7.1 DC & Single-Phase AC Network Analysis (Thevenin, Norton, Superposition)
Key Takeaways
- Thevenin and Norton phasor equivalents reduce linear single-phase AC networks to an open-circuit voltage source (V_th) in series with a complex Thevenin impedance (Z_th = R_th + jX_th) or a short-circuit current source (I_sc = I_N) in parallel with Z_th, where Z_th = V_th / I_sc.
- The Maximum Real Power Transfer Theorem for complex networks mandates a complex-conjugate load impedance (Z_L = Z_th* = R_th - jX_th), resulting in maximum power transfer P_max = |V_th|^2 / (4 R_th). For a purely resistive constrained load, R_L = |Z_th|.
- Superposition requires deactivating independent sources (independent voltage sources replaced by short circuits, independent current sources by open circuits) while keeping dependent sources fully active; circuits with sources at distinct frequencies cannot have their phasors summed directly and must be analyzed in the time/instantaneous power domain.
- Resonance in series RLC circuits occurs at omega_0 = 1/sqrt(LC) where circuit impedance is at a minimum (Z = R) and reactive voltages undergo magnification (V_L = V_C = Q * V_in), with quality factor Q = omega_0*L/R and fractional bandwidth BW = omega_0 / Q.
- Parallel RLC resonance exhibits maximum impedance (Z = R) at omega_0 with quality factor Q = R / (omega_0*L) = omega_0*C*R, acting as a band-stop current filter.
7.1 DC & Single-Phase AC Network Analysis (Thevenin, Norton, Superposition)
Executive Overview: Circuit analysis forms the quantitative backbone of the NCEES PE Electrical and Computer: Power examination. Questions in Domain 4 evaluate a candidate's ability to model complex single-phase AC networks in steady state using phasors, transform multi-loop networks into Thevenin/Norton equivalents, calculate maximum complex power transfer, apply superposition across linear networks, and evaluate RLC resonance phenomena. Mastering rigorous complex matrix algebra, conjugate arithmetic, and impedance transformations is essential for rapid, error-free problem solving under CBT time constraints.
1. Phasor Domain Fundamentals & Complex Impedance
In sinusoidal steady-state AC analysis, all voltages and currents oscillate at a constant angular frequency $\omega = 2\pi f\text{ rad/s}$. Time-domain signals $v(t) = V_{peak} \cos(\omega t + \theta)$ are mapped into the frequency domain as root-mean-square (RMS) phasors:
PE Exam Standard Note: Unless explicitly stated otherwise (such as in instantaneous waveform definitions), all voltage and current magnitudes on the PE Power exam are given in RMS values.
Constitutive Branch Impedance & Admittance Relationships
| Element | Time-Domain Relationship | Phasor Impedance ($\mathbf{Z}$) | Phasor Admittance ($\mathbf{Y} = 1/\mathbf{Z}$) |
|---|---|---|---|
| Resistor ($R$) | $v(t) = R , i(t)$ | $\mathbf{Z}_R = R \angle 0^\circ = R + j0$ | $\mathbf{Y}_R = G = \frac{1}{R}$ |
| Inductor ($L$) | $v(t) = L \frac{di(t)}{dt}$ | $\mathbf{Z}_L = j\omega L = \omega L \angle +90^\circ$ | $\mathbf{Y}_L = -j B_L = -j\frac{1}{\omega L} = \frac{1}{\omega L}\angle -90^\circ$ |
| Capacitor ($C$) | $i(t) = C \frac{dv(t)}{dt}$ | $\mathbf{Z}_C = \frac{1}{j\omega C} = -j\frac{1}{\omega C} = \frac{1}{\omega C}\angle -90^\circ$ | $\mathbf{Y}_C = +j B_C = +j\omega C = \omega C \angle +90^\circ$ |
For any general two-terminal passive branch:
- If $X > 0$ (inductive), the current lags voltage ($\theta_Z > 0$, susceptance $B < 0$).
- If $X < 0$ (capacitive), the current leads voltage ($\theta_Z < 0$, susceptance $B > 0$).
2. Thevenin & Norton Equivalent Circuits in the Phasor Domain
Any linear, active, two-terminal AC network containing independent sources, linear dependent sources, and linear passive impedances can be represented at its output terminals ($A-B$) by an equivalent circuit:
THEVENIN EQUIVALENT CIRCUIT: NORTON EQUIVALENT CIRCUIT:
Z_th +-----> A
+---[ZZZZ]---+-----> A |
| | |
+ | | +--+--+
(~) V_th | ^ | |
- | | I_sc ( ) [Z_th]|
| | | | |
+------------+-----> B +--+--+
|
+-----> B
Analytical Determination of Parameters
- Thevenin Open-Circuit Voltage ($\mathbf{V}_{th}$): The phasor voltage across terminals $A-B$ when the external load is disconnected ($I_{load} = 0$):
- Norton Short-Circuit Current ($\mathbf{I}_{sc}$): The phasor current flowing from terminal $A$ to terminal $B$ when terminals $A-B$ are connected by an ideal zero-impedance conductor:
- Thevenin Equivalent Impedance ($\mathbf{Z}_{th}$):
- General Definition (Works with all circuits, including those with dependent sources):
- Independent Sources Only Method (Deactivation/Look-in Method):
Deactivate all independent sources:
- Replace all independent voltage sources with short circuits ($\mathbf{V} = 0$).
- Replace all independent current sources with open circuits ($\mathbf{I} = 0$).
- Calculate looking-in equivalent impedance $\mathbf{Z}_{in}$ across terminals $A-B$.
- Test Source Method (For circuits containing Dependent Sources): Deactivate all independent sources, connect an external test source $\mathbf{V}{test}$ (or $\mathbf{I}{test}$) across terminals $A-B$, and calculate the resulting ratio:
Source Transformation Relationships
3. Maximum Power Transfer Theorem for Complex Impedances
When a linear AC source represented by Thevenin parameters $\mathbf{V}{th}$ and $\mathbf{Z}{th} = R_{th} + jX_{th}$ is connected to a variable load impedance $\mathbf{Z}_L = R_L + jX_L$, the complex power delivered to the load is:
The real active power dissipated in the load is:
+---------------------------------------------------------------------------------------------------+
| MAXIMUM REAL POWER TRANSFER CONDITIONS (PE POWER SUMMARY) |
+-----------------------+-----------------------------------------------+---------------------------+
| Load Constraint Type | Optimal Load Condition | Maximum Active Power |
+-----------------------+-----------------------------------------------+---------------------------+
| **Unconstrained** | **Complex Conjugate Match:** | |
| (Both R_L, X_L vary) | $\mathbf{Z}_L = \mathbf{Z}_{th}^* = R_{th} - jX_{th}$ | $P_{max} = \frac{|\mathbf{V}_{th}|^2}{4 R_{th}}$ |
+-----------------------+-----------------------------------------------+---------------------------+
| **Purely Resistive** | **Magnitude Match:** | |
| ($X_L = 0$, $R_L$ var)| $R_L = |\mathbf{Z}_{th}| = \sqrt{R_{th}^2 + X_{th}^2}$ | $P = \frac{|\mathbf{V}_{th}|^2 R_L}{(R_{th} + R_L)^2 + X_{th}^2}$ |
+-----------------------+-----------------------------------------------+---------------------------+
| **Fixed Load Angle** | **Magnitude Match:** | |
| ($\theta_L$ constant) | $|\mathbf{Z}_L| = |\mathbf{Z}_{th}|$ | Dependent on $\theta_L$ |
+-----------------------+-----------------------------------------------+---------------------------+
Critical Exam Pitfall: Under complex conjugate matching ($\mathbf{Z}L = \mathbf{Z}{th}^*$), the net reactance of the loop cancels entirely ($X_{th} + X_L = X_{th} - X_{th} = 0$), driving the circuit into resonance. The load current magnitude becomes $|\mathbf{I}L| = |\mathbf{V}{th}| / (2 R_{th})$. Note that total source power generated is $P_{source} = |\mathbf{I}L|^2 (R{th} + R_L) = 2 P_{max}$, representing an electrical efficiency of exactly $50%$.
4. Superposition Theorem & Multi-Frequency AC Networks
The Superposition Theorem states that in any linear bilateral electrical network, the current or voltage response across any branch is equal to the algebraic sum of the individual responses produced by each independent source acting alone.
Procedural Rules for Superposition
- Select one independent source at a time.
- Deactivate all other independent sources:
- Set independent voltage sources to zero $\rightarrow$ Replace with Short Circuit.
- Set independent current sources to zero $\rightarrow$ Replace with Open Circuit.
- Dependent sources (CCCS, CCVS, VCCS, VCVS) MUST REMAIN ACTIVE during every individual sub-circuit calculation.
- Calculate the partial response phasor (e.g., $\mathbf{V}_k'$ or $\mathbf{I}_k'$).
- Repeat for all independent sources.
The Frequency Superposition Boundary Rule
+---------------------------------------------------------------------------------------------------+
| SUPERPOSITION ACROSS FREQUENCIES: CRITICAL MATHEMATICAL RULES |
|
| CASE 1: All Sources at SAME Frequency (omega_1 = omega_2):
| -> Sum phasor quantities directly: V_total = V_1 + V_2 + ... + V_n
| -> Compute power from final total phasor: P_total = |I_total|^2 * R
|
| CASE 2: Sources at DIFFERENT Frequencies (omega_1 != omega_2, or DC + AC):
| -> NEVER ADD PHASORS OF DIFFERENT FREQUENCIES DIRECTLY! (Phasors rotate at different speeds).
| -> Sum responses in the TIME DOMAIN:
| v_total(t) = V_1_peak * cos(omega_1*t + theta_1) + V_2_peak * cos(omega_2*t + theta_2)
| -> Total RMS voltage across orthogonal frequencies:
| V_rms_total = sqrt( V_dc^2 + V_1_rms^2 + V_2_rms^2 + ... )
| -> Total average real power is the sum of independent average powers:
| P_total = P_1(omega_1) + P_2(omega_2) + ...
+---------------------------------------------------------------------------------------------------+
5. Systematic Nodal & Mesh Analysis in the Phasor Domain
Nodal Admittance Formulation ($[\mathbf{Y}]\mathbf{V} = \mathbf{I}$)
For an $N$-node network with one reference node, Kirchhoff's Current Law (KCL) at each non-reference node yields the matrix equation:
- Self-Admittance ($\mathbf{Y}_{ii}$): Sum of all branch admittances connected directly to node $i$ (always positive).
- Mutual Admittance ($\mathbf{Y}_{ij}$): Negative of the sum of all branch admittances connected directly between node $i$ and node $j$ ($\mathbf{Y}{ij} = -y{ij}$). For reciprocal networks, $\mathbf{Y}{ij} = \mathbf{Y}{ji}$.
- Injected Node Current ($\mathbf{I}_{si}$): Sum of all independent source currents entering node $i$ (positive if entering, negative if leaving).
Mesh Impedance Formulation ($[\mathbf{Z}]\mathbf{I} = \mathbf{V}$)
For an $M$-mesh planar network, Kirchhoff's Voltage Law (KVL) yields:
- Self-Impedance ($\mathbf{Z}_{kk}$): Sum of all impedances around mesh $k$.
- Mutual Impedance ($\mathbf{Z}_{kj}$): Negative of the impedance common to mesh $k$ and mesh $j$ (assuming consistent clockwise mesh current orientation).
6. Series & Parallel RLC Resonance
Resonance in an AC circuit occurs at the specific frequency $\omega_0$ where the input voltage and input current are in phase (i.e., the net reactive component of the driving-point impedance or admittance equals zero, $\text{Im}{\mathbf{Z}_{in}} = 0$, yielding a unity power factor).
SERIES RLC RESONANCE: PARALLEL RLC RESONANCE:
R L C +------+------+------+
+-[RRR]----[LLLL]----[ | ]----+ | | | |
| | +--+--++--+--++--+--+ |
(~) V_in | I_in( ) |[R]| |[L]| |[C]| |
| | +--+--++--+--++--+--+ |
+-----------------------------+ | | | |
+------+------+------+
Comprehensive Comparison of Series vs. Parallel RLC Resonance
| Parameter / Feature | Series RLC Circuit | Parallel RLC Circuit |
|---|---|---|
| Resonant Angular Frequency ($\omega_0$) | $\omega_0 = \frac{1}{\sqrt{LC}}\text{ rad/s}$ | $\omega_0 = \frac{1}{\sqrt{LC}}\text{ rad/s}$ |
| Resonant Frequency ($f_0$) | $f_0 = \frac{1}{2\pi\sqrt{LC}}\text{ Hz}$ | $f_0 = \frac{1}{2\pi\sqrt{LC}}\text{ Hz}$ |
| Input Impedance at $\omega_0$ | $\mathbf{Z}(\omega_0) = R$ (Minimum) | $\mathbf{Z}(\omega_0) = R$ (Maximum) |
| Input Current at $\omega_0$ | $\mathbf{I}(\omega_0) = \frac{\mathbf{V}_{in}}{R}$ (Maximum) | $\mathbf{I}(\omega_0) = \frac{\mathbf{V}_{in}}{R}$ (Minimum) |
| Quality Factor ($Q$) | $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}}$ | $Q = \frac{R}{\omega_0 L} = \omega_0 C R = R\sqrt{\frac{C}{L}}$ |
| Half-Power Bandwidth ($BW$) | $BW = \Delta\omega = \frac{\omega_0}{Q} = \frac{R}{L}\text{ rad/s}$ | $BW = \Delta\omega = \frac{\omega_0}{Q} = \frac{1}{RC}\text{ rad/s}$ |
| Bandwidth in Hertz ($\Delta f$) | $\Delta f = \frac{f_0}{Q} = \frac{R}{2\pi L}\text{ Hz}$ | $\Delta f = \frac{f_0}{Q} = \frac{1}{2\pi R C}\text{ Hz}$ |
| Half-Power Frequencies ($\omega_1, \omega_2$) | $\omega_{1,2} = \mp \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$ | $\omega_{1,2} = \mp \frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}}$ |
| High-$Q$ Approximation ($Q \ge 10$) | $\omega_{1,2} \approx \omega_0 \mp \frac{BW}{2} = \omega_0 \left(1 \mp \frac{1}{2Q}\right)$ | $\omega_{1,2} \approx \omega_0 \mp \frac{BW}{2} = \omega_0 \left(1 \mp \frac{1}{2Q}\right)$ |
| Voltage / Current Magnification | $V_L(\omega_0) = V_C(\omega_0) = Q \times V_{in}$ | $I_L(\omega_0) = I_C(\omega_0) = Q \times I_{in}$ |
7. Comprehensive Step-by-Step Worked AC Thevenin Example
Problem Statement
An AC circuit operates at steady state with a source voltage $\mathbf{V}_s = 120\angle 0^\circ\text{ V (RMS)}$. The circuit consists of:
- A source-connected series branch: $\mathbf{Z}_1 = 4 + j12,\Omega$
- A parallel shunt branch: $\mathbf{Z}_2 = -j10,\Omega$
- A series output branch leading to load terminals $A-B$: $\mathbf{Z}_3 = 2 + j4,\Omega$
Determine:
- The Thevenin equivalent voltage phasor $\mathbf{V}_{th}$ across terminals $A-B$.
- The Thevenin equivalent impedance $\mathbf{Z}_{th}$.
- The Norton short-circuit current $\mathbf{I}_{sc}$.
- The optimal load impedance $\mathbf{Z}L$ for maximum real power transfer, and the value of $P{max}$.
CIRCUIT SCHEMATIC:
Z_1 = 4 + j12 ohms Z_3 = 2 + j4 ohms
+-------[ZZZZZZ]-----------+-----------[ZZZZZZ]----------> A
| |
+ | |
(~) Vs = 120 /_ 0° V +--+--+
- | | | Z_2 = -j10 ohms
| [ C ]
| | |
| +--+--+
| |
+--------------------------+------------------------------> B
Step-by-Step Numerical Solution
=========================================================================================
CALCULATION WORKFLOW & SOLUTION:
=========================================================================================
Step 1: Compute Open-Circuit Thevenin Voltage (V_th)
With terminals A-B open, zero current flows through branch Z_3 (I_3 = 0).
The open-circuit voltage V_th = V_oc is the voltage across shunt impedance Z_2:
V_th = Vs * [ Z_2 / (Z_1 + Z_2) ]
Evaluate denominator:
Z_1 + Z_2 = (4 + j12) + (-j10) = 4 + j2 ohms
Polar form: |Z_sum| = sqrt(4^2 + 2^2) = sqrt(20) = 4.47214 ohms
theta = arctan(2 / 4) = 26.565°
Z_1 + Z_2 = 4.47214 /_ 26.565° ohms
Evaluate numerator:
Vs * Z_2 = (120 /_ 0°) * (10 /_ -90°) = 1200 /_ -90° V
Perform complex division:
V_th = (1200 /_ -90°) / (4.47214 /_ 26.565°)
= 268.328 /_ (-90° - 26.565°)
= 268.328 /_ -116.565° V
Rectangular Conversion:
V_th = 268.328 * cos(-116.565°) + j 268.328 * sin(-116.565°)
= 268.328 * (-0.44721) + j 268.328 * (-0.89443)
= -120.00 - j240.00 V
Step 2: Compute Looking-In Thevenin Impedance (Z_th)
Deactivate independent voltage source Vs (replace with short circuit).
Looking into terminals A-B:
Z_th = Z_3 + (Z_1 || Z_2)
Calculate parallel combination (Z_1 || Z_2):
Z_p = (Z_1 * Z_2) / (Z_1 + Z_2) = (4 + j12)(-j10) / (4 + j2)
= (120 - j40) / (4 + j2)
= [ (120 - j40) * (4 - j2) ] / [ 4^2 + 2^2 ]
= [ (480 - j240 - j160 - 80) ] / 20
= (400 - j400) / 20
= 20.00 - j20.00 ohms
Add series branch Z_3:
Z_th = (20.00 - j20.00) + (2.00 + j4.00)
= 22.00 - j16.00 ohms
Polar Conversion:
|Z_th| = sqrt(22^2 + (-16)^2) = sqrt(484 + 256) = sqrt(740) = 27.203 ohms
theta_Z = arctan(-16 / 22) = -36.027°
Z_th = 27.203 /_ -36.027° ohms
Step 3: Compute Norton Short-Circuit Current (I_sc)
I_sc = V_th / Z_th
= (268.328 /_ -116.565°) / (27.203 /_ -36.027°)
= 9.864 /_ (-116.565° - (-36.027°))
= 9.864 /_ -80.538° A
Rectangular Form:
I_sc = 9.864 * cos(-80.538°) + j 9.864 * sin(-80.538°)
= 1.622 - j9.730 A
Step 4: Determine Maximum Real Power Transfer Conditions
Optimal complex load impedance per conjugate matching:
Z_L = Z_th* = (22.00 - j16.00)* = 22.00 + j16.00 ohms (27.203 /_ +36.027° ohms)
Maximum Real Power delivered to load:
P_max = |V_th|^2 / (4 * R_th)
= (268.328)^2 / (4 * 22.00)
= 72,000.0 / 88.00
= 818.18 W
Cross-Verification via Load Current:
I_L = V_th / (Z_th + Z_L) = (268.328 /_ -116.565°) / (22 - j16 + 22 + j16)
= (268.328 /_ -116.565°) / 44.00 = 6.09836 /_ -116.565° A
P_L = |I_L|^2 * R_L = (6.09836)^2 * 22.00 = 37.190 * 22.00 = 818.18 W (CONFIRMED)
=========================================================================================
8. Common Exam Traps & Strategic Pitfalls
- Conjugate Load Reactance Sign Inversion: Selecting $\mathbf{Z}L = \mathbf{Z}{th}$ instead of $\mathbf{Z}L = \mathbf{Z}{th}^*$. If $\mathbf{Z}_{th} = 22 - j16,\Omega$ (capacitive), the matching load must be inductive ($\mathbf{Z}_L = 22 + j16,\Omega$) to cancel source reactance and achieve resonance.
- RMS vs. Peak Factor of 2 Confusion: Applying the textbook peak-voltage formula $P_{max} = V_{peak}^2 / (8 R_{th})$ when given RMS quantities. On the PE exam, where voltages are RMS, the denominator is $4 R_{th}$.
- Direct Phasor Summation Across Frequencies: Adding phasors from a $60\text{ Hz}$ source and a $180\text{ Hz}$ harmonic source directly. Phasors rotate at different angular velocities; their time waveforms must be combined or their powers summed orthogonally ($P_{total} = P_{60} + P_{180}$).
- Series vs. Parallel Quality Factor Inversion: Confusing $Q_{series} = \frac{\omega_0 L}{R}$ with $Q_{parallel} = \frac{R}{\omega_0 L}$. Note that in a series circuit, decreasing $R$ increases $Q$, whereas in a parallel circuit, increasing $R$ increases $Q$.
A linear AC source has an open-circuit Thevenin voltage of V_th = 200 /_ 0° V (RMS) and a Thevenin impedance of Z_th = 15 - j20 ohms. What load impedance Z_L will absorb the maximum active (real) power from this source, and what is that maximum power value?
A series RLC circuit connected to a 120 V (RMS), 60 Hz source consists of a resistor R = 4 ohms, an inductor L = 50 mH, and a capacitor C = 140.72 microfarads, placing the circuit in resonance. What is the quality factor Q of this circuit, and what is the magnitude of the RMS voltage drop across the capacitor?
A linear circuit resistor R = 10 ohms is subjected to two independent voltage sources operating simultaneously: a DC source producing V_dc = 30 V and an AC source producing v_ac(t) = 40 cos(377 t) V. What is the total average real power dissipated in the resistor?