected over the xx-axis. Using  information from Figure 6, we know the negative imaginary axis is −XC-X_C and the positive  imaginary axis is XLX_L. This means on the phasor diagram for parallel circuits, the expression  XL−XCX_L-X_C becomes −(XL−XC)-(X_L-X_C), resulting in XC−XLX_C-X_L. The argument, or phase angle, is the opposite of the same impedance as in a series circuit, where arg⁡(z)=2π−θ=−θ\arg(z)=2\pi-\theta=-\theta.
Figure 7: Reciprocals represented on an Argand diagram. Graphed on Desmos
Continuing from above, to have a real denominator, we multiply by the conjugate:
Ztot=LRCLC−jR(1ωC−ωL)×LC+jR(1ωC−ωL)LC+jR(1ωC−ωL)Z_{tot}=\frac{\frac {LR}C}{\frac LC -jR(\frac 1{\omega C}-\omega L)}\times \frac{\frac LC+jR(\frac 1{\omega C}-\omega L)}{\frac LC +jR(\frac 1{\omega C}-\omega L)} =LRC×LC+LRC×jR(1ωC−ωL)L2C2−j2R2(1ωC−ωL)2=\frac {\frac {LR}C\times \frac LC+\frac {LR}C\times jR(\frac 1 {\omega C}-\omega L)}{\frac {L^2}{C^2}-j^2R^2(\frac 1 {\omega C}-\omega L)^2}
=L2RC2+jLR2C(1ωC−ωL)L2C2+R2(1ωC−ωL)2=\frac{\frac{L^2R}{C^2}+j\frac {LR^2}C(\frac 1{\omega C}-\omega L)}{\frac {L^2}{C^2}+R^2(\frac 1 {\omega C}-\omega L)^2}
Separating these into imaginary and real components, we get:
Ztot=L2RC2L2C2+R2(1ωC−ωL)2+jLR2C(1ωC−ωL)L2C2+R2(1ωC−ωL)2Z_{tot}=\frac {\frac {L^2R}{C^2}}{\frac {L^2}{C^2}+R^2(\frac1{\omega C}-\omega L)^2}+j\frac{\frac{LR^2}C(\frac 1 {\omega C}-\omega L)}{\frac {L^2}{C^2}+R^2(\frac 1 {\omega C}-\omega L)^2}
While we could simplify this further, as the next step is simply substituting in our  example problem values this is unnecessary. We may now find the modulus of this to have a  formula for impedance:
Equation 9: ∣Zparallel∣=(L2RC2L2C2+R2(1ωC−ωL)2)2+(LR2C(1ωC−ωL)L2C2+R2(1ωC−ωL)2)2|Z_{parallel}|=\sqrt{(\frac {\frac{L^2R}{C^2}}{\frac{L^2}{C^2}+R^2(\frac 1 {\omega C}-\omega L)^2})^2+(\frac {\frac {LR^2}C(\frac 1{\omega C}-\omega L)}{\frac {L^2}{C^2}+R^2(\frac 1 {\omega C}-\omega L)^2})^2}
This appears more complicated than our previous equation, Equation 7, however, it is  still valuable to see whether we find the same result using this equation for our example  problem:
∣Zparallel∣=(52×200.102520.102+202(12π×0.10−2π×5)2)2+(5×2020.10(12π×0.10−2π×5)520.102+202(12π×0.10−2π×5)2)2|Z_{parallel}|=\sqrt{(\frac {\frac{5^2\times 20}{0.10^2}}{\frac{5^2}{0.10^2}+20^2(\frac 1 {2\pi\times 0.10}-2\pi\times 5)^2})^2+(\frac {\frac {5\times 20^2}{0.10}(\frac 1{2\pi\times 0.10}-2\pi\times 5)}{\frac {5^2}{0.10^2}+20^2(\frac 1 {2\pi\times0.10}-2\pi\times5)^2})^2}
∣Zparallel∣=1.67062|Z_{parallel}| = 1.67062 ohms, which is equal to the impedance we found with our previous  formula and verifies both my derivations.
I additionally wanted to explore how impedance varied between a series and parallel  circuit given the same values. Applying Equation 6, we have the answer:
∣Zseries∣=202+((5)(2π)−1(2π)(0.10))2|Z_{series}|=\sqrt{20^2+((5)(2\pi)-\frac 1{(2\pi)(0.10)})^2}
∣Zseries∣=35.9095|Z_{series}|=35.9095 ohms
This allows us to conclude that impedance is significantly higher when components are  connected in series than when they are connected in parallel. This was quite surprising, as I  had expected the impedance for series to be high, but did not expect such a large discrepancy. This doubt prompted me to compare my findings to the experimental values of an RLC circuit.

Comparison to Experimental Value

In order to determine the validity of my method in deriving these equations, I decided  to test these circuit components with experimental values on a simulated circuit (Figure 8). I  used the simulation software PhET Labs, which does not enable measurements of phase angle  so only the value of impedance will be experimentally determined. To be able to verify the  validity, I will use the same values as in the example problem. In parallel circuits, voltage is  constant across all components, while current varies (Tsokos). Hence, I had to find the current  through all components by connecting an ammeter in series with the individual components. By recording the simulation in slow-motion on my smartphone, I was able to estimate i0i_0 and V0V_0. These recorded measurements are below:
•
For all components: V0=20.0V_0=20.0 V
•
For resistor: i0=1.00i_0=1.00 A
•
For inductor: i0=1.17i_0=1.17 A
•
For capacitor: i0=12.17i_0=12.17 A
Figure 8: Experimentally testing impedance within a parallel circuit. Created with PhET Labs
By using Ohm’s Law for impedance, Z=V0i0Z=\frac{V_0}{i_0}, we can calculate the impedance of each component individually, and then combine them with the equation for resistance in parallel circuits:
ZR=201=20.0000Z_R=\frac {20}1=20.0000 ohms
ZL=201.17=17.0940Z_L=\frac{20}{1.17}=17.0940 ohms
ZR=2012.17=1.64339Z_R=\frac {20}{12.17}=1.64339 ohms
Now we combine these:
1Ztot=120.0000+117.0940+11.64339=0.716998\frac 1{Z_{tot}}=\frac 1{20.0000}+\frac 1{17.0940}+\frac 1{1.64339}=0.716998
Ztot=10.716998=1.39470Z_{tot}=\frac 1{0.716998}=1.39470 ohms
To verify this, I decided to substitute values of impedance into Equation 8 to determine the validity of my findings when using a different method:
Ztot=20.00000×17.0940×1.6433917.09400×1.64339+20.00000×1.64339+20.0000×17.09400=1.39470Z_{tot}=\frac {20.00000\times 17.0940\times 1.64339}{17.09400\times 1.64339+20.00000\times 1.64339+20.0000\times 17.09400}=1.39470 ohms
This is the same result I calculated from the previous equation. This confirms that both methods yield the same result. However, this is not quite the result we determined using Equations 7 and 9. There is a percentage error as follows:
percentage error=1.6702−1.394701.39470×100%=19.7832%\text{percentage error}= \frac {1.6702-1.39470}{1.39470}\times 100\%=19.7832\%
This is somewhat significant; however, it can be explained by the underlying aspects of the simulation. As aforementioned, I had to estimate the value of the peak current for all three components using slow-motion videotaping, so it is likely that there were some  inaccuracies in those areas. For example, the true value of i0i_0 for the inductor may have actually been 12.30 ohms, but I was simply unable to determine this due to the frequency being too fast  for the circuit simulation technology to display, or due to insufficiently slow slow-motion technology when recording values of i0i_0.

Conclusion, Limitations, and Extensions to Study

To reiterate, the aim of this exploration was exploring a variety of methodologies when analyzing AC RLC circuits, and to emphasize the critical value that complex numbers offer us for real-life applications of mathematics compared to the limitations of using only real numbers. Throughout this exploration, I have certainly achieved this aim. The contrast between using complicated second-order differential equations when utilizing only real numbers, compared to the linear relationships with phasors when using complex numbers, underlines  their value. Multiple methodologies have been demonstrated, for example when deriving the  formula for parallel circuits through two different methods, and then also experimentally verifying these results in two different methods.
These experimental verifications revealed that my derivation was more or less accurate, however, the percentage error of about 20% suggests there may have been inaccuracies when  measuring the values for peak current through individual components. This is a limitation of my exploration, as I could have chosen different software to record the peak current, or I could  have changed the value of frequency to make it more feasible to measure the peak current and hence have a more accurate comparison. Improving this limitation would allow me to be even more confident in stating that I have achieved my aim in using complex numbers to accurately analyze AC RLC circuits. Another limitation is the assumption of no internal resistance in the  inductors and capacitors; in reality, this is impossible, and slightly more complicated equations  would be required in realistic circuit analysis.
The main challenge I had throughout this exploration was being able to draw  information from essentially all domains of the Mathematics AA HL curriculum and  combining them in ways I was unfamiliar with. For example, I had not yet explored vectorial representations of complex numbers in depth, so independently exploring phasors was initially challenging. Another significant challenge I had was applying my knowledge of impedance in series circuits to parallel circuits. The confusion arose from realizing that there were quite a few different methodologies that one could use for the derivation, and wondering whether they would give the same result or not. This confusion was resolved by exploring multiple methods individually to see whether they gave the same result or not.
With newly gained knowledge from this IA, I can now comfortably approach  calculations regarding impedance and series and parallel circuits in my further studies, perhaps in studying physics in university. I am delighted to conclude that despite their name, complex  numbers often make applications of mathematics less complex through their versatility in  explaining many topics.
This exploration may be extended upon by examining impedance in a series-parallel combination circuit, or by taking away the assumption of zero internal resistance for inductors and capacitors.