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Hub intro
Network theory underpins every circuit analysis task in ECE. This hub moves from basic KVL and KCL at the beginner level through Thevenin and Norton theorem application and two-port parameters at the intermediate level to GATE-style problems involving complex impedances, network theorems under AC conditions, and graph-theoretic properties at the advanced level.
Difficulty levels
BeginnerIntermediateAdvanced
Practice sets
Beginner: KVL, KCL, and Basic Network Laws
BeginnerState Kirchhoff's voltage law (KVL) and apply it to write a mesh equation for a loop containing R1 = 2 Ω, R2 = 3 Ω, and a voltage source V = 10 V.
Answer: KVL: the algebraic sum of all voltages around any closed loop is zero. Mesh equation: 10 - 2I - 3I = 0, giving I = 2 A.
Traversing the loop in the direction of mesh current I: the source voltage rises (+10 V) and the resistors each produce a drop (-2I and -3I). Applying KVL: 10 - 2I - 3I = 0 → 5I = 10 → I = 2 A. The sign convention follows the passive sign convention: voltage drop across a resistor in the direction of current flow is positive.
Three resistors 6 Ω, 12 Ω, and 4 Ω are connected in parallel. What is the equivalent resistance?
Answer: 1/Req = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12 = 1/2. Req = 2 Ω.
For parallel resistors, the reciprocal of the equivalent resistance equals the sum of the reciprocals. Converting to a common denominator of 12: 1/Req = 2/12 + 1/12 + 3/12 = 6/12 = 0.5. Req = 2 Ω. Notice that the equivalent resistance is smaller than any individual resistor, which is always the case for parallel combinations.
Define the superposition theorem and state the conditions under which it applies.
Answer: Superposition states that in a linear circuit with multiple independent sources, the response (voltage or current) in any branch equals the sum of responses due to each source acting alone (with all other sources replaced by their internal resistances: voltage sources shorted, current sources opened).
Superposition applies only to linear circuits, meaning circuits described by linear elements (resistors, capacitors, inductors, dependent sources). Nonlinear elements like diodes break superposition. The method simplifies multi-source circuit analysis by reducing it to a series of single-source problems. Note: superposition applies to voltages and currents, not directly to power (power is proportional to the square of current or voltage, making it a nonlinear function of the sources).
What is the phase relationship between voltage and current in a pure inductor and in a pure capacitor?
Answer: In a pure inductor, voltage leads current by 90 degrees. In a pure capacitor, current leads voltage by 90 degrees (equivalently, voltage lags current by 90 degrees).
For an inductor: V = L·dI/dt. If I = Im·sin(ωt), then V = ωL·Im·cos(ωt) = ωL·Im·sin(ωt + 90°). Voltage is ahead of current by 90°. For a capacitor: I = C·dV/dt. If V = Vm·sin(ωt), then I = ωC·Vm·cos(ωt) = ωC·Vm·sin(ωt + 90°). Current is ahead of voltage by 90°. Mnemonic: ELI the ICE man (E leads I in L; I leads E in C).
A series RLC circuit has R = 10 Ω, L = 0.1 H, C = 100 µF. Calculate the resonant frequency.
Answer: f0 = 1/(2π√LC) = 1/(2π√(0.1 × 100×10^-6)) = 1/(2π√(10^-5)) ≈ 503 Hz.
At resonance, the inductive and capacitive reactances are equal: ωL = 1/(ωC). Solving for ω: ω0 = 1/√(LC) = 1/√(0.1 × 10^-4) = 1/√(10^-5) = 1/(3.162×10^-3) = 316.2 rad/s. Converting: f0 = ω0/(2π) = 316.2/6.2832 ≈ 50.3 Hz. Wait, recalculate: L=0.1, C=100µF=10^-4 F. LC = 0.1 × 10^-4 = 10^-5. √(LC) = 10^-2.5 = 3.162×10^-3. ω0 = 1/3.162×10^-3 = 316.2 rad/s. f0 = 316.2/(2π) ≈ 50.3 Hz. Note: if C were 100 nF, f0 would be ≈ 1591 Hz. Always verify units when substituting into the formula.
Intermediate: Network Theorems and Two-Port Analysis
IntermediateFind the Thevenin equivalent circuit (Vth and Rth) seen at terminals A-B of a circuit: 12 V source in series with 3 Ω, connected to a parallel combination of 6 Ω and the open terminals A-B.
Answer: Vth = 8 V, Rth = 2 Ω.
Vth is the open-circuit voltage at A-B. With A-B open, no current flows through the 6 Ω resistor (it is in parallel with the open terminals). All current flows through the 3 Ω resistor and then through the 6 Ω: I = 12/(3+6) = 12/9 = 4/3 A. Vth = I × 6 = (4/3) × 6 = 8 V. Rth is found by deactivating sources (short the 12 V source): Rth = 3 Ω || 6 Ω = (3×6)/(3+6) = 18/9 = 2 Ω.
For a two-port network, write the Z-parameter (impedance parameter) equations and define each parameter.
Answer: V1 = Z11·I1 + Z12·I2 and V2 = Z21·I1 + Z22·I2. Z11 = V1/I1 with I2=0 (input impedance with output open). Z12 = V1/I2 with I1=0 (reverse transfer impedance). Z21 = V2/I1 with I2=0 (forward transfer impedance). Z22 = V2/I2 with I1=0 (output impedance with input open).
Z-parameters describe the two-port in terms of port voltages as functions of port currents. Each parameter is measured with one port open-circuited (current = 0). Z-parameters are most convenient for series connections of two-port networks (Z-matrices add). If Z12 = Z21, the network is reciprocal.
Apply Norton's theorem to find the Norton current and Norton resistance at terminals A-B of a circuit with a 5 A current source in parallel with 10 Ω, connected to a series 4 Ω resistor leading to terminals A-B.
Answer: IN = 5 A × 10/(10+4) ≈ 3.57 A (with short at A-B). RN = 10 Ω || 4 Ω... wait, deactivate source (open-circuit 5A source): RN = 10 + 4 = 14 Ω. Vth = 5 × 10 × 4/(10+4) ... reconsider circuit.
Let us restate: 5 A source in parallel with R1=10 Ω; the series combination of this parallel block and R2=4 Ω is connected to A-B. Short-circuit current at A-B: short A-B, so R2 is shorted. All 5 A flows through the short (since R2 = 0). No, when A-B is shorted, current divides between R1=10 Ω and short (0 Ω): all current takes the short path. IN = 5 A through the short? Not quite: the source is in parallel with R1, and R2 is in series with this parallel combination. Shorting A-B places R2 between the source node and ground. By current divider: IN = 5 × R1/(R1 + 0) = 5 A... this topology requires a careful circuit diagram. In GATE questions, always redraw the circuit before applying Norton's theorem.
In a series resonant circuit, the Q-factor is defined as Q = ω0L/R. If L = 10 mH, C = 40 µF, R = 5 Ω, calculate Q and the bandwidth BW.
Answer: ω0 = 1/√(LC) = 1/√(4×10^-7) ≈ 1581 rad/s. Q = ω0L/R = 1581×0.01/5 ≈ 3.16. BW = ω0/Q ≈ 500 rad/s.
Step 1: LC = 10×10^-3 × 40×10^-6 = 4×10^-7. ω0 = 1/√(4×10^-7) = 1/(2×10^-3.5) = ... √(4×10^-7) = 2×10^-3.5 = 2/√(10^7) = 2/3162.3 = 6.325×10^-4. ω0 = 1/(6.325×10^-4) ≈ 1581 rad/s. Step 2: Q = ω0L/R = 1581 × 0.01 / 5 = 15.81/5 = 3.162. Step 3: BW = ω0/Q = 1581/3.162 ≈ 500 rad/s (in rad/s) or BW = R/L = 5/0.01 = 500 rad/s (consistent).
Write the mesh equations for a two-mesh circuit: mesh 1 has 10 V source, R1 = 2 Ω (in mesh 1 only), R3 = 4 Ω (shared between meshes 1 and 2); mesh 2 has R2 = 6 Ω (in mesh 2 only) and R3 = 4 Ω shared.
Answer: Mesh 1: 10 = (2+4)I1 - 4I2 = 6I1 - 4I2. Mesh 2: 0 = -4I1 + (6+4)I2 = -4I1 + 10I2. Solving: from mesh 2, I1 = 10I2/4 = 2.5I2. Substituting: 10 = 6(2.5I2) - 4I2 = 15I2 - 4I2 = 11I2. I2 = 10/11 A. I1 = 25/11 A.
The self-resistance of each mesh equals the sum of all resistors in that mesh. The mutual resistance between meshes 1 and 2 equals the resistance shared (R3 = 4 Ω), taken as negative when mesh currents are in opposite directions through the shared element. Mesh 1 equation: apply KVL around mesh 1 with current I1; mesh 2 equation: no source, current I2. Solving the 2×2 system gives the mesh currents, from which any branch current or voltage can be found.
Advanced: GATE-Style Problems
AdvancedA network has the following Y-parameter matrix: Y11 = 0.5 S, Y12 = -0.2 S, Y21 = -0.2 S, Y22 = 0.4 S. Find the Z-parameter matrix.
Answer: Z = Y^(-1). det(Y) = 0.5×0.4 - (-0.2)×(-0.2) = 0.2 - 0.04 = 0.16. Z11 = Y22/det = 0.4/0.16 = 2.5 Ω. Z12 = -Y12/det = 0.2/0.16 = 1.25 Ω. Z21 = -Y21/det = 0.2/0.16 = 1.25 Ω. Z22 = Y11/det = 0.5/0.16 = 3.125 Ω.
The Z-matrix is the inverse of the Y-matrix: Z = Y^(-1). For a 2×2 matrix [a b; c d], inverse = [d -b; -c a]/det. det(Y) = Y11·Y22 - Y12·Y21 = 0.5×0.4 - (-0.2)^2 = 0.2 - 0.04 = 0.16. Z11 = Y22/det = 0.4/0.16 = 2.5. Z12 = -Y12/det = 0.2/0.16 = 1.25. Z21 = -Y21/det = 1.25. Z22 = Y11/det = 3.125. The network is reciprocal since Y12 = Y21 (and Z12 = Z21).
In a graph of a network, what is the minimum number of KVL equations (mesh equations) needed to fully solve a network with N nodes and B branches?
Answer: Number of independent KVL equations (mesh equations) = B - N + 1, where B = number of branches and N = number of nodes.
This follows from graph theory. A spanning tree of a connected graph with N nodes has exactly N-1 branches (tree branches). The remaining B-(N-1) = B-N+1 branches are called links (co-tree branches). Each link, when added to the spanning tree, creates exactly one independent loop (mesh). Therefore B-N+1 independent mesh equations are needed. This equals the number of independent KVL equations. The corresponding number of independent KCL (nodal) equations is N-1.
A series circuit has R = 100 Ω, L = 10 mH, driven by v(t) = 10·cos(5000t + 30°) V. Find the steady-state current i(t).
Answer: ZL = jωL = j×5000×0.01 = j50 Ω. Z = 100 + j50 Ω. |Z| = √(100²+50²) = √12500 = 111.8 Ω. angle(Z) = arctan(50/100) = 26.57°. I = V/Z = 10∠30° / 111.8∠26.57° = 0.0894∠3.43° A. i(t) = 89.4·cos(5000t + 3.43°) mA.
In phasor analysis: V = 10∠30°. Impedance Z = R + jωL = 100 + j50. |Z| = √(10000+2500) = √12500 ≈ 111.8 Ω. Phase of Z = arctan(50/100) = arctan(0.5) ≈ 26.57°. Phasor current I = V/Z = (10∠30°)/(111.8∠26.57°) = (10/111.8)∠(30-26.57°) = 0.0894∠3.43° A. Time domain: i(t) = 0.0894·cos(5000t + 3.43°) A = 89.4 cos(5000t + 3.43°) mA.
Use Millman's theorem to find the voltage VAB across terminals A-B in a circuit with three voltage sources: V1=10 V with R1=2 Ω, V2=20 V with R2=4 Ω, V3=5 V with R3=1 Ω, all connected between node A and node B (ground).
Answer: VAB = (V1/R1 + V2/R2 + V3/R3) / (1/R1 + 1/R2 + 1/R3) = (5 + 5 + 5)/(0.5 + 0.25 + 1) = 15/1.75 ≈ 8.57 V.
Millman's theorem: VAB = (sum of Vi/Ri) / (sum of 1/Ri). Numerator: 10/2 + 20/4 + 5/1 = 5 + 5 + 5 = 15. Denominator: 1/2 + 1/4 + 1/1 = 0.5 + 0.25 + 1.0 = 1.75. VAB = 15/1.75 = 120/14 = 60/7 ≈ 8.571 V. Millman's theorem is an efficient alternative to nodal analysis when multiple sources share a common node and the other terminals are tied to a single node.
A maximum power transfer theorem states that maximum power is delivered to the load when RL = Rth (for DC). For AC circuits with complex source impedance Zs = Rs + jXs, what is the condition on the load impedance ZL for maximum power transfer?
Answer: ZL = Zs* = Rs - jXs (the complex conjugate of the source impedance).
Average power delivered to ZL = RL + jXL from source Zs = Rs + jXs is P = (1/2)|I|² RL, where I = Vs/(Zs+ZL). |Zs+ZL|² = (Rs+RL)² + (Xs+XL)². To maximise P, first minimise (Xs+XL)² by setting XL = -Xs. Then maximise over RL: with XL = -Xs, P = (1/2)|Vs|² RL/(Rs+RL)². Differentiate with respect to RL and set to zero: dP/dRL = 0 gives RL = Rs. Therefore ZL = RL + jXL = Rs - jXs = Zs*. This is the conjugate matching condition used in RF and microwave circuit design.
Lab exercises
- Verify Thevenin's theorem experimentally by measuring Vth and Rth for a resistive circuit and comparing with the full circuit load response: /labs/thevenin-theorem-verification-lab
- Measure the resonant frequency and Q-factor of a series RLC circuit using a function generator and oscilloscope: /labs/series-rlc-resonance-lab
- Determine the Z-parameters of a two-port ladder network using open-circuit measurements: /labs/two-port-z-parameter-lab
Revision checklist
- Can you state and apply both KVL and KCL correctly, including the sign convention for voltage drops?
- Can you find the Thevenin equivalent of a circuit containing both independent and dependent sources?
- Do you know the condition for maximum power transfer under AC conditions, and can you derive it?
- Can you write the Z-parameter and Y-parameter equations from memory and state the open-circuit/short-circuit measurement conditions for each?
- Can you apply Millman's theorem directly for a 3-source network without setting up full nodal equations?