Network Elements, Network Theorems, Transients and AC Power
1. Elements, sources, and node and mesh analysis
An ideal voltage source holds its terminal voltage whatever current flows — zero internal resistance; an ideal current source delivers its current whatever the voltage — infinite internal resistance. A dependent source is set by a voltage or current elsewhere in the circuit: VCVS, VCCS, CCVS and CCCS. The passive elements obey v = Ri, v = L di/dt and i = C dv/dt; the inductor stores ½Li² and the capacitor ½Cv². Mutual inductance M couples two coils, v₂ = M di₁/dt, with coupling coefficient k = M/√(L₁L₂) ≤ 1; in series the total inductance is L₁ + L₂ + 2M aiding and L₁ + L₂ − 2M opposing, the sense set by the dot convention.
KCL: the currents leaving a node sum to zero. KVL: the voltages around a loop sum to zero. Node analysis takes one node as reference and writes KCL at the other n − 1 in terms of node voltages; a voltage source between two non-reference nodes is handled as a supernode. Mesh analysis writes KVL around each of the b − n + 1 meshes of a planar network in terms of mesh currents; a current source shared by two meshes gives a supermesh. Mesh analysis needs a planar network; node analysis does not.
2. Thevenin, Norton, superposition and maximum power transfer
Any linear two-terminal network is equivalent to a source V_th (the open-circuit voltage) in series with R_th (the resistance seen with the independent sources deactivated), or to a Norton source I_N = V_th/R_th (the short-circuit current) in parallel with the same resistance; R_th = V_oc/I_sc also works when dependent sources are present. Superposition adds the responses to each independent source acting alone, with the others deactivated — a voltage source replaced by a short, a current source by an open — and dependent sources left in place. It applies to voltages and currents, never directly to power.
Maximum power reaches a resistive load when R_L = R_th, and it is P_max = V_th²/(4R_th), at 50% efficiency; for an AC source the load that takes maximum power is the conjugate, Z_L = Z_th*. Worked: a 12 V source with 4 Ω in series feeds a 12 Ω shunt resistor at the terminals. V_th = 12 × 12/16 = 9 V, R_th = 4 ∥ 12 = 3 Ω, I_N = 3 A, and a 3 Ω load takes 81/12 = 6.75 W.
3. Transient response of DC and AC networks
An inductor’s current and a capacitor’s voltage cannot change instantaneously — both would need infinite power — so they carry the circuit across a switching instant. Every first-order response is x(t) = x(∞) + [x(0⁺) − x(∞)] e−t/τ, with τ = RC or τ = L/R, R being the Thevenin resistance seen by the storage element. A 10 V source charging 100 μF through 10 kΩ has τ = 1 s, and v(1 s) = 10(1 − e−1) ≈ 6.32 V: 63.2% of the way in one time constant, and within 1% after about 5τ.
A series RLC circuit is second order, with ω_n = 1/√(LC) and damping ratio ζ = (R/2)√(C/L): overdamped for ζ > 1, critically damped at ζ = 1, underdamped and ringing at ω_d = ω_n√(1 − ζ²) for ζ < 1. When an AC source is switched on, the response is the sinusoidal steady state plus a transient that decays with the same τ or ζ; the transient’s size depends on the switching instant, which is why closing onto an inductive load at a voltage zero gives the largest current offset.
4. Sinusoidal steady state, resonance and two-port networks
In the sinusoidal steady state every quantity is a phasor and every element an impedance: Z_R = R, Z_L = jωL, Z_C = 1/(jωC). A series RLC circuit is resonant when the reactances cancel, at ω₀ = 1/√(LC); the impedance is then its minimum, R, the current its maximum, and the quality factor Q = ω₀L/R = (1/R)√(L/C) sets the bandwidth, BW = ω₀/Q. With L = 10 mH, C = 1 μF and R = 10 Ω: ω₀ = 10⁴ rad/s (f₀ ≈ 1592 Hz) and Q = (1/10)√(10⁻²/10⁻⁶) = 10. A parallel LC tank is the dual: impedance a maximum at resonance.
| Set | Defining equations | Reciprocal network |
|---|---|---|
| z (impedance) | V₁ = z₁₁I₁ + z₁₂I₂, V₂ = z₂₁I₁ + z₂₂I₂ | z₁₂ = z₂₁ |
| y (admittance) | I₁ = y₁₁V₁ + y₁₂V₂, I₂ = y₂₁V₁ + y₂₂V₂ | y₁₂ = y₂₁ |
| h (hybrid) | V₁ = h₁₁I₁ + h₁₂V₂, I₂ = h₂₁I₁ + h₂₂V₂ | h₁₂ = −h₂₁ |
| ABCD (transmission) | V₁ = AV₂ − BI₂, I₁ = CV₂ − DI₂ | AD − BC = 1 |
A network is symmetric as well when z₁₁ = z₂₂ (or A = D) — a stronger property than reciprocity, which every network of R, L, C and M elements has. The h parameters are the natural description of a transistor’s small-signal model, and cascaded two-ports multiply their ABCD matrices.
5. Balanced three-phase circuits, complex power and power factor
A balanced three-phase supply has three equal phasors 120° apart. In star (Y), V_L = √3 V_ph and I_L = I_ph; in delta (Δ), V_L = V_ph and I_L = √3 I_ph. Either way the total power is P = √3 V_L I_L cos φ, φ the angle of the per-phase impedance. A 400 V line supply feeding a balanced star of 10 Ω resistors per phase: V_ph = 400/√3 ≈ 230.9 V, I = 23.1 A, and P = 3 × 230.9²/10 = 400²/10 = 16 kW. The same resistors in delta see the full 400 V and take three times as much, 48 kW.
Complex power S = V I* = P + jQ: P in watts is the real power, Q in volt-ampere reactive (VAR) the reactive power, positive for an inductive load, and |S| in VA the apparent power. The power factor is cos φ = P/|S|. A 10 kW load at 0.8 lagging draws |S| = 12.5 kVA and Q = √(12.5² − 10²) = 7.5 kVAR; a shunt capacitor supplying that 7.5 kVAR raises the power factor to 1 and cuts the line current by a fifth, with the load’s own P unchanged.
Key takeaways
- An ideal voltage source has zero internal resistance and an ideal current source infinite; M couples coils with k = M/√(L₁L₂) and L = L₁ + L₂ ± 2M in series.
- Deactivate only independent sources: a voltage source becomes a short, a current source an open. P_max = V_th²/(4R_th) at R_L = R_th.
- First-order: x(t) = x(∞) + [x(0⁺) − x(∞)]e−t/τ; inductor current and capacitor voltage are continuous.
- Series resonance at ω₀ = 1/√(LC), Q = (1/R)√(L/C), BW = ω₀/Q; reciprocity is z₁₂ = z₂₁, h₁₂ = −h₂₁, AD − BC = 1.
- Star: V_L = √3 V_ph; delta: I_L = √3 I_ph; P = √3 V_L I_L cos φ. S = P + jQ and pf = P/|S|.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The internal resistance of an ideal current source is:
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Answer: A — infinite
An ideal current source delivers its current into any load, which requires that none of it be diverted through an internal shunt — an infinite parallel resistance. The ideal voltage source is the dual, with zero series resistance.A 12 V ideal source in series with 4 Ω feeds terminals a–b, across which a 12 Ω resistor is connected. The Thevenin voltage at a–b is ____ V.
Numerical answer — type the value.
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Answer: 9
With a–b open, the 4 Ω and 12 Ω form a divider: V_th = 12 × 12/(4 + 12) = 9 V. Reporting 12 V ignores the drop across the series 4 Ω.A 12 V ideal source in series with 4 Ω feeds terminals a–b, across which a fixed 12 Ω resistor is connected. A variable resistive load is now added across a–b as well. The maximum power that can be delivered to this variable load is ____ W.
Numerical answer — type the value.
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Answer: 6.75
R_th = 4 ∥ 12 = 48/16 = 3 Ω with the source shorted, so the load is 3 Ω and P_max = V_th²/(4R_th) = 81/12 = 6.75 W. Using 4 Ω as R_th forgets the 12 Ω in parallel and gives 81/16 ≈ 5.06 W.While applying superposition to a circuit containing independent and dependent sources, a source not under consideration is handled by:
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Answer: A — shorting independent voltage sources, opening independent current sources, and leaving dependent sources in place
A zero-valued voltage source is a short and a zero-valued current source an open. Dependent sources are not inputs; they describe how the circuit itself behaves, so they stay. Opening a voltage source would break the circuit rather than zero the source.An uncharged 100 μF capacitor is charged from a 10 V DC source through 10 kΩ, the switch closing at t = 0. The capacitor voltage at t = 1 s, correct to two decimal places, is ____ V.
Numerical answer — type the value.
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Answer: 6.32
τ = RC = 10⁴ × 10⁻⁴ = 1 s, and v = 10(1 − e−t/τ) = 10(1 − e−1) = 10 × 0.6321 = 6.32 V. Writing 10e−1 ≈ 3.68 V gives the discharge curve instead.At the instant a switch operates in an RLC circuit with finite source voltages, which quantities cannot change instantaneously?
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Answer: A — The inductor current and the capacitor voltage
A step in inductor current needs infinite v = L di/dt, and a step in capacitor voltage needs infinite i = C dv/dt; the stored energies ½Li² and ½Cv² are continuous. The inductor voltage and the capacitor current can, and usually do, jump.A series RLC circuit has R = 10 Ω, L = 10 mH and C = 1 μF. Its quality factor at resonance is ____.
Numerical answer — type the value.
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Answer: 10
ω₀ = 1/√(LC) = 1/√(10⁻⁸) = 10⁴ rad/s, and Q = ω₀L/R = 10⁴ × 0.01/10 = 10, the same as (1/R)√(L/C) = (1/10)√(10⁴) = 10. The bandwidth is ω₀/Q = 1000 rad/s.Two coupled coils of 4 mH and 9 mH with coupling coefficient 0.5 are connected in series opposing. The total inductance is:
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Answer: A — 7 mH
M = k√(L₁L₂) = 0.5 × √36 = 3 mH. Opposing: L = L₁ + L₂ − 2M = 13 − 6 = 7 mH. Aiding would give 19 mH; 13 mH ignores the coupling, and 10 mH subtracts M only once.A balanced star-connected load of 10 Ω resistance per phase is supplied from a 400 V (line), three-phase supply. The total power drawn is ____ kW.
Numerical answer — type the value.
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Answer: 16
V_ph = 400/√3 ≈ 230.9 V, so each phase takes V_ph²/R = (160000/3)/10 ≈ 5333 W and the total is 16 000 W = 16 kW; equivalently √3 V_L I_L = √3 × 400 × 23.09 ≈ 16 kW. Using 400 V across each resistor is the delta connection and gives 48 kW.For a reciprocal two-port network, which relations always hold?
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Answer: A — z₁₂ = z₂₁; B — AD − BC = 1; C — h₁₂ = −h₂₁
Reciprocity means the transfer ratio is the same in both directions, which reads as z₁₂ = z₂₁, y₁₂ = y₂₁, h₁₂ = −h₂₁ and AD − BC = 1 in the four sets. y₁₁ = y₂₂ is symmetry, a further condition: a T-network with unequal arms is reciprocal but not symmetric.The reactive power drawn by a load is expressed in:
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Answer: A — volt-ampere reactive (VAR)
S = P + jQ: real power P is in watts, reactive power Q in VAR and the apparent power |S| in VA. All three are dimensionally V × A; the different names mark which part of S is meant.A load takes 10 kW at a power factor of 0.8 lagging. Its reactive power is ____ kVAR.
Numerical answer — type the value.
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Answer: 7.5
|S| = P/cos φ = 10/0.8 = 12.5 kVA, and Q = |S| sin φ = 12.5 × 0.6 = 7.5 kVAR (or √(12.5² − 10²)). 12.5 is the apparent power, not the reactive part.A connected planar network has n nodes and b branches. Which statements about node and mesh analysis are true?
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Answer: A — It has n − 1 independent KCL equations; B — It has b − n + 1 independent mesh equations; C — A voltage source between two non-reference nodes is treated as a supernode
(A) The n KCL equations sum to 0 = 0, so one is dependent. (B) Branch unknowns less node equations: b − (n − 1). (C) The source fixes the difference of the two node voltages, so KCL is written around both together. (D) False: meshes are defined only for a planar drawing; a non-planar network needs loop or node analysis.