When studying for electrical exams or diagnosing bench prototypes, abstract formulas rarely stick until you apply them to a concrete scenario. A classic example of alternating current analysis involves calculating the behavior of a series Resistor-Inductor (RL) circuit. Unlike DC circuits where resistance is the only opposition to current flow, AC circuits introduce frequency-dependent reactance, shifting the phase between voltage and current.

In this walkthrough, we will solve a standard exam-style problem. We will apply AC Ohm's Law using Phasor Domain Analysis to find the total impedance, RMS current, real power, and power factor. Every algebraic step is shown, followed by an independent verification method to ensure your exam answers are bulletproof.

The Problem: A Practical Example of Alternating Current Analysis

Problem Statement:

A 120V RMS, 60Hz AC voltage source is connected in series with a 40Ω resistor and a 79.58mH inductor. Calculate:

  1. The total circuit impedance (Z).
  2. The RMS current (IRMS).
  3. The real power dissipated (P) in watts.
  4. The power factor (PF).

The Method and The Trap

Which theorem applies and why? We use AC Ohm's Law in the phasor domain. Because the circuit contains an inductor, the opposition to current is a complex number (impedance) rather than a simple scalar (resistance). We must calculate the inductive reactance (XL) and combine it vectorially with the resistance.

⚠ The Trap in This Problem: The most common mistake students make here is calculating Apparent Power (S = VRMS × IRMS) and labeling it as Real Power (Watts). In an AC circuit with reactance, voltage and current are out of phase. Multiplying RMS voltage by RMS current gives you Volt-Amps (VA), not Watts. To find Real Power, you must either multiply Apparent Power by the Power Factor, or use the formula P = IRMS² × R, which inherently isolates the resistive (heat-dissipating) component.

Step-by-Step Solution: Applying AC Ohm's Law

Step 1: Calculate the angular frequency (ω)
Alternating current frequency (f) is given in Hertz. We need angular frequency in radians per second for reactance formulas.

  • ω = 2πf
  • ω = 2 × 3.14159 × 60 Hz
  • ω ≈ 377 rad/s

Step 2: Calculate the inductive reactance (XL)
Inductors oppose changes in current. This opposition scales linearly with frequency.

  • XL = ωL
  • XL = 377 rad/s × 0.07958 H (converting millihenries to henries)
  • XL = 30 Ω

Step 3: Calculate total impedance (Z)
Resistance and reactance are 90 degrees out of phase. We cannot simply add them (40 + 30 = 70 is wrong). We must use the Pythagorean theorem to find the magnitude of the impedance vector.

  • Z = √(R² + XL²)
  • Z = √(40² + 30²)
  • Z = √(1600 + 900)
  • Z = √2500
  • Z = 50 Ω

Step 4: Calculate the RMS current (IRMS)
Now we apply Ohm's Law using the RMS voltage and the total impedance magnitude.

  • IRMS = VRMS / Z
  • IRMS = 120V / 50Ω
  • IRMS = 2.4 A

Step 5: Calculate the real power (P)
Only the resistor dissipates real power (as heat). The inductor stores and releases energy but dissipates zero net real power.

  • P = IRMS² × R
  • P = (2.4 A)² × 40Ω
  • P = 5.76 × 40
  • P = 230.4 W

Step 6: Calculate the power factor (PF)
The power factor is the ratio of real power to apparent power, which geometrically equals the cosine of the phase angle (θ). In a series circuit, it is simply R / Z.

  • PF = R / Z
  • PF = 40Ω / 50Ω
  • PF = 0.8 (or 80% lagging)

Sanity Check & Independent Verification

How do you verify this answer independently on an exam without just re-doing the same math? Use the Power Triangle. According to Georgia State University's HyperPhysics, the relationship between Apparent Power (S), Real Power (P), and Reactive Power (Q) must satisfy the Pythagorean theorem: S² = P² + Q².

Let's test our numbers:

  1. Apparent Power (S): VRMS × IRMS = 120V × 2.4A = 288 VA
  2. Reactive Power (Q): IRMS² × XL = (2.4)² × 30 = 5.76 × 30 = 172.8 VAR
  3. Real Power (P): We calculated this as 230.4 W

The Verification Equation:

  • S² = P² + Q²
  • 288² = 230.4² + 172.8²
  • 82,944 = 53,084.16 + 29,859.84
  • 82,944 = 82,944

The equation balances perfectly. This confirms that our impedance magnitude, current, and power calculations are mathematically sound and phase-aligned. If you had mistakenly used Z = 70Ω in Step 3, this power triangle check would immediately fail, alerting you to the error before you handed in your exam.

Frequently Asked Questions (FAQ)

What is a real-world example of alternating current in a home?

The most common example of alternating current in a residential setting is the power delivered to your wall outlets. In North America, this is a 120V RMS, 60Hz sine wave. When you plug in a device with an electric motor (like a refrigerator compressor or a HVAC blower), you are creating a real-world RL (Resistive-Inductive) circuit. The motor windings act as the inductor, which is why heavy appliances often have a lower power factor and require careful breaker sizing to handle the apparent power (VA) rather than just the real running wattage.

Why do we use RMS values in an example of alternating current calculations?

RMS (Root Mean Square) is used because it provides the DC-equivalent heating value of an AC waveform. If a 120V RMS AC source is applied to a 40Ω resistor, it will generate the exact same amount of heat (230.4W) as a 120V DC battery connected to that same resistor. If we used the peak voltage of the AC sine wave (which is roughly 170V for a 120V RMS system), our power calculations would be artificially inflated and would not reflect the actual thermal reality of the circuit. For a deeper dive into the calculus behind this, refer to Electronics Tutorials on RMS Voltage.

How does an example of alternating current differ from DC when hitting a resistor?

Fundamentally, a resistor behaves the same way in both domains: it opposes current flow and dissipates heat according to I²R. The difference lies in the surrounding circuit context. In a DC circuit, a resistor limits current based purely on its ohmic value. In an AC circuit, the resistor's voltage drop is perfectly in phase with the current, but if there are capacitors or inductors in the same loop, the *total* current becomes phase-shifted relative to the source voltage. Therefore, while the resistor's internal physics don't change, the mathematical framework required to analyze its voltage drop shifts from simple scalar algebra to complex vector (phasor) math.