3.2 AC Electrical Fundamentals, Reactance & Impedance

FAA Aviation Mechanic — General AMG

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3 Chapter 3: Basic Electricity

3.1 DC Electrical Circuits, Ohm's Law & Power 3.2 AC Electrical Fundamentals, Reactance & Impedance 3.3 Aircraft Lead-Acid & Nickel-Cadmium Storage Batteries 3.4 DC Generators, Alternators & Electric Motors 3.5 Aircraft Wiring, Bonding & Circuit Protection Devices 3.6 Resistor Color Codes, Semiconductors, Digital Logic & Electrostatic Discharge

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3.2 AC Electrical Fundamentals, Reactance & Impedance

Key Takeaways Explain This Topic AI

Alternating current (AC) periodically reverses polarity and varies in magnitude; standard transport aircraft AC distribution operates at 115V / 200V, 3-phase, 400 Hz, achieving a 70% to 80% reduction in magnetic core iron weight compared to 60 Hz systems.

AC values include peak ( V p k V_{pk} V p k ​ ), peak-to-peak ( V p − p = 2 V p k V_{p-p} = 2V_{pk} V p − p ​ = 2 V p k ​ ), average ( V a v g = 0.637 V p k V_{avg} = 0.637V_{pk} V a vg ​ = 0.637 V p k ​ ), and Root-Mean-Square/Effective ( V R M S = 0.707 V p k V_{RMS} = 0.707V_{pk} V RMS ​ = 0.707 V p k ​ ), where RMS represents the equivalent DC thermal heating value in a resistor.

Inductive reactance ( X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L ) opposes current change with voltage leading current by 90° ('ELI'); capacitive reactance ( X C = 1 2 π f C X_C = \frac{1}{2\pi f C} X C ​ = 2 π f C 1 ​ ) opposes voltage change with current leading voltage by 90° ('ICE').

Total AC circuit opposition is Impedance ( Z = R 2 + ( X L − X C ) 2 Z = \sqrt{R^2 + (X_L - X_C)^2} Z = R 2 + ( X L ​ − X C ​ ) 2 ​ ); the phase angle θ \theta θ establishes the Power Factor ( PF = cos ⁡ θ = R / Z = True Power (W) / Apparent Power (VA) \text{PF} = \cos\theta = R/Z = \text{True Power (W)} / \text{Apparent Power (VA)} PF = cos θ = R / Z = True Power (W) / Apparent Power (VA) ).

Series resonance occurs when inductive reactance equals capacitive reactance ( X L = X C X_L = X_C X L ​ = X C ​ ), reducing circuit impedance to pure resistance ( Z = R Z = R Z = R ) and producing maximum current flow at resonant frequency f r = 1 / ( 2 π L C ) f_r = 1 / (2\pi\sqrt{LC}) f r ​ = 1/ ( 2 π L C ​ ) .

Last updated: August 2026

3.2 AC Electrical Fundamentals, Reactance & Impedance

Alternating current (AC) powers transport category aircraft electrical distribution systems, driving radar transmitters, flight guidance computers, windshield anti-ice heating grids, motor-driven hydraulic pumps, and cabin environmental systems. Unlike direct current, AC periodic waveforms produce dynamic magnetic and electrostatic fields that introduce inductive and capacitive reactances.

1. AC Sine Wave Characteristics & Aviation Frequency Standards

An alternating voltage or current changes continuously in magnitude and periodically reverses polarity. When a single conductor loop rotates at constant angular velocity through a uniform magnetic field, the induced instantaneous electromotive force forms a sinusoidal wave: ≠ = E p k sin ⁡ ( θ ) = E p k sin ⁡ ( 2 π f t ) \ne = E_{pk} \sin(\theta) = E_{pk} \sin(2\pi f t)  = = E p k ​ sin ( θ ) = E p k ​ sin ( 2 π f t ) Sinusoidal AC Waveform Anatomy:

• Voltage
• +Vpk | * * *
• | * *
• | * *
• 0 --+--o-----------------o-----------------o--> Time (or Phase Angle θ)
• | 0° 180° 360° (1 complete cycle)
• | * *
• | * *
• Vpk | * * *
• |<------------ Period (T) ------------>|
• Sine Wave Terminology

Cycle: One complete 360 ∘ 360^\circ 36 0 ∘ ( 2 π radians 2\pi\text{ radians} 2 π radians ) sequence of positive and negative variations.

Alternation: One half-cycle ( 180 ∘ 180^\circ 18 0 ∘ ), representing either the positive or negative half-wave.

Period ( T T T ): The time in seconds required to complete one full cycle: T = 1 / f T = 1/f T = 1/ f .

Frequency ( f f f ): The number of complete cycles per second, measured in Hertz (Hz) : f = 1 / T f = 1/T f = 1/ T .

Wavelength ( λ \lambda λ ): The physical distance traveled by the electromagnetic wave in free space during the time of one cycle: λ = c / f \lambda = c/f λ = c / f (where speed of light c ≈ 3 × 10 8 m/s c \approx 3 \times 10^8\text{ m/s} c ≈ 3 × 1 0 8 m/s ). In aircraft weather radar operating at 9.375 GHz 9.375\text{ GHz} 9.375 GHz , λ ≈ 3.2 cm \lambda \approx 3.2\text{ cm} λ ≈ 3.2 cm .

The 400 Hz Aircraft Standard vs. 60 Hz Terrestrial Power

MIL-STD-704 and commercial transport specifications mandate 400 Hz 400\text{ Hz} 400 Hz as the primary aircraft AC power frequency (typically 115 V / 200 V 115\text{V} / 200\text{V} 115 V /200 V , 3-phase, 400 Hz 400\text{ Hz} 400 Hz ):

Mass Reduction Benefit: The physical cross-section and weight of transformer magnetic iron cores, motor armatures, and filter inductors are inversely proportional to operating frequency. Operating at 400 Hz 400\text{ Hz} 400 Hz reduces magnetic core iron weight by approximately 70% to 80% compared to 60 Hz 60\text{ Hz} 60 Hz terrestrial equipment. A 400 Hz 400\text{ Hz} 400 Hz transformer weighing 2.5 lbs 2.5\text{ lbs} 2.5 lbs would require approximately 16 lbs 16\text{ lbs} 16 lbs of core laminations to process the equivalent power at 60 Hz 60\text{ Hz} 60 Hz .

Aviation Trade-off: Higher frequency increases inductive line reactance ( X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L ) and creates conductor skin effect (high-frequency electron migration toward the outer perimeter of conductors), making 400 Hz 400\text{ Hz} 400 Hz impractical for long-distance municipal grids, but ideal for compact airframe routing.

Loading diagram... AC Sine Wave Voltage Value Relationships

2. AC Voltage and Current Measurement Values

Because AC amplitude changes continuously throughout each cycle, technicians must distinguish between instantaneous, peak, peak-to-peak, average, and effective (RMS) values:

Mathematical Value Relationships

Peak Value ( V p k V_{pk} V p k ​ or I p k I_{pk} I p k ​ ): The maximum instantaneous amplitude attained during a waveform alternation ( 90 ∘ 90^\circ 9 0 ∘ and 270 ∘ 270^\circ 27 0 ∘ ).

Peak-to-Peak Value ( V p − p V_{p-p} V p − p ​ or I p − p I_{p-p} I p − p ​ ): The total voltage difference between the positive peak and negative peak: V p − p = 2 × V p k V_{p-p} = 2 \times V_{pk} V p − p ​ = 2 × V p k ​

Average Value ( V a v g V_{avg} V a vg ​ or I a v g I_{avg} I a vg ​ ): The mathematical mean of all instantaneous values over one half-cycle alternation ( 180 ∘ 180^\circ 18 0 ∘ ): V a v g = 2 π × V p k ≈ 0.637 × V p k V_{avg} = \frac{2}{\pi} \times V_{pk} \approx 0.637 \times V_{pk} V a vg ​ = π 2 ​ × V p k ​ ≈ 0.637 × V p k ​

Root-Mean-Square (RMS) / Effective Value ( V R M S V_{RMS} V RMS ​ or I R M S I_{RMS} I RMS ​ ): The value of alternating voltage or current that produces the exact same thermal power dissipation (heating effect) in a pure resistive load as an equivalent direct current: V R M S = V p k 2 = V p k 1.4142 ≈ 0.7071 × V p k V_{RMS} = \frac{V_{pk}}{\sqrt{2}} = \frac{V_{pk}}{1.4142} \approx 0.7071 \times V_{pk} V RMS ​ = 2 ​ V p k ​ ​ = 1.4142 V p k ​ ​ ≈ 0.7071 × V p k ​ V p k = 2 × V R M S ≈ 1.4142 × V R M S V_{pk} = \sqrt{2} \times V_{RMS} \approx 1.4142 \times V_{RMS} V p k ​ = 2 ​ × V RMS ​ ≈ 1.4142 × V RMS ​ V p − p = 2 × 1.4142 × V R M S = 2.8284 × V R M S V_{p-p} = 2 \times 1.4142 \times V_{RMS} = 2.8284 \times V_{RMS} V p − p ​ = 2 × 1.4142 × V RMS ​ = 2.8284 × V RMS ​

Standard AC Voltmeter Calibration: All standard aviation AC voltmeters and ammeters display RMS (effective) values. For standard aircraft 115 V R M S 115\text{V}_{RMS} 115 V RMS ​ single-phase power, the peak voltage is V p k = 115 × 1.4142 = 162.63 V V_{pk} = 115 \times 1.4142 = 162.63\text{ V} V p k ​ = 115 × 1.4142 = 162.63 V , and the peak-to-peak voltage is V p − p = 325.26 V V_{p-p} = 325.26\text{ V} V p − p ​ = 325.26 V .

• 3. Inductance, Capacitance, and Reactance
• Inductance ( L L L ) & Inductive Reactance ( X L X_L X L ​ )

Inductance ( L L L ): The property of an electric circuit that opposes any change in current flow. When alternating current flows through a wire coil, the expanding and collapsing magnetic flux cuts adjacent conductor loops, inducing a counter-electromotive force (CEMF or back-EMF) in accordance with Lenz's Law ( e L = − L d i d t e_L = -L \frac{di}{dt} e L ​ = − L d t d i ​ ). Inductance is measured in Henrys (H) .

Inductive Reactance ( X L X_L X L ​ ): The opposition offered to alternating current by an inductor, measured in Ohms ( Ω \Omega Ω ) : X L = 2 π f L X_L = 2\pi f L X L ​ = 2 π f L

Inductive reactance is directly proportional to frequency ( f f f ) and inductance ( L L L ). In a pure DC circuit ( f = 0 f = 0 f = 0 ), an ideal inductor offers zero reactance ( X L = 0 Ω X_L = 0\,\Omega X L ​ = 0 Ω ).

Phase Angle in a Pure Inductor: Voltage LEADS current by 90 ∘ 90^\circ 9 0 ∘ ( π / 2 radians \pi/2\text{ radians} π /2 radians ). Current lags voltage because CEMF opposes the initial buildup of current.

Capacitance ( C C C ) & Capacitive Reactance ( X C X_C X C ​ )

Capacitance ( C C C ): The ability of two conductive plates separated by a dielectric insulator to store an electrical charge in an electrostatic field. Capacitance is measured in Farads (F) , with charge Q = C × V Q = C \times V Q = C × V . Physical factors: C = ϵ A d C = \frac{\epsilon A}{d} C = d ϵ A ​ (where ϵ \epsilon ϵ is dielectric permittivity, A A A is plate surface area, and d d d is separation distance).

Capacitive Reactance ( X C X_C X C ​ ): The opposition offered to alternating current by a capacitor, measured in Ohms ( Ω \Omega Ω ) : X C = 1 2 π f C X_C = \frac{1}{2\pi f C} X C ​ = 2 π f C 1 ​

Capacitive reactance is inversely proportional to frequency ( f f f ) and capacitance ( C C C ). At direct current ( f = 0 f = 0 f = 0 ), a capacitor presents infinite reactance ( X C = ∞ X_C = \infty X C ​ = ∞ ), completely blocking DC current. As frequency increases, X C X_C X C ​ decreases.

Phase Angle in a Pure Capacitor: Current LEADS voltage by 90 ∘ 90^\circ 9 0 ∘ ( π / 2 radians \pi/2\text{ radians} π /2 radians ). Current must flow onto the capacitor plates before an electrostatic charge and potential difference can build up.

Phase Relationship Mnemonic: "ELI the ICE man"

ELI: In an Inductive circuit ( L ), Voltage ( E ) Leads Current ( I ).

ICE: In a Capacitive circuit ( C ), Current ( I ) Leads Voltage ( E ). Phase Vector Relationships:

Pure Inductive Circuit (ELI): Pure Capacitive Circuit (ICE):

• Voltage (E) Current (I)
• ^ ^
• | |
• | 90° Phase Lead | 90° Phase Lead
• +---------> Current (I) +---------> Voltage (E)

4. Impedance, Power Factor & Resonance in AC Circuits

In practical AC circuits containing resistance ( R R R ), inductance ( L L L ), and capacitance ( C C C ), the total combined opposition to alternating current flow is the vector sum called Impedance ( Z Z Z ) , measured in Ohms ( Ω \Omega Ω ) .

Series RLC Circuit Impedance

In a series RLC circuit, inductive and capacitive reactances are 180 ∘ 180^\circ 18 0 ∘ out of phase and directly oppose each other. The net reactance is X = X L − X C X = X_L - X_C X = X L ​ − X C ​ : Z = R 2 + ( X L − X C ) 2 Z = \sqrt{R^2 + (X_L - X_C)^2} Z = R 2 + ( X L ​ − X C ​ ) 2 ​

AC Ohm's Law: E = I Z ⟹ I = E Z ⟹ Z = E I E = I Z \implies I = \frac{E}{Z} \implies Z = \frac{E}{I} E = I Z ⟹ I = Z E ​ ⟹ Z = I E ​

The AC Power Triangle and Power Factor AC Power Triangle:

• /| Apparent Power (S) in Volt-Amperes (VA)
• / | S = E_RMS × I_RMS
• / |
• / | Reactive Power (Q) in VAR
• / θ | Q = E_RMS × I_RMS × sin(θ)
• /_____|
• True Power (P) in Watts (W)
• P = E_RMS × I_RMS × cos(θ) = I²R

True Power ( P P P ): The actual power consumed by circuit resistance and converted into heat, light, or mechanical shaft work. Measured in Watts (W) or Kilowatts (kW): P = E R M S × I R M S × cos ⁡ ( θ ) = I 2 R P = E_{RMS} \times I_{RMS} \times \cos(\theta) = I^2 R P = E RMS ​ × I RMS ​ × cos ( θ ) = I 2 R

Apparent Power ( S S S ): The total power delivered to the circuit, calculated as the product of measured RMS voltage and RMS current. Measured in Volt-Amperes (VA) or Kilovolt-Amperes (kVA): S = E R M S × I R M S S = E_{RMS} \times I_{RMS} S = E RMS ​ × I RMS ​

Reactive Power ( Q Q Q ): The "wattless" power stored in magnetic or electrostatic fields and returned to the source each cycle. Measured in Volt-Amperes Reactive (VAR) : Q = E R M S × I R M S × sin ⁡ ( θ ) Q = E_{RMS} \times I_{RMS} \times \sin(\theta) Q = E RMS ​ × I RMS ​ × sin ( θ )

Power Factor (PF): The ratio of true power dissipated to total apparent power delivered: PF = cos ⁡ ( θ ) = True Power (W) Apparent Power (VA) = R Z \text{PF} = \cos(\theta) = \frac{\text{True Power (W)}}{\text{Apparent Power (VA)}} = \frac{R}{Z} PF = cos ( θ ) = Apparent Power (VA) True Power (W) ​ = Z R ​

In a purely resistive circuit, θ = 0 ∘ \theta = 0^\circ θ = 0 ∘ , cos ⁡ ( 0 ∘ ) = 1.0 \cos(0^\circ) = 1.0 cos ( 0 ∘ ) = 1.0 (unity power factor).

In a purely reactive circuit, θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ , cos ⁡ ( 90 ∘ ) = 0 \cos(90^\circ) = 0 cos ( 9 0 ∘ ) = 0 (zero true power consumed).

Series Resonance

When circuit frequency reaches the point where inductive reactance equals capacitive reactance ( X L = X C X_L = X_C X L ​ = X C ​ ), the reactances cancel each other completely ( X L − X C = 0 X_L - X_C = 0 X L ​ − X C ​ = 0 ). At this resonant frequency ( f r f_r f r ​ ) , total circuit impedance drops to pure resistance ( Z = R Z = R Z = R ), and circuit current reaches its absolute theoretical maximum: 2 π f r L = 1 2 π f r C ⟹ f r = 1 2 π L C 2\pi f_r L = \frac{1}{2\pi f_r C} \implies f_r = \frac{1}{2\pi \sqrt{LC}} 2 π f r ​ L = 2 π f r ​ C 1 ​ ⟹ f r ​ = 2 π L C ​ 1 ​

5. Aircraft Transformers & Three-Phase AC Systems

Transformers transfer alternating electrical energy between circuits via electromagnetic mutual induction without moving parts or changes in frequency.

Transformer Operation & Turns Ratio Formulas

Primary / Secondary Voltage and Turns Ratio: V p V s = N p N s \frac{V_p}{V_s} = \frac{N_p}{N_s} V s ​ V p ​ ​ = N s ​ N p ​ ​

Current and Turns Ratio (Assuming 100% efficiency, V p I p = V s I s V_p I_p = V_s I_s V p ​ I p ​ = V s ​ I s ​ ): I s I p = N p N s = V p V s \frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{V_p}{V_s} I p ​ I s ​ ​ = N s ​ N p ​ ​ = V s ​ V p ​ ​

Step-Up Transformer: Secondary has more turns than primary ( N s > N p N_s > N_p N s ​ > N p ​ ), stepping voltage up ( V s > V p V_s > V_p V s ​ > V p ​ ) while stepping current down ( I s < I p I_s < I_p I s ​ < I p ​ ).

Step-Down Transformer: Secondary has fewer turns than primary ( N s < N p N_s < N_p N s ​ < N p ​ ), stepping voltage down ( V s < V p V_s < V_p V s ​ < V p ​ ) while stepping current up ( I s > I p I_s > I_p I s ​ > I p ​ ).

Transformer Core Losses

Copper Losses ( I 2 R I^2R I 2 R ): Heat generated by electrical resistance in the copper windings.

Eddy Current Losses: Induced circulating currents in the iron core. Minimized by constructing the core from thin, varnished silicon-steel sheets ( laminations ).

Hysteresis Losses: Molecular magnetic friction created by alternating magnetic domain alignment. Minimized using magnetically soft silicon-steel alloys.

Three-Phase AC Power Systems

Modern transport aircraft generate three-phase AC power consisting of three sinusoidal waveforms displaced by 120 ∘ 120^\circ 12 0 ∘ of phase angle .

Wye (Star / Y) Configuration: The universal aircraft alternator stator connection. Three phase windings join at a common central neutral point (grounded to the airframe structure):

Line-to-Neutral Voltage ( V L N V_{LN} V L N ​ ): 115 V AC 115\text{V AC} 115 V AC RMS (powers single-phase loads).

Line-to-Line Voltage ( V L L V_{LL} V LL ​ ): Potential difference measured between any two phase lines: V L L = 3 × V L N = 1.732 × 115 V = 200 V AC V_{LL} = \sqrt{3} \times V_{LN} = 1.732 \times 115\text{ V} = 200\text{V AC} V LL ​ = 3 ​ × V L N ​ = 1.732 × 115 V = 200 V AC

Line current equals phase winding current ( I l i n e = I p h a s e I_{line} = I_{phase} I l in e ​ = I p ha se ​ ).

Delta ( Δ \Delta Δ ) Configuration: Phase windings connect end-to-end in a closed triangle. V l i n e = V p h a s e V_{line} = V_{phase} V l in e ​ = V p ha se ​ , and line current is I l i n e = 3 × I p h a s e = 1.732 × I p h a s e I_{line} = \sqrt{3} \times I_{phase} = 1.732 \times I_{phase} I l in e ​ = 3 ​ × I p ha se ​ = 1.732 × I p ha se ​ .

• 6. Worked Numerical Examples
• Example 1: AC Sine Wave Conversion

Problem: A cockpit AC voltmeter indicates 115.0 V R M S 115.0\text{V}_{RMS} 115.0 V RMS ​ across a 400 Hz 400\text{ Hz} 400 Hz avionics instrument bus.

Calculate peak voltage ( V p k V_{pk} V p k ​ ), peak-to-peak voltage ( V p − p V_{p-p} V p − p ​ ), average half-cycle voltage ( V a v g V_{avg} V a vg ​ ), and the period ( T T T ) of one cycle.

Solution:

Calculate Peak Voltage: V p k = 1.4142 × V R M S = 1.4142 × 115.0 V = 162.63 Volts V_{pk} = 1.4142 \times V_{RMS} = 1.4142 \times 115.0\text{ V} = 162.63\text{ Volts} V p k ​ = 1.4142 × V RMS ​ = 1.4142 × 115.0 V = 162.63 Volts

Calculate Peak-to-Peak Voltage: V p − p = 2 × V p k = 2 × 162.63 V = 325.26 Volts V_{p-p} = 2 \times V_{pk} = 2 \times 162.63\text{ V} = 325.26\text{ Volts} V p − p ​ = 2 × V p k ​ = 2 × 162.63 V = 325.26 Volts

Calculate Average Voltage: V a v g = 0.637 × V p k = 0.637 × 162.63 V = 103.60 Volts V_{avg} = 0.637 \times V_{pk} = 0.637 \times 162.63\text{ V} = 103.60\text{ Volts} V a vg ​ = 0.637 × V p k ​ = 0.637 × 162.63 V = 103.60 Volts

Calculate Waveform Period: T = 1 f = 1 400 Hz = 0.0025 seconds = 2.50 milliseconds T = \frac{1}{f} = \frac{1}{400\text{ Hz}} = 0.0025\text{ seconds} = 2.50\text{ milliseconds} T = f 1 ​ = 400 Hz 1 ​ = 0.0025 seconds = 2.50 milliseconds

Example 2: Series RLC Circuit Impedance and Power Factor

Problem: An aircraft radar power supply on a 115.0 V AC 115.0\text{V AC} 115.0 V AC , 400 Hz 400\text{ Hz} 400 Hz bus contains a resistance R = 40.0 Ω R = 40.0\,\Omega R = 40.0 Ω , an inductance L = 31.83 mH L = 31.83\text{ mH} L = 31.83 mH ( 0.03183 H 0.03183\text{ H} 0.03183 H ), and a capacitance C = 26.53 μ F C = 26.53\,\mu\text{F} C = 26.53 μ F ( 2.653 × 10 − 5 F 2.653 \times 10^{-5}\text{ F} 2.653 × 1 0 − 5 F ).

Calculate inductive reactance ( X L X_L X L ​ ), capacitive reactance ( X C X_C X C ​ ), total circuit impedance ( Z Z Z ), total current ( I I I ), power factor (PF), and true power consumed ( P P P ).

Solution:

Calculate Inductive Reactance ( X L X_L X L ​ ): X L = 2 π f L = 2 × 3.1416 × 400 Hz × 0.03183 H = 80.0 Ω X_L = 2\pi f L = 2 \times 3.1416 \times 400\text{ Hz} \times 0.03183\text{ H} = 80.0\,\Omega X L ​ = 2 π f L = 2 × 3.1416 × 400 Hz × 0.03183 H = 80.0 Ω

Calculate Capacitive Reactance ( X C X_C X C ​ ): X C = 1 2 π f C = 1 2 × 3.1416 × 400 × ( 2.653 × 10 − 5 ) = 1 0.06668 = 15.0 Ω X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.1416 \times 400 \times (2.653 \times 10^{-5})} = \frac{1}{0.06668} = 15.0\,\Omega X C ​ = 2 π f C 1 ​ = 2 × 3.1416 × 400 × ( 2.653 × 1 0 − 5 ) 1 ​ = 0.06668 1 ​ = 15.0 Ω

Calculate Total Circuit Impedance ( Z Z Z ): Z = R 2 + ( X L − X C ) 2 = 40.0 2 + ( 80.0 − 15.0 ) 2 = 40.0 2 + 65.0 2 Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40.0^2 + (80.0 - 15.0)^2} = \sqrt{40.0^2 + 65.0^2} Z = R 2 + ( X L ​ − X C ​ ) 2 ​ = 40. 0 2 + ( 80.0 − 15.0 ) 2 ​ = 40. 0 2 + 65. 0 2 ​ Z = 1600 + 4225 = 5825 = 76.32 Ω Z = \sqrt{1600 + 4225} = \sqrt{5825} = 76.32\,\Omega Z = 1600 + 4225 ​ = 5825 ​ = 76.32 Ω

Calculate Circuit Current ( I I I ): I = E Z = 115.0 V 76.32 Ω = 1.507 Amperes I = \frac{E}{Z} = \frac{115.0\text{ V}}{76.32\,\Omega} = 1.507\text{ Amperes} I = Z E ​ = 76.32 Ω 115.0 V ​ = 1.507 Amperes

Calculate Power Factor (PF): PF = cos ⁡ ( θ ) = R Z = 40.0 Ω 76.32 Ω = 0.524 ( or 52.4 % lagging ) \text{PF} = \cos(\theta) = \frac{R}{Z} = \frac{40.0\,\Omega}{76.32\,\Omega} = 0.524 \quad (\text{or } 52.4\% \text{ lagging}) PF = cos ( θ ) = Z R ​ = 76.32 Ω 40.0 Ω ​ = 0.524 ( or 52.4% lagging )

Calculate True Power ( P P P ): P = E × I × PF = 115.0 V × 1.507 A × 0.524 = 90.81 Watts P = E \times I \times \text{PF} = 115.0\text{ V} \times 1.507\text{ A} \times 0.524 = 90.81\text{ Watts} P = E × I × PF = 115.0 V × 1.507 A × 0.524 = 90.81 Watts Verify via I 2 R : P = ( 1.507 A ) 2 × 40.0 Ω = 2.271 × 40.0 = 90.84 Watts \text{Verify via } I^2 R: P = (1.507\text{ A})^2 \times 40.0\,\Omega = 2.271 \times 40.0 = 90.84\text{ Watts} Verify via I 2 R : P = ( 1.507 A ) 2 × 40.0 Ω = 2.271 × 40.0 = 90.84 Watts

Example 3: Step-Down Instrument Transformer

Problem: An aircraft step-down transformer connected to a 115.0 V AC 115.0\text{V AC} 115.0 V AC primary bus delivers 28.0 V AC 28.0\text{V AC} 28.0 V AC to an instrument lighting circuit drawing 15.0 Amperes 15.0\text{ Amperes} 15.0 Amperes . The primary winding has 690 turns 690\text{ turns} 690 turns .

Calculate the number of secondary turns ( N s N_s N s ​ ), primary winding current ( I p I_p I p ​ ), and apparent power processed ( S S S ).

Solution:

Calculate Secondary Turns ( N s N_s N s ​ ): N s = N p × ( V s V p ) = 690 × ( 28.0 V 115.0 V ) = 690 × 0.24348 = 168 turns N_s = N_p \times \left(\frac{V_s}{V_p}\right) = 690 \times \left(\frac{28.0\text{ V}}{115.0\text{ V}}\right) = 690 \times 0.24348 = 168\text{ turns} N s ​ = N p ​ × ( V p ​ V s ​ ​ ) = 690 × ( 115.0 V 28.0 V ​ ) = 690 × 0.24348 = 168 turns

Calculate Primary Current ( I p I_p I p ​ ): I p = I s × ( V s V p ) = 15.0 A × ( 28.0 V 115.0 V ) = 3.652 Amperes I_p = I_s \times \left(\frac{V_s}{V_p}\right) = 15.0\text{ A} \times \left(\frac{28.0\text{ V}}{115.0\text{ V}}\right) = 3.652\text{ Amperes} I p ​ = I s ​ × ( V p ​ V s ​ ​ ) = 15.0 A × ( 115.0 V 28.0 V ​ ) = 3.652 Amperes

Calculate Apparent Power ( S S S ): S = V s × I s = 28.0 V × 15.0 A = 420.0 VA S = V_s \times I_s = 28.0\text{ V} \times 15.0\text{ A} = 420.0\text{ VA} S = V s ​ × I s ​ = 28.0 V × 15.0 A = 420.0 VA Verify Primary Apparent Power: S = V p × I p = 115.0 V × 3.652 A = 420.0 VA \text{Verify Primary Apparent Power: } S = V_p \times I_p = 115.0\text{ V} \times 3.652\text{ A} = 420.0\text{ VA} Verify Primary Apparent Power: S = V p ​ × I p ​ = 115.0 V × 3.652 A = 420.0 VA