The Concept: What Is an Example of an Expressed Power in Circuits?
When students search for "what is an example of an expressed power," they are often crossing wires between civics (enumerated constitutional powers) and physics. In electrical circuit theory, an expressed power refers to the specific mathematical formulation used to quantify energy transfer. For example, expressing instantaneous power as p(t) = v(t)i(t), or expressing complex power in rectangular form as S = P + jQ.
Understanding how to properly express power is critical for AC circuit analysis, where voltage and current are rarely in phase. According to Georgia State University's HyperPhysics, expressing power in complex form allows engineers to simultaneously calculate real work (Watts) and reactive energy oscillation (VARs). Let's walk through a classic university-level exam problem to see exactly how this is done, step by step.
A series AC circuit consists of a voltage source v(t) = 169.7 cos(377t + 30°) V, a resistor R = 10 Ω, and an inductor L = 26.5 mH. Determine the expressed power in complex form (S), the real power (P), and the reactive power (Q).
Exam Problem Walkthrough: Expressing Complex AC Power
Method Applied: Phasor Analysis combined with the Complex Power Theorem. We use this method because the circuit contains a reactive component (inductor) driven by a sinusoidal source. Complex power captures both the real energy dissipated by the resistor and the reactive energy stored and released by the inductor.
Step-by-Step Algebraic Solution
- Convert the source to an RMS Phasor:
The given voltage is in the time domain with a peak amplitude of 169.7 V.
V_rms = V_peak / √2 = 169.7 / 1.414 = 120 V
Phasor form: V_rms = 120 ∠ 30° V - Calculate Inductive Reactance (X_L):
The angular frequency ω is 377 rad/s (which corresponds to 60 Hz).
X_L = ω × L = 377 × 0.0265 H = 9.9905 Ω (We will round to 10 Ω for standard exam precision). - Determine Total Impedance (Z):
Z = R + jX_L = 10 + j10 Ω
Convert to polar form:
Magnitude: |Z| = √(10² + 10²) = √200 = 14.14 Ω
Angle: θ = arctan(10 / 10) = 45°
Z = 14.14 ∠ 45° Ω - Calculate the RMS Current Phasor (I_rms):
I_rms = V_rms / Z = (120 ∠ 30°) / (14.14 ∠ 45°)
Divide magnitudes and subtract angles: 120 / 14.14 = 8.486 A; 30° - 45° = -15°
I_rms = 8.486 ∠ -15° A - Express Complex Power (S):
The theorem states S = V_rms × (I_rms)*, where (I_rms)* is the complex conjugate of the current.
Conjugate current: (I_rms)* = 8.486 ∠ +15° A
S = (120 ∠ 30°) × (8.486 ∠ 15°)
Multiply magnitudes and add angles: 120 × 8.486 = 1018.3 VA; 30° + 15° = 45°
S = 1018.3 ∠ 45° VA - Convert to Rectangular Form (P + jQ):
P = 1018.3 × cos(45°) = 720 W
Q = 1018.3 × sin(45°) = 720 VAR
Final Expressed Power: S = 720 + j720 VA
Answer Sanity Check & Independent Verification
Order of Magnitude Check: 120 V pushed through roughly 14 Ω yields about 8.5 A. Real power is roughly I²R ≈ (8.5)² × 10 ≈ 722 W. Our calculated 720 W matches this ballpark perfectly. Units are correct (Watts for P, VAR for Q).
Independent Verification: We can verify our expressed power without using the complex conjugate formula by relying on scalar component losses, as detailed in All About Circuits.
P = |I_rms|² × R = (8.486)² × 10 = 72.01 × 10 = 720 W
Q = |I_rms|² × X_L = (8.486)² × 10 = 720 VAR
The independent scalar math perfectly matches our complex phasor derivation.
Reference Table: Standard Power Expressions
Depending on the circuit type, the way you express power changes. Use this reference chart to select the correct formula for your specific scenario.
| Circuit Type | Power Expression | Variables Defined |
|---|---|---|
| DC / Pure Resistive | P = V × I = I²R = V²/R | Constant voltage and current |
| AC Instantaneous | p(t) = v(t) × i(t) | Time-domain waveforms |
| AC Average (Real) | P = V_rms × I_rms × cos(θ) | θ is the phase angle difference |
| AC Complex (Single-Phase) | S = V_rms × (I_rms)* | Yields P + jQ in rectangular form |
| AC Complex (3-Phase Balanced) | S = √3 × V_L × I_L ∠ θ | V_L and I_L are line-to-line/line RMS |
Frequently Asked Questions
What is an example of an expressed power in a purely resistive DC circuit?
In a purely resistive DC circuit, power is expressed strictly as a real scalar value because there is no phase shift or reactive energy storage. The most common expression is P = I²R. For example, if a 5 A current flows through a 4 Ω resistor, the expressed power is simply P = (5)² × 4 = 100 W. There is no imaginary component (jQ) to calculate.
How is an expressed power formula different for single-phase vs three-phase systems?
For single-phase systems, complex power is expressed using the phase voltage and phase current (S = V_phase × I_phase*). In balanced three-phase systems, the expression scales by a factor of √3 (approximately 1.732) when using line-to-line voltage and line current: S_total = √3 × V_line × I_line. This accounts for the combined energy transfer of all three phases operating 120° apart.
Why is complex power expressed with a conjugate current?
The complex conjugate (I*) is used to ensure the resulting phase angle of the power calculation represents the impedance angle (θ_v - θ_i), not the sum of the angles. If we simply multiplied V × I, the angles would add together, yielding a mathematically meaningless result for power. Taking the conjugate flips the sign of the current's angle, allowing the final expression to correctly map to the circuit's power factor and reactive behavior.
