When studying for electrical exams or debugging bench prototypes, working through practical alternating current examples is the fastest way to internalize phasor math. Unlike DC circuits, AC analysis requires tracking both magnitude and phase angle. A common stumbling block is power factor correction in inductive loads. Below is a complete, decision-forward walkthrough of a classic AC problem: analyzing an RL load and sizing a parallel capacitor to correct the power factor.
The Problem Statement
Given: A 120V RMS, 60Hz AC source feeds a series RL load (modeling a small induction motor) with a resistance R = 15 Ω and an inductance L = 40 mH.
Tasks:
- Calculate the RMS current (
I_RMS), Real Power (P), and initial Power Factor (PF). - Determine the exact capacitance required in parallel to correct the power factor to 0.95 lagging.
- Select a real-world, off-the-shelf capacitor part number.
Method Selection and The "Radians vs. Degrees" Trap
Which theorem applies? We use Phasor Analysis (complex impedance). Because the source is sinusoidal and the circuit is in steady-state, phasors allow us to treat inductors and capacitors as complex resistors, turning differential equations into simple algebra.
The Trap: The most frequent error in alternating current examples like this is mixing up degrees and radians on your calculator. When calculating angular frequency (ω), you must use radians. When calculating the phase angle (θ) via the arctangent function, your calculator must output degrees to match standard power factor conventions. Furthermore, never use peak voltage (170V) when calculating RMS power; stick strictly to the 120V RMS value provided.
Step-by-Step Algebraic Solution
Let us break the math down without skipping any steps.
Step 1: Angular Frequency and Reactance
First, find the angular frequency (ω):
ω = 2πf = 2 × π × 60 = 376.99 rad/s
Next, calculate the inductive reactance (XL):
X_L = ωL = 376.99 × 0.040 H = 15.08 Ω
Step 2: Complex Impedance and RMS Current
The total series impedance (Z) is:
Z = R + jX_L = 15 + j15.08 Ω
Convert Z to polar form to find magnitude and phase:
|Z| = √(15² + 15.08²) = √(225 + 227.4) = √452.4 = 21.27 Ω
θ = tan¹(15.08 / 15) = tan¹(1.0053) = 45.15°
Now, apply Ohm's Law for AC to find the RMS current:
I_RMS = V_RMS / |Z| = 120V / 21.27 Ω = 5.64 A
Step 3: Real Power and Initial Power Factor
Real power (P) is only dissipated by the resistor:
P = I_RMS² × R = (5.64)² × 15 = 31.81 × 15 = 477.2 W
The initial power factor is the cosine of the impedance angle:
PF_initial = cos(45.15°) = 0.705 (lagging)
Decision Tree: Selecting the Correction Capacitor
To correct the power factor to 0.95, we must add a parallel capacitor to supply reactive power (Qc), reducing the burden on the grid. The real power (P) remains exactly 477.2 W.
Calculate Required Reactive Power:
Old Reactive Power (Q_old) = P × tan(45.15°) = 477.2 × 1.005 = 479.6 VAR
New Target Angle (θ_new) = cos¹(0.95) = 18.19°
New Reactive Power (Q_new) = P × tan(18.19°) = 477.2 × 0.3285 = 156.8 VAR
Required Capacitor Reactive Power (Q_c) = Q_old - Q_new = 479.6 - 156.8 = 322.8 VAR
Calculate Capacitance:
Q_c = V_RMS² / X_c = V_RMS² × ωC
C = Q_c / (V_RMS² × ω) = 322.8 / (120² × 376.99) = 322.8 / 5,428,656 = 59.4 μF
Now, we map this calculated value to a physical component using the decision matrix below.
| Decision Criteria | Condition | Component Choice |
|---|---|---|
| Capacitance Range | C ≈ 60 μF | Metallized Polypropylene Film (Motor Run) |
| Voltage Rating | 120V RMS × 1.414 = 170V Peak. Add 20% margin = 204V min. | Standard 370VAC Rating |
| Dielectric Type | Continuous AC waveform (bipolar) | Film (Never use polarized Electrolytic) |
| Final Concrete Pick | Closest standard E12 value | 60μF, 370VAC Genteq 97F9002 |
Sanity Checks and Independent Verification
Before wiring this up, we must verify the math.
- Order of Magnitude Check: 120V divided by roughly 20 Ω should yield around 6A. Our 5.64A answer is perfectly in range.
- Power Check: Apparent power (S) is 120V × 5.64A = 676.8 VA. Real power (477.2 W) must be less than apparent power. 477.2 < 676.8. The math holds.
- Units Check: We calculated VAR for reactive power, which correctly maps to Farads when divided by V² and rad/s.
How to verify independently on the bench:
Do not just trust the math. Wire the 15Ω power resistor and 40mH inductor in series. Plug the circuit into a 120V source via a digital power meter (like a Kill-A-Watt or a Yokogawa WT310). Read the initial PF (it should read ~0.71). Next, wire the 60μF Genteq capacitor in parallel with the RL series combination. The power meter should now display a PF between 0.94 and 0.96. If it reads leading (e.g., -0.95), you accidentally over-corrected; drop to a 50μF capacitor.
FAQ: Common Alternating Current Examples Pitfalls
Q: Can I just use a larger capacitor to get a power factor of exactly 1.0?
A: You can, but it is rarely recommended in practice. Correcting to exactly 1.0 requires a highly specific capacitance that changes if the motor load fluctuates. If the load drops and the capacitor remains, you push the circuit into a "leading" power factor, which can cause voltage swells and resonance issues with the utility grid. Targeting 0.95 lagging is the industry standard default.
Q: Does the resistor value change when I add the capacitor?
A: No. The resistor represents the mechanical work and heat losses of the motor. The parallel capacitor only alters the phase angle of the total current drawn from the source; it does not change the real power consumed by the load itself.
For further reading on AC power theory, consult the All About Circuits chapter on Power Factor or review the MIT OpenCourseWare Circuits and Electronics lecture notes on steady-state AC analysis.
