Mastering the AC Impedance Trig Function Formula: Worked Examples and Sizing

The core trig function formula for determining the phase angle in an AC circuit is θ = arctan(X / R). This single relationship bridges the gap between the physical components on your workbench (resistors, inductors, capacitors) and the power triangle that dictates how efficiently your system transfers energy. Whether you are sizing a run capacitor for a compressor motor or calculating the voltage drop across a solar inverter output choke, this formula is your primary diagnostic tool.

The Core AC Impedance Trig Function Formula and Symbol Table

In alternating current (AC) theory, impedance (Z) is a vector quantity. It has a real component (Resistance, R) and an imaginary component (Reactance, X). The phase angle (θ) represents the time delay between the voltage waveform and the current waveform. The primary trig function formula that defines this relationship is:

θ = arctan(X / R)

Where total reactance X is the net sum of inductive and capacitive reactance: X = X_L - X_C.

Symbol	Parameter	Unit	Definition & Calculation
θ	Phase Angle	Degrees (°) or Radians	The angular displacement between voltage and current. Positive for inductive, negative for capacitive.
R	Resistance	Ohms (Ω)	Real power dissipation. Measured directly with a DC ohmmeter.
X	Net Reactance	Ohms (Ω)	X = XL - XC. The opposition to AC current change.
XL	Inductive Reactance	Ohms (Ω)	XL = 2πfL. Increases with frequency.
XC	Capacitive Reactance	Ohms (Ω)	XC = 1 / (2πfC). Decreases with frequency.
Z	Impedance Magnitude	Ohms (Ω)	Z = √(R² + X²). The total AC opposition.

Rearranged Forms and Variable Isolation

On the bench, you rarely have all the variables handed to you. You will need to isolate specific parameters based on what your multimeter or oscilloscope can measure. Here are the algebraic rearrangements of the impedance triangle:

• Solving for Resistance (R): R = X / tan(θ) or R = Z × cos(θ)
• Solving for Reactance (X): X = R × tan(θ) or X = Z × sin(θ)
• Solving for Impedance (Z): Z = R / cos(θ) or Z = √(R² + X²)
• Solving for Phase Angle (θ): θ = arctan(X / R) or θ = arccos(R / Z)
• Solving for Inductance (L): L = (R × tan(θ)) / (2πf) (Assuming purely inductive reactance)
• Solving for Capacitance (C): C = 1 / (2πf × R × |tan(θ)|) (Assuming purely capacitive reactance)

Application Boundaries, Assumptions, and Unit Traps

This trig function formula assumes steady-state sinusoidal AC and linear components. It breaks down if you are analyzing non-linear loads (like uncorrected LED drivers or VFDs generating heavy harmonic distortion) because the waveforms are no longer pure sine waves, rendering a single phase angle mathematically incomplete. For those, you must use true power factor calculations via a power analyzer.

Unit Mistakes That Will Break Your Calculation

1. Radians vs. Degrees: Your calculator must be in Degree mode when finding θ for power factor analysis. If your calculator is in Radians, arctan(1) yields 0.785 instead of 45°, which will completely destroy your subsequent power triangle math.
2. Prefix Blindness: Inductors are rated in milliHenries (mH) and capacitors in microFarads (μF). The formula requires base units (Henries and Farads). Plugging '15' into the X_L formula for a 15mH choke instead of '0.015' will inflate your reactance by 1,000%.
3. Peak vs. RMS Voltage: When using impedance to find current (I = V/Z), ensure V is in RMS (e.g., 120V or 240V), not peak voltage (170V or 340V), unless you specifically want peak current.

Worked Example 1: Power Factor Correction for an Inductive Motor

Scenario: You have a 120V, 60Hz single-phase compressor motor drawing 8A. A teardown measurement shows the main winding resistance is 10Ω. The utility is penalizing your shop for poor power factor, so you need to calculate the current phase angle and size a parallel run capacitor to bring the Power Factor (PF) up to 0.95.

Step 1: Find current Impedance (Z) and Reactance (X_L)

• Z = V / I = 120V / 8A = 15Ω
• X_L = √(Z² - R²) = √(15² - 10²) = √(225 - 100) = √125 = 11.18Ω

Step 2: Calculate the existing Phase Angle (θ)

• θ = arctan(X_L / R) = arctan(11.18 / 10) = arctan(1.118)
• θ = 48.19° (This yields a poor initial PF of cos(48.19°) = 0.66)

Step 3: Determine Required Reactive Power (Q) Compensation

• Real Power (P) = I² × R = 8² × 10 = 640W
• Current Reactive Power (Q_initial) = P × tan(48.19°) = 640 × 1.118 = 715.5 VAR
• Target Angle for PF 0.95: θ_target = arccos(0.95) = 18.19°
• Target Reactive Power (Q_target) = P × tan(18.19°) = 640 × 0.3286 = 210.3 VAR
• Required Capacitive VARs (Q_C) = 715.5 - 210.3 = 505.2 VAR

Step 4: Size the Capacitor

• Required X_C = V² / Q_C = 120² / 505.2 = 14400 / 505.2 = 28.50Ω
• C = 1 / (2πf × X_C) = 1 / (377 × 28.50) = 1 / 10744.5 = 0.00009307 F
• C = 93.1 μF

Worked Example 2: Phase Shift and Voltage Drop in an Inverter Choke

Scenario: You are building an LC low-pass filter for a 240V, 60Hz off-grid solar inverter. The load is a 24Ω resistive heating element. You place a 15mH series choke (inductor) on the output to smooth the PWM waveform. You need to know the phase shift introduced and the voltage actually reaching the load.

Step 1: Calculate Inductive Reactance (X_L)

• Convert 15mH to base units: 0.015 H
• X_L = 2πfL = 2 × π × 60 × 0.015 = 5.655Ω

Step 2: Calculate Total Impedance (Z) and Phase Angle (θ)

• Z = √(R² + X_L²) = √(24² + 5.655²) = √(576 + 31.98) = √607.98 = 24.66Ω
• θ = arctan(X_L / R) = arctan(5.655 / 24) = arctan(0.2356) = 13.26°

Step 3: Calculate Current and Load Voltage Drop

• Total Current (I) = V_source / Z = 240V / 24.66Ω = 9.73A
• Voltage dropped across the choke (V_L) = I × X_L = 9.73A × 5.655Ω = 55.0V
• Voltage reaching the load (V_R) = I × R = 9.73A × 24Ω = 233.5V

Bench Note: The choke is dropping 55V (nearly 23% of your source voltage). While the 13.26° phase shift is perfectly acceptable for a heating element, the voltage drop is too high. You must reduce the inductance to maintain proper load voltage.

Decision Path: Selecting the Right Compensation Component

Use this decision tree to terminate your trig function formula calculations into a concrete hardware selection. Do not guess component values; let the math dictate the bill of materials.

Measured / Calculated Condition	Diagnosis	Concrete Action & Part Pick
θ > +45° (PF < 0.70 Lagging)	Severe inductive lag. Motor is drawing excessive magnetizing current, wasting wire ampacity and tripping breakers on startup.	Add Parallel Run Capacitor. Pick: Genteq 90μF, 370V Round Run Capacitor (or calculated nearest standard value). Wire directly across the motor contactor load side.
θ is Negative (Leading PF)	Capacitive dominance. Common in lightly loaded underground cable runs or over-corrected motor banks. Risks voltage magnification.	Add Series Reactor or Remove Caps. Pick: Hammond 100mH AC Line Reactor to absorb leading VARs, or physically disconnect the lowest-step capacitor bank.
Inverter Choke VL Drop > 10% of Vsource	Filter inductance is too high for the load current. The trig formula proves the reactance is choking the real voltage.	Downsize Choke Inductance. Pick: 5mH Toroidal Iron Powder Core Choke (rewound or replaced) to drop XL to ~1.8Ω, restoring load voltage above 220V.
θ ≈ 0° but Current is High	Purely resistive overload. Reactance is zero, but real power exceeds circuit limits.	Upgrade Wire & Breaker. Pick: 10 AWG THHN copper wire on a 30A QO breaker. No capacitors or inductors will fix a pure resistive overload.

For further reading on the relationship between the impedance triangle and real-world power billing, refer to the Fluke guide on Power Factor and the Electronics Tutorials AC Power Triangle breakdown. Always verify your calculated component ratings against the physical nameplate voltage and temperature limits before energizing the circuit.