AC Circuit Examples: Solving Parallel RLC Networks with Phasors

When studying for electrical engineering exams or preparing for journeyman calculations, abstract theory only gets you so far. You need to grind through real ac circuit examples to build muscle memory for complex algebra and phasor manipulation. In this walkthrough, we will dissect a classic parallel RL and C network. We will not skip steps, we will highlight the exact trap that causes 80% of exam failures, and we will verify the final answer using a completely independent method.

Reference Data: 60Hz Reactance & Admittance Baselines

Before touching the algebra, we need to establish our baseline component values. In practical 60Hz AC systems (standard in North America), inductors and capacitors are often specified in Henrys and Farads, but exam problems frequently hand you the pre-calculated reactances. The table below maps the physical components to their complex impedance ($Z$) and admittance ($Y$) equivalents. Keep this reference handy; admittance is the secret weapon for parallel AC circuit examples.

Component	Physical Value	Reactance ($X$)	Impedance ($Z$)	Admittance ($Y = 1/Z$)
AC Voltage Source	120V RMS, 60Hz	N/A	$120\angle 0^\circ$ V	N/A
Resistor (Branch 1)	$3.0\ \Omega$	$0\ \Omega$	$3 + j0\ \Omega$	$0.333 + j0\ \text{S}$
Inductor (Branch 1)	$10.61\ \text{mH}$	$+4.0\ \Omega$	$0 + j4\ \Omega$	$0 - j0.25\ \text{S}$
Capacitor (Branch 2)	$663.1\ \mu\text{F}$	$-4.0\ \Omega$	$0 - j4\ \Omega$	$0 + j0.25\ \text{S}$

Note: Reactance values assume $\omega = 2\pi(60) \approx 377\ \text{rad/s}$. Data sourced from standard All About Circuits AC theory references.

Problem Statement & Method Selection

Which Method Applies and Why?

For parallel AC circuit examples, Phasor Domain Analysis using Impedance Reduction is the most direct method. While nodal analysis works, it is overkill for a two-branch parallel circuit. We will use the product-over-sum formula for parallel impedances: $Z_{eq} = \frac{Z_1 Z_2}{Z_1 + Z_2}$. We choose this over admittance summation ($Y_{eq} = Y_1 + Y_2$) simply because the problem provides clean integer impedances, making the complex fraction straightforward to resolve without dealing with repeating decimals early in the calculation.

Step-by-Step Phasor Algebra

We will solve this by defining branch impedances, reducing them to a single equivalent impedance, and then applying Ohm's Law in the phasor domain.

Step 1: Define Branch Impedances

Branch 1 (RL Series): $Z_1 = R_1 + jX_L = 3 + j4\ \Omega$

Branch 2 (C Only): $Z_2 = -jX_C = 0 - j4 = -j4\ \Omega$

Step 2: Calculate Equivalent Impedance ($Z_{eq}$)

Apply the parallel impedance formula:

$Z_{eq} = \frac{Z_1 \cdot Z_2}{Z_1 + Z_2}$

Numerator Calculation (Product):

$Z_1 \cdot Z_2 = (3 + j4)(-j4)$

$= 3(-j4) + (j4)(-j4)$

$= -j12 - j^2(16)$

Since $j^2 = -1$, this becomes: $-j12 - (-1)(16) = 16 - j12$

Denominator Calculation (Sum):

$Z_1 + Z_2 = (3 + j4) + (-j4)$

$= 3 + j(4 - 4) = 3 + j0 = 3$

Division:

$Z_{eq} = \frac{16 - j12}{3} = 5.333 - j4\ \Omega$

Convert $Z_{eq}$ to Polar Form:

Magnitude: $|Z_{eq}| = \sqrt{5.333^2 + (-4)^2} = \sqrt{28.444 + 16} = \sqrt{44.444} = 6.667\ \Omega$

Angle: $\theta = \arctan\left(\frac{-4}{5.333}\right) = \arctan(-0.75) = -36.87^\circ$

$Z_{eq} = 6.667\angle -36.87^\circ\ \Omega$

Step 3: Calculate Total Current ($I_{tot}$)

Using Ohm's Law ($I = \frac{V}{Z}$):

$I_{tot} = \frac{120\angle 0^\circ}{6.667\angle -36.87^\circ}$

Magnitude: $\frac{120}{6.667} = 18.0\ \text{A}$

Angle: $0^\circ - (-36.87^\circ) = +36.87^\circ$

$I_{tot} = 18.0\angle +36.87^\circ\ \text{A}$

Step 4: Determine Power Factor

Power Factor (PF) = $\cos(\theta_Z)$ or $\cos(\theta_V - \theta_I)$.

$PF = \cos(-36.87^\circ) = 0.80$

Because the total current angle ($+36.87^\circ$) is positive relative to the voltage reference ($0^\circ$), the current leads the voltage. Therefore, the power factor is 0.80 Leading.

Sanity Checks, Traps, & Independent Verification

Order of Magnitude & Unit Sanity Check

Does 18A make physical sense? Let's check the individual branch magnitudes. Branch 1 impedance is $5\ \Omega$ ($\sqrt{3^2+4^2}$), so $I_1 = \frac{120}{5} = 24\ \text{A}$. Branch 2 impedance is $4\ \Omega$, so $I_2 = \frac{120}{4} = 30\ \text{A}$. If we simply added the scalar magnitudes (24 + 30), we would get 54A. However, because the inductor and capacitor currents are 180 degrees out of phase with each other, they partially cancel. A total current of 18A is perfectly reasonable and mathematically sound. The unit is Amperes, matching the RMS voltage and Ohms.

Independent Verification via KCL

To prove our answer without relying on the equivalent impedance method, we will calculate branch currents individually and sum them using Kirchhoff's Current Law (KCL). This method is heavily endorsed in MIT's OpenCourseWare circuits curriculum as a secondary check.

1. Calculate Branch 1 Current ($I_1$):

$I_1 = \frac{120\angle 0^\circ}{3 + j4} = \frac{120\angle 0^\circ}{5\angle 53.13^\circ} = 24\angle -53.13^\circ\ \text{A}$

Convert to rectangular: $24\cos(-53.13^\circ) + j24\sin(-53.13^\circ) = 14.4 - j19.2\ \text{A}$

2. Calculate Branch 2 Current ($I_2$):

$I_2 = \frac{120\angle 0^\circ}{-j4} = \frac{120\angle 0^\circ}{4\angle -90^\circ} = 30\angle +90^\circ\ \text{A}$

Convert to rectangular: $0 + j30\ \text{A}$

3. Sum via KCL ($I_{tot} = I_1 + I_2$):

$I_{tot} = (14.4 - j19.2) + (0 + j30)$

$I_{tot} = 14.4 + j10.8\ \text{A}$

4. Convert back to Polar:

Magnitude: $\sqrt{14.4^2 + 10.8^2} = \sqrt{207.36 + 116.64} = \sqrt{324} = 18.0\ \text{A}$

Angle: $\arctan\left(\frac{10.8}{14.4}\right) = \arctan(0.75) = +36.87^\circ$

Result: $18.0\angle +36.87^\circ\ \text{A}$

The KCL verification perfectly matches the impedance reduction method. When you are practicing AC circuit examples for an exam, always use the impedance method for speed, but keep the KCL branch-summation method in your back pocket to verify your complex arithmetic if the numbers feel off.