The Problem Statement and Given Parameters

When studying AC circuit theory, textbook problems often isolate variables in ways that don't reflect real-world bench conditions. However, mastering these foundational ac examples is critical before you start wiring actual motor loads or sizing capacitor banks. Below is a classic, high-yield exam problem that tests your understanding of parallel impedance, the power triangle, and power factor correction.

Problem Statement:
A 120V RMS, 60Hz AC voltage source supplies a parallel circuit containing a 40Ω heating resistor and a 159.15mH induction motor coil (assume pure inductance for the coil model). Calculate:
1) Total impedance ($Z_T$)
2) Real Power ($P$), Reactive Power ($Q$), and Apparent Power ($S$)
3) The exact capacitance ($C$) required in parallel to correct the power factor to 0.95 lagging.

Which Method Applies and Why?

For series circuits, we simply add impedances ($Z_T = Z_1 + Z_2$). But for parallel AC circuits, adding impedances directly is a mathematical trap. Instead, we use the admittance method ($Y = 1/Z$). Admittance allows us to sum the parallel branches algebraically in rectangular form ($Y_T = Y_R + Y_L$) before converting back to impedance. This avoids the messy complex-number fraction arithmetic required by the product-over-sum formula.

Table 1: Component Parameters and Initial Admittance Values
Component Given Value Impedance ($Z$) Admittance ($Y = 1/Z$)
AC Source 120V RMS, 60Hz N/A N/A
Resistor (Branch 1) $R = 40 \Omega$ $40 + j0 \Omega$ $0.025 + j0$ S
Inductor (Branch 2) $L = 159.15$ mH $0 + j60 \Omega$ $0 - j0.01667$ S
Target PF 0.95 Lagging N/A Angle: $18.19^\circ$

Step-by-Step Algebraic Solution

Let's break down the math without skipping the intermediate algebra. We will use $j$ as the imaginary operator (standard in electrical engineering to avoid confusion with current $i$).

Step 1: Calculate Inductive Reactance ($X_L$)

The problem gives inductance in millihenries. We must convert to Henrys and apply the reactance formula:

  • $X_L = 2 \pi f L$
  • $X_L = 2 \cdot \pi \cdot 60 \text{ Hz} \cdot 0.15915 \text{ H}$
  • $X_L = 376.99 \cdot 0.15915$
  • $X_L = 60 \Omega$

The inductor's impedance is $Z_L = j60 \Omega$.

Step 2: Total Admittance and Total Impedance ($Z_T$)

Convert each branch's impedance to admittance (Siemens, S):

  • $Y_R = 1 / 40 = 0.025$ S
  • $Y_L = 1 / (j60) = -j(1/60) = -j0.01667$ S

Sum the admittances to find total admittance ($Y_T$):

  • $Y_T = 0.025 - j0.01667$ S

Convert $Y_T$ to polar form to find its magnitude:

  • $|Y_T| = \sqrt{0.025^2 + (-0.01667)^2}$
  • $|Y_T| = \sqrt{0.000625 + 0.0002778} = \sqrt{0.0009028}$
  • $|Y_T| = 0.03005$ S

Now, invert the magnitude to get total impedance:

  • $|Z_T| = 1 / |Y_T| = 1 / 0.03005 = \mathbf{33.28 \Omega}$

Step 3: The Power Triangle ($P$, $Q$, $S$)

Because this is a parallel circuit, the full 120V RMS is applied across both branches independently. This makes power calculations straightforward:

  • Real Power ($P$): $P = V^2 / R = 120^2 / 40 = 14400 / 40 = \mathbf{360 \text{ W}}$
  • Reactive Power ($Q_L$): $Q_L = V^2 / X_L = 14400 / 60 = \mathbf{240 \text{ VAR}}$
  • Apparent Power ($S$): $S = \sqrt{P^2 + Q_L^2} = \sqrt{360^2 + 240^2} = \sqrt{129600 + 57600} = \sqrt{187200} = \mathbf{432.67 \text{ VA}}$

Step 4: Power Factor Correction Capacitor Sizing

The initial power factor is $P / S = 360 / 432.67 = 0.832$. We need to reach 0.95. According to All About Circuits, adding a parallel capacitor supplies negative reactive power ($Q_C$) to offset the inductor's positive reactive power, without altering the real power consumed by the resistor.

  • Target angle: $\theta_{new} = \arccos(0.95) = 18.19^\circ$
  • Target reactive power: $Q_{new} = P \cdot \tan(18.19^\circ) = 360 \cdot 0.3287 = 118.33 \text{ VAR}$
  • Required capacitive reactive power: $Q_C = Q_L - Q_{new} = 240 - 118.33 = 121.67 \text{ VAR}$

Now, find the required capacitive reactance ($X_C$) and capacitance ($C$):

  • $X_C = V^2 / Q_C = 14400 / 121.67 = 118.35 \Omega$
  • $C = 1 / (2 \pi f X_C) = 1 / (376.99 \cdot 118.35) = 1 / 44616$
  • $C = 0.0000224 \text{ F} = \mathbf{22.4 \mu F}$

The Trap, Sanity Checks, and Independent Verification

The Trap in this Problem: The most common mistake students make in these ac examples is trying to calculate total impedance using $Z_T = Z_R + Z_L$. That formula only works for series circuits. If you add $40 + j60$, you get an impedance magnitude of $72.1 \Omega$, which is physically impossible for a parallel circuit (total parallel impedance must always be lower than the smallest branch resistance). Always use admittance for parallel branches.

Answer Sanity Check

Before moving to the next exam question, run a quick order-of-magnitude and unit check:

  1. Impedance Check: Is $Z_T$ (33.28 Ω) less than the smallest branch impedance (40 Ω)? Yes. The math holds up.
  2. Power Check: Does $S$ (432.67 VA) equal $V_{RMS} \cdot I_{total}$? Total current $I = V / Z_T = 120 / 33.28 = 3.6$ A. $120 \text{ V} \cdot 3.6 \text{ A} = 432$ VA. Matches perfectly.
  3. Capacitance Units: Power factor correction capacitors for 120V/60Hz line voltage typically fall in the 10 μF to 100 μF range for fractional horsepower loads. Our answer of 22.4 μF is highly realistic for a benchtop motor model.

How to Verify the Answer Independently on the Bench

Theory is great, but verifying with hardware builds true intuition. To verify this independently, wire the 40Ω power resistor and the 159mH inductor in parallel across a 120V variac (set to exactly 120V RMS).

Clamp a true-RMS AC clamp meter (like a Fluke 375) around the main feeder wire. You should read approximately 3.6 A. Next, measure the real power using a digital wattmeter; it should read 360 W. Now, wire a 22 μF, 250VAC-rated film capacitor in parallel with the existing load. The wattmeter will still read 360 W (the capacitor consumes zero real power), but your clamp meter will drop to roughly 3.15 A. As noted by Fluke's engineering blog, this reduction in total line current for the same real work output is the exact physical proof that your power factor correction math was correct.

Frequently Asked Questions on AC Power Examples

Why do we use RMS voltage instead of peak voltage for these calculations?

Root Mean Square (RMS) voltage represents the equivalent DC voltage that would deliver the same average heating power to a resistive load. If we used the peak voltage ($120 \cdot \sqrt{2} = 169.7$ V) in the $P = V^2/R$ formula, we would overstate the real power by a factor of two. Standard AC multimeters and utility meters all read and bill based on RMS values.

What happens if I overcorrect the power factor to 'leading'?

If you install a 40 μF capacitor instead of 22.4 μF, you will inject more capacitive VARs than the inductor consumes. The circuit becomes net-capacitive (leading PF). While this still reduces the phase angle magnitude, it can cause voltage regulation issues, leading to voltage swells at the source and potentially damaging sensitive electronics connected to the same branch circuit. Always target slightly lagging (e.g., 0.95) rather than unity (1.0) or leading.