Solving a Complex Alternating Current Example: RLC Series Circuit Walkthrough

The Alternating Current Example: RLC Series Problem Statement

When studying AC circuit theory, textbooks often isolate resistors, inductors, or capacitors. But on the bench—and on your exams—they interact. To build real intuition, we need to look at a complete network where phase angles and reactances collide. Below is a classic, high-yield alternating current example that tests your grasp of complex impedance, phasor math, and power calculations.

Before touching a calculator, we map the given values to their impedance equivalents. In AC analysis, resistance is a real number, while inductive and capacitive reactances are imaginary components that shift the phase of the current. According to All About Circuits, treating these as vectors (phasors) is mandatory for accurate results.

Parameter	Symbol	Given Value	Calculated Reactance (Ω)	Phase Shift
Source Voltage	VRMS	120 V @ 60 Hz	N/A	0° (Reference)
Resistor	R	40 Ω	40.00 (Real)	0°
Inductor	L	150 mH	XL = 56.55	+90°
Capacitor	C	50 µF	XC = 53.05	-90°

Step-by-Step Solution: Phasor Domain Analysis

Which method applies and why? We use Phasor Analysis (Complex Ohm's Law). Because the circuit contains energy-storing components (L and C), the voltage and current are not in phase. Standard DC Ohm's law (V=IR) fails here. We must convert the frequency-dependent components into reactances and combine them geometrically using vector math. HyperPhysics provides an excellent foundational breakdown of why the RLC series impedance forms a right triangle in the complex plane.

Step 1: Calculate Angular Frequency (ω)

Reactance depends on the angular frequency, not just the Hertz value.

• ω = 2 × π × f
• ω = 2 × 3.14159 × 60 Hz
• ω = 377 rad/s

Step 2: Calculate Individual Reactances

Inductive reactance opposes changes in current, while capacitive reactance opposes changes in voltage.

• X_L = ω × L = 377 × 0.150 H = 56.55 Ω
• X_C = 1 / (ω × C) = 1 / (377 × 0.000050 F) = 1 / 0.01885 = 53.05 Ω

Step 3: Determine Total Impedance (Z)

Because X_L and X_C are 180° out of phase with each other, they subtract. The net reactance (X) is then combined with the resistance (R) using the Pythagorean theorem.

• Net Reactance (X) = X_L - X_C = 56.55 - 53.05 = 3.50 Ω (Inductive)
• Z = √(R² + X²)
• Z = √(40² + 3.50²)
• Z = √(1600 + 12.25)
• Z = √(1612.25)
• Z = 40.15 Ω

Step 4: Calculate RMS Current and Phase Angle

Now we apply AC Ohm's Law to find the current, and trigonometry to find the phase angle (θ).

• I_RMS = V_RMS / Z = 120 V / 40.15 Ω = 2.989 A
• θ = arctan(X / R) = arctan(3.50 / 40) = arctan(0.0875)
• θ = 5.0° (Current lags voltage, as the circuit is net inductive)

Step 5: Calculate Power Factor and Real Power

Real power (measured in Watts) is only consumed by the resistor. The inductor and capacitor merely exchange reactive power (VARs) with the source.

• Power Factor (PF) = cos(θ) = cos(5.0°) = 0.996 lagging
• Real Power (P) = V_RMS × I_RMS × PF
• P = 120 × 2.989 × 0.996 = 357.2 W

Sanity Checks and Independent Verification

Never hand in an exam or finalize a design without a sanity check. Here is how to verify the answer independently.

1. The Order of Magnitude & Resonance Check

Look at the reactances: X_L (56.55 Ω) and X_C (53.05 Ω) are very close to each other. This means the circuit is operating near its resonant frequency. At exact resonance, X_L and X_C cancel entirely, leaving Z = R. Because they almost cancel here (leaving only 3.5 Ω of net reactance), the total impedance Z must be just slightly larger than the 40 Ω resistor. Our calculated Z of 40.15 Ω perfectly aligns with this physical reality. If you had calculated a Z of 96 Ω, you would instantly know you added the reactances instead of subtracting them.

2. Independent Power Verification

We calculated Real Power using the source parameters (P = V × I × PF). We can independently verify this by looking strictly at the resistive component, since inductors and capacitors consume zero net real power.

• P_verify = I_RMS² × R
• P_verify = (2.989 A)² × 40 Ω
• P_verify = 8.934 × 40 = 357.3 W

The 0.1 W difference is purely due to rounding in the third decimal place during intermediate steps. The math holds up perfectly.

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