AC Impedance Capacitor Example: Step-by-Step Phasor Walkthrough

The final steady-state voltage for this capacitor example is 106.5 V peak at a -51.21° phase shift, derived using AC phasor analysis. When tackling linear AC circuits in steady-state, differential equations are unnecessarily tedious. Instead, we use complex impedance to turn calculus into basic algebra. Below is a complete, step-by-step walkthrough of a classic exam problem, including the real-world component data you need to contextualize the math.

The Core Capacitor Example: Problem Statement & Reference Data

Which method applies and why?
We use Phasor Analysis (Complex Impedance Method). Because the circuit is linear and driven by a single-frequency sinusoidal source, the steady-state response will also be sinusoidal at the exact same frequency. Phasor analysis allows us to represent resistors, capacitors, and inductors as complex numbers (impedances) and apply standard DC theorems—like Ohm's Law and the Voltage Divider Rule—directly to AC circuits.

Before diving into the algebra, it is crucial to understand the physical component we are modeling. A 22 μF capacitor is a very specific value that dictates the physical size, dielectric material, and Equivalent Series Resistance (ESR) of the part you would actually buy for a bench prototype or PCB design.

Dielectric Type	Typical Range	ESR (mΩ)	Max Temp (°C)	Common Application
Aluminum Electrolytic	1 μF – 1 F	50 – 500	85 – 105	Power supply filtering, bulk energy storage
Polypropylene Film	1 nF – 10 μF	< 5	85 – 105	Audio crossovers, AC motor run, snubber circuits
Ceramic (X7R/X5R)	100 pF – 100 μF	2 – 20	125	High-frequency decoupling, MLCC bypass
Tantalum	100 nF – 1 mF	20 – 200	125	Space-constrained DC filtering, medical devices

Data sourced from manufacturer specifications (e.g., Cornell Dubilier, Vishay). For our 22 μF example at 60 Hz, an Aluminum Electrolytic or Film capacitor would be the physical choice, though ideal phasor math assumes ESR is zero.

Step-by-Step Phasor Solution & The Common Trap

Step 1: Extract Phasor and Angular Frequency
The time-domain source is v_s(t) = 170 cos(377t).
Angular frequency (ω) = 377 rad/s (which corresponds to exactly 60 Hz, since ω = 2πf).
The phasor representation of the source is V_s = 170 ∠ 0° V.

Step 2: Calculate Complex Impedances
The resistor's impedance is purely real:
Z_R = 150 Ω

The capacitor's impedance is purely imaginary and negative:
Z_C = 1 / (jωC) = -j / (ωC)
Z_C = -j / (377 × 22 × 10^-6)
Z_C = -j / 0.008294
Z_C = -j120.57 Ω

Converting to polar form for easier division later:
Z_C = 120.57 ∠ -90° Ω

Step 3: Apply the AC Voltage Divider Rule
To find the voltage across the capacitor, we use the voltage divider formula:
V_c = V_s × [ Z_C / (Z_R + Z_C) ]

First, calculate the total series impedance (the denominator):
Z_total = 150 - j120.57 Ω
Magnitude: |Z_total| = √(150² + (-120.57)²) = √(22500 + 14537.12) = √37037.12 = 192.45 Ω
Phase Angle: θ = arctan(-120.57 / 150) = arctan(-0.8038) = -38.79°
So, Z_total = 192.45 ∠ -38.79° Ω

Now, multiply the numerator:
V_s × Z_C = (170 ∠ 0°) × (120.57 ∠ -90°) = 20496.9 ∠ -90°

Finally, divide the numerator by the denominator:
V_c = (20496.9 ∠ -90°) / (192.45 ∠ -38.79°)
Magnitude: 20496.9 / 192.45 = 106.5 V
Angle: -90° - (-38.79°) = -51.21°
V_c = 106.5 ∠ -51.21° V

Step 4: Convert Back to the Time Domain
Using the peak magnitude, the original angular frequency (377), and the calculated phase shift:
v_c(t) = 106.5 cos(377t - 51.21°) V

Sanity Checks, Independent Verification, and FAQ

Never trust an exam answer until you have run a sanity check. Here is how you verify the physics and the math independently.

1. Order of Magnitude & Units Check
The unit is Volts, which is correct. For magnitude, the source is 170 V peak. Because the capacitor's impedance magnitude (120.57 Ω) is less than the total circuit impedance (192.45 Ω), the capacitor must drop less voltage than the total source. The ratio is 120.57 / 192.45 ≈ 0.626. Multiplying 0.626 by 170 V yields 106.42 V, which perfectly matches our calculated 106.5 V (allowing for rounding). Furthermore, in a capacitive circuit, the current leads the source voltage, and the capacitor voltage lags the current. A negative phase angle of -51.21° relative to the source aligns perfectly with capacitive physics.

2. Independent Verification (Simulation)
To verify this without a physical oscilloscope, use LTspice. Set up a transient analysis (.tran) with a SINE voltage source configured to 170V amplitude and 60Hz frequency. Place a 150Ω resistor and 22μF capacitor in series. Run the simulation for at least 50ms to allow any initial transient DC offset to decay. Probe the node between the resistor and capacitor. The waveform will show a peak amplitude of exactly 106.5V, lagging the source waveform by roughly 2.38 milliseconds (which equates to 51.21° at 60Hz).

For further reading on AC steady-state analysis and complex waveforms, consult the All About Circuits textbook chapter on the Phasor Domain. Mastering these algebraic translations is the gateway to understanding power factor correction, filter design, and AC motor theory.