Study guide
7.1 DC & AC Circuit Analysis & Network Theorems
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7.1 DC & AC Circuit Analysis & Network Theorems
Key Takeaways Explain This Topic AI
Kirchhoff's Current Law (KCL) based on charge conservation ( ∑ I = 0 \sum I = 0 ∑ I = 0 ) and Kirchhoff's Voltage Law (KVL) based on energy conservation ( ∑ V = 0 \sum V = 0 ∑ V = 0 ) form the foundational nodal and mesh equations for network analysis.
Thevenin's Theorem reduces any linear two-terminal circuit to an open-circuit voltage source (V_{th}) in series with equivalent resistance (R_{th}), while Norton's Theorem yields a short-circuit current source (I_N = V_{th}/R_{th}) in parallel with R_N = R_{th}.
Maximum Power Transfer Theorem dictates that maximum power is delivered to a load when load impedance matches the complex conjugate of source impedance (Z_L = Z_s^*), reducing to R_L = R_{th} in purely resistive DC networks with 50% maximum efficiency.
RLC Resonance in series circuits produces minimum impedance (Z = R), maximum current, and high voltage magnification, whereas parallel resonance produces maximum impedance and minimum current; both defined by resonant frequency f 0 = 1 2 π L C f_0 = \frac{1}{2\pi \sqrt{LC}} f 0 = 2 π L C 1 .
Quality Factor (Q) quantifies energy storage versus dissipation ( Q = ω 0 L R Q = \frac{\omega_0 L}{R} Q = R ω 0 L for series RLC), directly determining circuit selectivity and bandwidth ( B W = f 0 Q = f 2 − f 1 BW = \frac{f_0}{Q} = f_2 - f_1 B W = Q f 0 = f 2 − f 1 ).
Last updated: July 2026
7.1 DC & AC Circuit Analysis & Network Theorems
Quick Answer: Network analysis methods—Kirchhoff's Laws (KVL/KCL), Nodal/Mesh Analysis, Thevenin/Norton Equivalent Circuits, and Superposition—allow engineers to simplify and solve complex DC and AC electrical networks. In AC circuits, impedance ( Z = R + j X Z = R + jX Z = R + j X ) replaces resistance. Maximum Power Transfer occurs when Z L = Z s ∗ Z_L = Z_s^* Z L = Z s ∗ , achieving 50% efficiency. RLC Resonance occurs at frequency f 0 = 1 2 π L C f_0 = \frac{1}{2\pi\sqrt{LC}} f 0 = 2 π L C 1 , where reactive components cancel ( X L = X C X_L = X_C X L = X C ), yielding a Quality Factor Q = ω 0 L R Q = \frac{\omega_0 L}{R} Q = R ω 0 L and Bandwidth B W = f 0 Q BW = \frac{f_0}{Q} B W = Q f 0 .
Fundamental Circuit Laws & Network Theorems
Linear circuit analysis relies on fundamental conservation laws and network theorems that simplify complex multi-loop circuits into solvable algebraic or complex phasor equations.
1. Kirchhoff's Laws (KCL & KVL)
Kirchhoff's Current Law (KCL): Based on the conservation of electric charge, the algebraic sum of all currents entering any node must equal zero: ∑ k = 1 n I k = 0 \sum_{k=1}^{n} I_k = 0 k = 1 ∑ n I k = 0 In nodal analysis, KCL equations are written in terms of node voltages relative to a reference node (ground).
Kirchhoff's Voltage Law (KVL): Based on the conservation of energy, the algebraic sum of all potential differences (voltages) around any closed loop must equal zero: ∑ k = 1 m V k = 0 \sum_{k=1}^{m} V_k = 0 k = 1 ∑ m V k = 0 In mesh analysis, KVL equations are written around independent loops using fictitious mesh currents.
2. Thevenin's and Norton's Theorems
Thevenin's Theorem: Any linear, two-terminal network containing voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source V t h V_{th} V t h in series with a resistor R t h R_{th} R t h .
V t h V_{th} V t h : Open-circuit voltage measured across terminals A − B A-B A − B with the load disconnected ( V t h = V o c V_{th} = V_{oc} V t h = V oc ).
R t h R_{th} R t h : Equivalent resistance across terminals A − B A-B A − B with all independent voltage sources shorted and independent current sources opened.
Norton's Theorem: Dual of Thevenin's theorem. Replaces the network with a single current source I N I_N I N in parallel with resistance R N R_N R N .
I N I_N I N : Short-circuit current flowing between terminals A − B A-B A − B ( I N = I s c = V t h R t h I_N = I_{sc} = \frac{V_{th}}{R_{th}} I N = I sc = R t h V t h ).
R N = R t h R_N = R_{th} R N = R t h .
Network Parameter Thevenin Equivalent Norton Equivalent Source Transformation
Source Parameter Voltage V t h = V o c V_{th} = V_{oc} V t h = V oc Current I N = I s c I_N = I_{sc} I N = I sc V t h = I N ⋅ R t h V_{th} = I_N \cdot R_{th} V t h = I N ⋅ R t h
Internal Resistance Series R t h R_{th} R t h Parallel R N = R t h R_N = R_{th} R N = R t h R N = R t h R_N = R_{th} R N = R t h
Load Current ( I L I_L I L ) I L = V t h R t h + R L I_L = \frac{V_{th}}{R_{th} + R_L} I L = R t h + R L V t h I L = I N R N R N + R L I_L = I_N \frac{R_N}{R_N + R_L} I L = I N R N + R L R N Identical load output
Open-Circuit Voltage V o c = V t h V_{oc} = V_{th} V oc = V t h V o c = I N ⋅ R N V_{oc} = I_N \cdot R_N V oc = I N ⋅ R N Identical open-circuit voltage
3. Superposition Theorem
In any linear bilateral network containing multiple independent sources, the response (voltage or current) in any branch is the algebraic sum of the responses caused by each independent source acting alone, with all other independent voltage sources short-circuited and current sources open-circuited. Note: Power calculation cannot directly use superposition because power is proportional to the square of current ( P = I 2 R P = I^2 R P = I 2 R ).
4. Maximum Power Transfer Theorem
For a DC network, maximum power is transferred from a source to a load resistor R L R_L R L when R L = R t h R_L = R_{th} R L = R t h . Under this condition, the maximum power transferred is: P m a x = V t h 2 4 R t h P_{max} = \frac{V_{th}^2}{4 R_{th}} P ma x = 4 R t h V t h 2
The efficiency η \eta η under maximum power transfer condition is exactly 50%: η = P l o a d P t o t a l × 100 % = 50 % \eta = \frac{P_{load}}{P_{total}} \times 100\% = 50\% η = P t o t a l P l o a d × 100% = 50%
For AC networks with complex impedances ( Z s = R s + j X s Z_s = R_s + jX_s Z s = R s + j X s and Z L = R L + j X L Z_L = R_L + jX_L Z L = R L + j X L ), maximum power transfer requires complex conjugate matching: Z L = Z s ∗ ⟹ R L = R s and X L = − X s Z_L = Z_s^* \implies R_L = R_s \quad \text{and} \quad X_L = -X_s Z L = Z s ∗ ⟹ R L = R s and X L = − X s
RLC Resonance Phenomena & AC Analysis
In AC circuits driven by sinusoidal sources, reactive components (inductors and capacitors) exhibit frequency-dependent reactances: X L = ω L = 2 π f L X_L = \omega L = 2\pi f L X L = ω L = 2 π f L and X C = 1 ω C = 1 2 π f C X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} X C = ω C 1 = 2 π f C 1 .
Series RLC Resonance
A series RLC circuit achieves resonance at the frequency f 0 f_0 f 0 where capacitive reactance cancels inductive reactance ( X L = X C X_L = X_C X L = X C ): ω 0 = 1 L C ⟹ f 0 = 1 2 π L C \omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}} ω 0 = L C 1 ⟹ f 0 = 2 π L C 1
At series resonance:
Total impedance is purely resistive and minimum: Z 0 = R Z_0 = R Z 0 = R .
Circuit current is maximum and in phase with source voltage: I 0 = V s R I_0 = \frac{V_s}{R} I 0 = R V s .
Voltages across inductor and capacitor are equal in magnitude but 180 ∘ 180^\circ 18 0 ∘ out of phase: V L = V C = Q ⋅ V s V_L = V_C = Q \cdot V_s V L = V C = Q ⋅ V s .
Parallel RLC Resonance (Tank Circuit)
In an ideal parallel RLC circuit, resonance occurs when total susceptibility is zero ( B L = B C B_L = B_C B L = B C ): f 0 = 1 2 π L C f_0 = \frac{1}{2\pi\sqrt{LC}} f 0 = 2 π L C 1
At parallel resonance:
Total impedance is maximum (purely resistive): Z 0 = R p Z_0 = R_p Z 0 = R p or dynamic resistance R D = L C R s R_D = \frac{L}{C R_s} R D = C R s L for a practical inductor with series resistance R s R_s R s .
Total line current is minimum and in phase with source voltage.
Circulating currents inside the LC tank loop can be significantly higher than the line current ( I t a n k = Q ⋅ I l i n e I_{tank} = Q \cdot I_{line} I t ank = Q ⋅ I l in e ).
Quality Factor (Q) & Bandwidth (BW)
The Quality Factor ( Q Q Q ) measures the sharpness of resonance and energy efficiency: Q = Reactive Power Average Power = ω 0 L R = 1 ω 0 C R = 1 R L C ( Series RLC ) Q = \frac{\text{Reactive Power}}{\text{Average Power}} = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}} \quad (\text{Series RLC}) Q = Average Power Reactive Power = R ω 0 L = ω 0 CR 1 = R 1 C L ( Series RLC )
The half-power bandwidth ( B W BW B W ) is defined by upper ( f 2 f_2 f 2 ) and lower ( f 1 f_1 f 1 ) cut-off frequencies where power drops to half maximum (current drops to 1 / 2 ≈ 0.707 1/\sqrt{2} \approx 0.707 1/ 2 ≈ 0.707 of peak): B W = f 2 − f 1 = f 0 Q = R 2 π L BW = f_2 - f_1 = \frac{f_0}{Q} = \frac{R}{2\pi L} B W = f 2 − f 1 = Q f 0 = 2 π L R
The half-power cut-off frequencies for high-Q ( Q ≥ 10 Q \ge 10 Q ≥ 10 ) circuits are symmetrical around f 0 f_0 f 0 : f 1 = f 0 − B W 2 , f 2 = f 0 + B W 2 f_1 = f_0 - \frac{BW}{2}, \quad f_2 = f_0 + \frac{BW}{2} f 1 = f 0 − 2 B W , f 2 = f 0 + 2 B W
- Step-by-Step Worked Examples
- Example 1: Thevenin Equivalent & Maximum Power Calculation
Problem: A DC circuit connected across terminals A − B A-B A − B has an open-circuit voltage of 24 V 24\text{ V} 24 V . When a 12 Ω 12\ \Omega 12 Ω load is connected, the terminal voltage drops to 16 V 16\text{ V} 16 V . Find: (a) Thevenin resistance R t h R_{th} R t h , (b) Norton current I N I_N I N , and (c) Maximum power P m a x P_{max} P ma x that can be delivered to a variable load.
Solution:
Identify Open-Circuit Voltage: V t h = V o c = 24 V V_{th} = V_{oc} = 24\text{ V} V t h = V oc = 24 V
Determine R t h R_{th} R t h using Load Test: V L = V t h ⋅ R L R t h + R L ⟹ 16 = 24 ⋅ 12 R t h + 12 V_L = V_{th} \cdot \frac{R_L}{R_{th} + R_L} \implies 16 = 24 \cdot \frac{12}{R_{th} + 12} V L = V t h ⋅ R t h + R L R L ⟹ 16 = 24 ⋅ R t h + 12 12 16 ( R t h + 12 ) = 288 ⟹ 16 R t h + 192 = 288 ⟹ 16 R t h = 96 ⟹ R t h = 6 Ω 16(R_{th} + 12) = 288 \implies 16 R_{th} + 192 = 288 \implies 16 R_{th} = 96 \implies R_{th} = 6\ \Omega 16 ( R t h + 12 ) = 288 ⟹ 16 R t h + 192 = 288 ⟹ 16 R t h = 96 ⟹ R t h = 6 Ω
Calculate Norton Current I N I_N I N : I N = V t h R t h = 24 V 6 Ω = 4 A I_N = \frac{V_{th}}{R_{th}} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A} I N = R t h V t h = 6 Ω 24 V = 4 A
Calculate Maximum Power P m a x P_{max} P ma x : Set R L = R t h = 6 Ω R_L = R_{th} = 6\ \Omega R L = R t h = 6 Ω : P m a x = V t h 2 4 R t h = ( 24 ) 2 4 × 6 = 576 24 = 24 W P_{max} = \frac{V_{th}^2}{4 R_{th}} = \frac{(24)^2}{4 \times 6} = \frac{576}{24} = 24\text{ W} P ma x = 4 R t h V t h 2 = 4 × 6 ( 24 ) 2 = 24 576 = 24 W
Example 2: Series RLC Resonant Frequency and Bandwidth
Problem: A series RLC circuit consists of R = 10 Ω R = 10\ \Omega R = 10 Ω , L = 100 μ H L = 100\ \mu\text{H} L = 100 μ H , and C = 100 pF C = 100\text{ pF} C = 100 pF connected to a 10 V 10\text{ V} 10 V RMS AC source. Calculate: (a) Resonant frequency f 0 f_0 f 0 , (b) Quality factor Q Q Q , and (c) Bandwidth B W BW B W .
Solution:
Calculate f 0 f_0 f 0 : f 0 = 1 2 π L C = 1 2 π 100 × 10 − 6 × 100 × 10 − 12 = 1 2 π 10 − 14 = 1 2 π × 10 − 7 ≈ 1.5915 MHz f_0 = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{100 \times 10^{-6} \times 100 \times 10^{-12}}} = \frac{1}{2\pi \sqrt{10^{-14}}} = \frac{1}{2\pi \times 10^{-7}} \approx 1.5915\text{ MHz} f 0 = 2 π L C 1 = 2 π 100 × 1 0 − 6 × 100 × 1 0 − 12 1 = 2 π 1 0 − 14 1 = 2 π × 1 0 − 7 1 ≈ 1.5915 MHz
Calculate Q Q Q : ω 0 = 2 π f 0 = 10 7 rad/s \omega_0 = 2\pi f_0 = 10^7\text{ rad/s} ω 0 = 2 π f 0 = 1 0 7 rad/s Q = ω 0 L R = 10 7 × 100 × 10 − 6 10 = 1000 10 = 100 Q = \frac{\omega_0 L}{R} = \frac{10^7 \times 100 \times 10^{-6}}{10} = \frac{1000}{10} = 100 Q = R ω 0 L = 10 1 0 7 × 100 × 1 0 − 6 = 10 1000 = 100
Calculate Bandwidth B W BW B W : B W = f 0 Q = 1.5915 MHz 100 = 15.915 kHz BW = \frac{f_0}{Q} = \frac{1.5915\text{ MHz}}{100} = 15.915\text{ kHz} B W = Q f 0 = 100 1.5915 MHz = 15.915 kHz