A connected resistive network has 6 nodes and 9 branches. Three of its ideal independent voltage sources each have one terminal on the reference node, and one ideal independent current source lies wholly inside a single mesh. After those sources have been used to delete unknowns, which method is left with fewer unknowns, and by how many?

Both methods are left with the same number of unknowns

A two-port network is built from resistors, capacitors, an ideal transformer and one voltage-controlled current source. An experimenter drives port 1 with a voltage source and measures the short-circuit current at port 2, then interchanges the source and the ammeter and finds a different reading. What accounts for the change?

The voltage-controlled current source: a dependent source of any kind destroys reciprocity

The capacitors, because reciprocity holds only in a purely resistive network

The ideal transformer, which passes power in one direction only

The two ports presenting different driving-point impedances, so the network is unsymmetric

A network is made of linear resistors, several independent sources and one current-controlled voltage source. Which of the following procedures give correct results for this network?

Obtaining the Thevenin resistance at a port as the open-circuit voltage divided by the short-circuit current

Adding the branch voltages produced by each independent source acting alone, with the controlled source left active throughout

Obtaining the Thevenin resistance at a port as the open-circuit voltage divided by the short-circuit current

Adding the branch voltages produced by each independent source acting alone, with the controlled source left active throughout

Obtaining the Thevenin resistance by shorting every voltage source, opening every current source and reducing the remaining resistors

Adding the powers dissipated in a branch by each independent source acting alone to obtain the total power in that branch

Circuit Analysis and the Sinusoidal Steady State

Section 2 of the Electronics & Communication paper is called Networks, Signals and Systems, and it opens with the only part of the section where nothing is transformed: a network of resistors and sources, sitting still, obeying two laws. Everything later in the section — the Laplace solution, the two-port matrix, the transfer function, the filter — is built on the habit this chapter teaches, which is writing the smallest correct set of equations and then not solving them by hand any longer than necessary. Kirchhoff's current law at a node and voltage law round a loop are the only physics; the skill is choosing between them. Node analysis needs N - 1 equations for N nodes and loves voltage sources, because each one whose terminal sits on the reference node deletes an unknown outright. Mesh analysis needs B - N + 1 equations and loves current sources, for the mirror-image reason. Count both before you write anything, because the count is the whole decision and it takes ten seconds. On top of that sit four theorems that GATE returns to every year. Superposition lets you switch sources on one at a time — and refuses, absolutely, to work on power, because power is quadratic in the response and quadratic functions do not add. Thevenin and Norton collapse an arbitrary linear box into one source and one impedance, and the safe route to both is V_oc and I_sc, because that route survives dependent sources whereas killing the sources and combining resistors does not. Reciprocity says a passive network cannot tell which way you are driving it: swap the source and the ammeter and the reading does not change, even though nothing else about the circuit is symmetric. Then the chapter turns the sources on and leaves them on. In the sinusoidal steady state every derivative becomes a multiplication by jω, differential equations become algebra, and a circuit becomes a complex-number division. Complex power S = V I* splits into the P that heats things and the Q that only sloshes back and forth, and the split is what a power factor measures. Maximum power transfer closes the chapter with the single most misquoted result in the section: the load must be the conjugate of the source impedance, not a copy of it, and the difference between the two is 36 per cent of the available power.

Node and Mesh Analysis: Count Before You Compute

Both methods are the same two laws applied in opposite orders. Node analysis takes the node voltages as unknowns and writes KCL at each of them, using Ohm's law to turn every branch current into a difference of node voltages. Mesh analysis takes the mesh currents as unknowns and writes KVL round each mesh. For a connected network with N nodes and B branches, node analysis produces N - 1 equations and mesh analysis produces B - N + 1, and those two numbers are almost never equal — so the first thing to do with a circuit is count them. Then apply the two adjustments that actually decide the contest. An ideal voltage source with one terminal on the reference node removes a node unknown, because that node voltage is simply known; if neither terminal is on the reference you merge the two nodes into a supernode and write one KCL round the pair plus the source's own voltage constraint. An ideal current source belonging to exactly one mesh removes a mesh unknown, because that mesh current is known; if it is shared between two meshes you form a supermesh and write KVL round the outside of the pair.

There is also a structural reason to prefer node analysis as a default habit rather than a coin-toss. Mesh analysis needs the circuit to be planar — you must be able to draw it without crossings for meshes to exist at all — while node analysis works on any topology whatever, which is why every circuit simulator you will ever use solves nodal equations internally. So the working rule is: count the two totals, subtract the sources that delete unknowns, and take the smaller; on a tie, take nodes. The remaining discipline is bookkeeping. Fix the reference node once and never move it, choose all mesh currents in the same rotational sense so that the off-diagonal terms of the resistance matrix come out negative and a sign error stands out, and label every unknown before writing a single equation.

🎯 The network this chapter keeps returning to, solved once properly

Here is the circuit that carries the rest of this chapter, described so that you can draw it. A 12 V independent voltage source has its negative terminal on the reference node and its positive terminal feeding a 4 Ω resistor whose far end is node 1. From node 1 a 12 Ω resistor returns to the reference node, and a 2 Ω resistor crosses to node 2. From node 2 a second 2 Ω resistor returns to the reference node, and a 1 A independent current source pushes current from the reference node into node 2. Count first: N = 4 nodes and B = 6 branches, so node analysis wants three equations less one for the grounded 12 V source, and mesh analysis wants 6 - 4 + 1 = 3 less one for the current source, which sits in a single mesh. Two equations either way — a genuine tie, so take nodes. KCL at node 1 gives (V1 - 12)/4 + V1/12 + (V1 - V2)/2 = 0, which clears to 5V1 - 3V2 = 18. KCL at node 2 gives (V2 - V1)/2 + V2/2 = 1, which clears to -V1 + 2V2 = 2. Solving, V1 = 6 V and V2 = 4 V. Now spend the ten seconds that catches most slips: the 4 Ω carries (12 - 6)/4 = 1.5 A into node 1, the 12 Ω carries 0.5 A away, the 2 Ω bridge carries (6 - 4)/2 = 1 A towards node 2, and 1.5 = 0.5 + 1 ✓; at node 2, 1 A arrives through the bridge and 1 A from the source while 4/2 = 2 A leaves through the shunt ✓. Both KCLs close exactly. That check is not optional politeness — a sign error in a nodal equation produces a perfectly plausible pair of voltages, and the only thing that exposes it is putting the currents back.

The two-node network: every branch, and the two KCLs closing
Branch	Voltage across	Current	Power
4 Ω, source to node 1	6 V	1.5 A into node 1	9 W absorbed
12 Ω, node 1 to reference	6 V	0.5 A to reference	3 W absorbed
2 Ω bridge, node 1 to node 2	2 V	1 A towards node 2	2 W absorbed
2 Ω, node 2 to reference	4 V	2 A to reference	8 W absorbed
12 V source	12 V	1.5 A delivered	18 W delivered
1 A source	4 V	1 A delivered	4 W delivered
Totals	not additive	KCL closes at both nodes	22 W in = 22 W out

The last row is worth pausing on, because it is a second independent check and a different one from KCL. Tellegen's theorem says the total power delivered by the sources equals the total absorbed by the passive elements, always, in any network obeying Kirchhoff's laws. Here the sources supply 18 + 4 = 22 W and the four resistors absorb 9 + 3 + 2 + 8 = 22 W. If your node voltages satisfy KCL but not the power balance you have made an arithmetic error in the powers; if they satisfy neither you have made an error in the equations. Two cheap checks that fail in different ways are worth more than one expensive one.

Superposition, and the One Quantity That Refuses to Superpose

Superposition is a consequence of linearity and nothing more. In a network of linear elements and independent sources, the response to all sources acting together is the sum of the responses to each acting alone, with the others deactivated — voltage sources replaced by short circuits and current sources by open circuits, because those are the states in which each contributes nothing. Two conditions and one restriction. The elements must be linear, so a network containing a diode or a saturating core is out of scope. Dependent sources are never deactivated: they are not inputs, they are part of the network's description, and switching one off changes the circuit into a different circuit. And the response you superpose must itself be a linear function of the sources — a voltage or a current, never a power.

Run it on the two-node network. With the 12 V source alone the current source becomes an open circuit, node 2 sees only the 2 Ω bridge and the 2 Ω shunt, and solving the same pair of equations with the right-hand side (3, 0) gives V1 = 36/7 V and V2 = 18/7 V. With the 1 A source alone the 12 V source becomes a short, so the 4 Ω now runs from node 1 to the reference, and the right-hand side (0, 1) gives V1 = 6/7 V and V2 = 10/7 V. Add them: V1 = 36/7 + 6/7 = 42/7 = 6 V and V2 = 18/7 + 10/7 = 28/7 = 4 V, exactly the values found by solving the full circuit. Notice that neither partial answer is a round number and the sum is. Superposition guarantees nothing about the parts, only about the total — a candidate who abandons a correct partial result because it came out in sevenths has thrown away a correct method.

⚠️ Superposing the powers gets 4.33 W where the answer is 8 W

Ask for the power dissipated in the 2 Ω resistor at node 2. Done correctly it is trivial: V2 = 4 V, so P = V2²/R = 16/2 = 8 W. Now do it the way superposition tempts you to. The 12 V source alone puts 18/7 V across that resistor, worth (18/7)²/2 = 3.306 W. The 1 A source alone puts 10/7 V across it, worth (10/7)²/2 = 1.020 W. Their sum is 4.327 W, and the true answer is 8 W — the method has lost 46 per cent of the power. The reason is exact and worth seeing rather than remembering. Power is v²/R, and (a + b)² = a² + 2ab + b²: the cross term 2ab/R = 2(18/7)(10/7)/2 = 3.673 W is precisely what superposing the two powers throws away, and 3.306 + 1.020 + 3.673 = 8.000 ✓. So the rule has a sharp form. Superpose the voltage or the current, then square once at the end. The cross term can also be negative — if the two sources drive the branch in opposite senses, superposed powers come out larger than the truth, so you cannot even treat the error as a safe under-estimate. The same warning covers RMS values, and for the same algebraic reason: RMS values add in quadrature only when the components are orthogonal, which two sinusoids at the same frequency are not.

The positive use of superposition in GATE is rarely arithmetic economy — solving the same circuit twice is usually more work than solving it once. It is structural insight. It tells you at a glance that a response is a weighted sum of the source values, so a question that gives you two measurements and asks for a third is answered by linear algebra rather than by circuit analysis: if V = αV_s + βI_s and you know V for two source pairs, you know α and β and therefore V for every pair. It also underlies every result later in the section — the transfer function, the two-port matrix, the convolution integral — all of which are superposition applied to a continuum of inputs rather than to two of them.

Thevenin and Norton: One Source, One Resistance

Seen from any two terminals, a linear network of sources and resistors is indistinguishable from a single voltage source V_th in series with a single resistance R_th — that is Thevenin's theorem — and equally indistinguishable from a single current source I_N in parallel with R_N, which is Norton's. The two are the same statement, related by V_th = I_N R_N and R_th = R_N, so finding either gives you the other for free. Three routes to R_th exist, and knowing which is legal when is the examinable part:

Deactivate and combine. Short every independent voltage source, open every independent current source, and reduce the resistor network by series and parallel. Fast, and legal only if there are no dependent sources.
Apply a test source. Deactivate the independent sources only, drive the port with 1 V (or 1 A), find the resulting current (or voltage), and take the ratio. Works with dependent sources present.
Open circuit over short circuit. R_th = V_oc / I_sc, with every source left exactly as it is. This is the safest route of the three, needs no judgement about what may be switched off, and fails only in the pathological case I_sc = 0.

Do all of it on the two-node network, looking in at node 2 with the 2 Ω shunt removed as the load. Open circuit: with that resistor gone the 1 A source has nowhere to go but through the 2 Ω bridge, so node 1 rises to 12 V, node 2 sits 2 V above it, and V_th = 14 V. Deactivate and combine: short the 12 V source and open the 1 A source, and from node 2 you see the 2 Ω bridge in series with 4 ∥ 12 = 3 Ω, so R_th = 5 Ω. Short circuit: ground node 2 and node 1 falls to 3.6 V, sending 1.8 A down the bridge, to which the source adds its 1 A, so I_sc = 2.8 A. Three independent computations, and they agree: V_oc/I_sc = 14/2.8 = 5 Ω ✓. Finally put the load back and check against the original solution — 14 × 2/(2 + 5) = 4 V, which is exactly the V2 we found the long way ✓. The Norton form is I_N = 2.8 A in parallel with 5 Ω, and 2.8 × 5/7 = 2 A through the load, 4 V across it, the same again.

⚠️ A dependent source, and the 33 per cent error from killing it

This is the configuration that separates candidates who learned the theorem from candidates who learned one recipe for it, and it is a small-signal amplifier in disguise. A 10 V source drives a 2 kΩ resistor into node a, which is the port. From node a a 4 kΩ resistor goes to the reference. And a dependent current source of value 0.5 i pulls current out of node a into the reference, where i is the current flowing from the 10 V source through the 2 kΩ into node a. Work in volts, milliamps and kilohms so no powers of ten appear. Open circuit: i = (10 - V_a)/2 and KCL at a reads i = V_a/4 + 0.5i, so 0.5i = V_a/4, giving (10 - V_a)/4 = V_a/4 and hence V_a = 5 V. Check it: i = 2.5 mA, the dependent source takes 1.25 mA, the 4 kΩ takes 5/4 = 1.25 mA, and 2.5 = 1.25 + 1.25 ✓. So V_th = 5 V. Short circuit: with V_a = 0 we get i = 5 mA, the 4 kΩ carries nothing, and the current into the short is 5 - 2.5 = 2.5 mA. So R_th = 5/2.5 = 2 kΩ. Confirm it independently with a test source: short the 10 V, apply V_a at the port, and I_t = V_a/4 - 0.5i with i = -V_a/2, which gives I_t = V_a/4 + V_a/4 = V_a/2, so R_th = 2 kΩ again ✓. Now the wrong route. Kill everything in sight, including the dependent source, and you get 2 ∥ 4 = 1.333 kΩ — low by a third. The dependent source is not an excitation; it is a description of how this circuit responds to its own state, and removing it describes a different circuit. The practical sting is that a wrong R_th gives a wrong matched load: choosing R_L = 1.333 kΩ instead of 2 kΩ delivers 3.00 mW instead of the available 3.125 mW. Only four per cent lost — which is exactly why the error survives undetected in coursework and is examined directly instead.

ℹ️ What the marking scheme does to a question like that

GATE is 65 questions, 100 marks, 180 minutes, of which General Aptitude is 10 questions for 15 marks and the rest belongs to your chosen paper. The mechanics that matter for a circuit question are in the negative marking. A wrong MCQ costs one third of the question's value — a third of a mark on a one-marker, two thirds on a two-marker — while MSQ and NAT questions carry no negative marking at all. Read that against the trap above. A Thevenin resistance question set as MCQ will list 1.33 kΩ among the options precisely because it is the answer produced by the common wrong method, so a half-remembered recipe does not merely fail to earn the mark, it costs you a third of one. The same question set as NAT is free to attempt: your best value is strictly better than a blank. And because the MCQ penalty is exactly one third of the value, a blind guess among four options is priced at exactly zero, so every option you can eliminate makes the guess worth taking — which for this topic usually means eliminating on units and magnitude before doing any circuit work at all.

Reciprocity: Swap the Source and the Ammeter

Reciprocity is the least computational theorem in the syllabus and the one most often misread as obvious. Its statement is precise: in a linear, passive, time-invariant network containing a single independent source and no dependent sources, if a voltage source at port 1 produces a short-circuit current at port 2, then moving that same voltage source to port 2 produces the same current at a short across port 1. The ratio of excitation to response — a transfer impedance, in ohms — is unchanged by the swap. The conditions are where the marks are. Any dependent source destroys reciprocity, which is why a transistor amplifier is not reciprocal and why an isolator or a circulator, both of which rely on non-reciprocal magnetic materials, exists at all.

🎯 A deliberately lopsided T-network reads 1 A in both directions

Take a T-network with nothing symmetric about it. A 4 Ω resistor runs from port 1 to an internal node x; a 6 Ω resistor runs from x to port 2; a 12 Ω resistor runs from x to the common reference. Both ports return to that reference. Drive port 1 with 12 V and short port 2. Looking in from port 1 you see 4 Ω in series with 12 ∥ 6 = 4 Ω, so the source delivers 12/8 = 1.5 A, the internal node sits at V_x = 1.5 × 4 = 6 V, and the current through the 6 Ω into the short is 6/6 = 1 A. Now swap: drive port 2 with 12 V and short port 1. Looking in from port 2 you see 6 Ω in series with 12 ∥ 4 = 3 Ω, so the source delivers 12/9 = 1.333 A, the internal node sits at V_x = 1.333 × 3 = 4 V, and the current through the 4 Ω into the short is 4/4 = 1 A. Identical. Look at what did not stay the same: the input resistance went from 8 Ω to 9 Ω, the source current from 1.5 A to 1.333 A, the internal node voltage from 6 V to 4 V, and the power drawn from the source from 18 W to 16 W. Every single one of those changed. The transfer impedance, 12 V / 1 A = 12 Ω, is the one quantity that did not, and it is no accident that 12 Ω is the value of the shunt arm — for a T-network the transfer impedance is the shunt element. This is why reciprocity is examined as a one-line reasoning question rather than a calculation: you are given the response one way round and asked for it the other, and the correct answer is the same number, while the distractors are the plausible-looking values you would get by re-solving the circuit and accidentally reporting the input current or the node voltage instead.

Reciprocity has a second form that is the same theorem with the roles of voltage and current exchanged: a current source at port 1 producing an open-circuit voltage at port 2 gives the same open-circuit voltage at port 1 when moved to port 2. And it has a matrix form that the next chapter will use directly — a reciprocal two-port satisfies z12 = z21, equivalently y12 = y21, equivalently h12 = -h21, equivalently AD - BC = 1. Note carefully what reciprocity does not claim. It does not say the network is symmetric: symmetry is the stronger condition z11 = z22, meaning the two ports are interchangeable in every respect, and the T-network above is reciprocal (z12 = z21 = 12 Ω) while being thoroughly unsymmetric (z11 = 16 Ω, z22 = 18 Ω). Every network of resistors, inductors, capacitors and ideal transformers is reciprocal; only a few of them are symmetric.

Phasors: Turning Calculus into Algebra

In the sinusoidal steady state every voltage and current in a linear circuit is a sinusoid at the source frequency, differing only in amplitude and phase. That single fact is what makes the phasor method exact rather than approximate. Represent v(t) = V_m cos(ωt + θ) by the complex number V = V_m∠θ (or, as power engineering prefers, by its RMS value V_m/√2 at the same angle), and every differentiation becomes multiplication by jω. So v = L di/dt becomes V = jωL · I, and i = C dv/dt becomes I = jωC · V, and the whole apparatus of resistive circuit analysis transfers unchanged: series and parallel combination, node and mesh analysis, superposition, Thevenin, Norton and reciprocity all hold with resistance replaced by complex impedance Z. The differential equations have not been solved; they have been sidestepped, because the steady-state solution was assumed at the start.

Element impedances, and what each one does to the phase
Element	Impedance Z	Current relative to voltage	As ω rises
Resistance R	`R`	in phase	no change
Inductance L	`jωL`	lags by 90°	blocks more
Capacitance C	`1/(jωC) = -j/(ωC)`	leads by 90°	blocks less
Series R-L	`R + jωL`	lags by `arctan(ωL/R)`	lag grows towards 90°
Series R-C	`R - j/(ωC)`	leads by `arctan(1/ωRC)`	lead falls towards 0°

Two habits keep phasor work clean. First, decide once whether your magnitudes are peak or RMS and never mix them; GATE questions on power are almost always in RMS, and a factor of √2 misplaced in a power answer is a factor of 2. Second, keep impedances in rectangular form for addition and in polar form for multiplication and division, converting only when the operation demands it. A series R-L load of R = 8 Ω and X_L = 6 Ω is Z = 8 + j6 Ω for the purpose of adding a source impedance, and Z = 10∠36.87° Ω for the purpose of dividing a voltage by it. Both descriptions are exact and neither is preferable in general; the arithmetic decides. Driven by V = 230∠0° V RMS, that load draws I = 230∠0°/10∠36.87° = 23∠-36.87° A, a current of 23 A lagging the voltage by 36.87 degrees — and the fact that the current lags is not a detail, it is the whole subject of the next section.

Complex Power: What Heats, What Only Sloshes

For a load carrying phasor current I at phasor voltage V, both RMS, the complex power is S = V I* — voltage times the conjugate of the current, and the conjugate is what makes the sign of the imaginary part mean something. Write S = P + jQ. The real part P = |V||I| cos φ, in watts, is active power: the average rate at which energy leaves the circuit and does not come back. The imaginary part Q = |V||I| sin φ, in volt-amperes reactive, is reactive power: energy that flows into the reactance for a quarter cycle and flows straight back out again, averaging to zero over a cycle while being entirely real while it happens. The magnitude |S| = |V||I|, in volt-amperes, is apparent power, and the ratio P/|S| = cos φ is the power factor, described as lagging when the current lags (an inductive load, Q > 0) and leading when it leads (a capacitive load, Q < 0).

Continue with the 8 + j6 Ω load on 230 V RMS. We had I = 23∠-36.87° A, so S = 230 × 23∠+36.87° = 5290∠36.87° = 4232 + j3174 VA. Read that off: P = 4232 W, Q = 3174 VAr inductive, |S| = 5290 VA, power factor 4232/5290 = 0.8 lagging. Then check it the other way, from the elements: P = |I|²R = 529 × 8 = 4232 W ✓ and Q = |I|²X = 529 × 6 = 3174 VAr ✓. That second route is the one to trust, because it is immune to the commonest sign error in the subject — writing S = V*I instead of S = V I*, which flips the sign of Q and turns a lagging load into a leading one.

🧠 Why 191 microfarads is worth more than it looks, and the last third costs the most

The 4232 W load above draws 23 A from the supply. A resistive load of the same 4232 W would draw only 4232/230 = 18.4 A. The extra 4.6 A is doing no work whatever — it is charging and discharging the reactance — and yet the cable, the switchgear and the transformer must all carry it, and the cable heats on the full I²R. That is the entire commercial motive for power factor correction: put a capacitor in parallel with the load, so it supplies the reactive power locally and the supply no longer has to. To reach unity you need Q_C = 3174 VAr, and since Q_C = ωCV², at 50 Hz C = 3174/(2π × 50 × 230²) = 190.98 μF, call it 191 μF. Two things then happen at once. The supply current falls from 23 A to 18.4 A, a factor of exactly the old power factor, 0.8; and the transmission loss, which goes as I², falls by a factor 0.8² = 0.64, that is by 36 per cent. The load still consumes 4232 W and still draws 23 A through its own terminals — nothing about the load changed, only what the supply sees. Now the part that decides real designs. Correcting only as far as 0.9 needs Q_C = 3174 - 4232 tan(cos⁻¹ 0.9) = 3174 - 2050 = 1124 VAr, which is 67.7 μF. So 68 μF buys you 0.8 → 0.9, and the remaining 123 μF buys you only 0.9 → 1.0: nearly twice the capacitance for a tenth of the improvement in current. The reason is geometric — Q = P tan φ, and tan φ collapses slowly at first and then steeply as φ → 0. This is exactly why utility tariffs set a target power factor around 0.9 to 0.95 rather than unity, and why overcorrecting into a leading power factor is a real hazard: the capacitor does not know the load has switched off.

The 4232 W, 0.8 lagging load: what correction actually buys
Target power factor	Q supplied by capacitor	C at 50 Hz, 230 V	Supply current
0.8, uncorrected	0 VAr	no capacitor fitted	23.0 A
0.9 lagging	1124 VAr	67.7 μF	20.4 A
1.0, unity	3174 VAr	191 μF	18.4 A
Loss at unity against uncorrected	same 3174 VAr	same 191 μF	36% less `I²R`

Maximum Power Transfer: Conjugate, Not Copy

Take the Thevenin equivalent of any linear source and ask what load extracts the most power from it. For a DC or purely resistive case with V_th and R_th, the power in R_L is P = V_th² R_L/(R_L + R_th)²; differentiate and the maximum sits at R_L = R_th, where P_max = V_th²/(4R_th) and the efficiency is exactly 50 per cent — half the power is burned inside the source. That efficiency figure is why maximum power transfer is a signal criterion and never a power-delivery criterion. A radio receiver front end is matched because it has microvolts to work with and cannot afford to lose half of them; a distribution transformer is emphatically not matched, because a grid running at 50 per cent efficiency is unthinkable.

Return to the Thevenin equivalent we built earlier: V_th = 14 V behind R_th = 5 Ω. The best possible load is 5 Ω, delivering P_max = 196/20 = 9.8 W. The actual load in the original circuit was 2 Ω, delivering (14/7)² × 2 = 8 W — 81.6 per cent of the maximum from a load 2.5 times too small. That flatness is the practically important property of the result, and it has an exact and rather elegant form. Writing R_L = k R_th, the fraction of the available power delivered is 4k/(1 + k)², which is invariant under k → 1/k. So being a factor 2.5 too low and a factor 2.5 too high cost identically: R_L = 2 Ω and R_L = 12.5 Ω both deliver exactly 8 W from this source. A factor of ten either way still delivers 33 per cent. The lesson is that the value of matching is modest, while the symmetry is exact — which is why matching questions are usually about identifying the correct condition rather than about the arithmetic of the optimum.

⚠️ Copying the source impedance throws away 36 per cent of the power

In the sinusoidal steady state the condition changes shape, and this is where the marks go. With a source V_th behind Z_th = R_th + jX_th, the load that extracts maximum power is the complex conjugate, Z_L = Z_th* = R_th - jX_th. The reason is worth one line of reasoning rather than memory: the reactances cancel, the circuit becomes purely resistive with total resistance 2R_th, and the reactive power sloshing uselessly between source and load is eliminated. Then P_max = |V_th|²/(4R_th), exactly the DC formula with R_th alone in it. Put numbers on it. Let Z_th = 8 + j6 Ω and V_th = 230 V RMS. Conjugate match, Z_L = 8 - j6: the total impedance is 16 Ω purely real, I = 230/16 = 14.375 A, and P = 14.375² × 8 = 1653.1 W, which matches 52900/32 exactly ✓. Now the trap. A candidate who remembers match the impedance and sets Z_L = 8 + j6 — a copy, not a conjugate — gets a total of 16 + j12, magnitude 20 Ω, so I = 11.5 A and P = 11.5² × 8 = 1058 W. That is 64 per cent of the available power; 36 per cent thrown away by a sign. And there is a third case GATE likes, because it is the physically realistic one: if the load is constrained to be a pure resistance R, the optimum is not R_th but R = |Z_th|, here 10 Ω, giving P = 230² × 10/(18² + 6²) = 529000/360 = 1469.4 W, or 8/9 of the conjugate-matched maximum. Check that R = 10 really beats R = 8: at R = 8, P = 423200/292 = 1449.3 W, which is less ✓. So three conditions, three different answers, and the one that applies depends entirely on what the load is allowed to be — read that clause in the question before reaching for a formula.

Vth = 230 V RMS behind Zth = 8 + j6 Ω: four candidate loads
Load	Current	Power delivered	Fraction of the maximum
`8 - j6 Ω`, the conjugate	14.375 A	1653.1 W	1.000
`10 Ω`, best pure resistance	12.122 A	1469.4 W	0.889
`8 Ω`, pure resistance equal to R_th	13.460 A	1449.3 W	0.877
`8 + j6 Ω`, a copy and not a conjugate	11.500 A	1058.0 W	0.640

Key takeaways

Count before you compute. Node analysis needs N - 1 equations, mesh needs B - N + 1; a grounded voltage source deletes a node unknown and a single-mesh current source deletes a mesh unknown. On a tie take nodes, because mesh analysis needs planarity and nodal does not.
Superpose voltages and currents, then square once at the end. On the two-node network the true power in the 2 Ω is 8 W while superposed powers give 4.33 W — the missing 3.67 W is the cross term 2ab/R, and it can be negative, so the error is not even a safe bound.
R_th = V_oc/I_sc is the route that always works. Never deactivate a dependent source — in the 10 V, 2 kΩ, 4 kΩ example the true R_th is 2 kΩ while killing the dependent source gives 1.33 kΩ, and that wrong value is exactly what an MCQ will offer you.
Reciprocity fixes the transfer impedance and nothing else. In the lopsided 4-6-12 Ω T-network the swapped drive gives the same 1 A both ways while the input resistance, source current, node voltage and source power all change. It fails the instant a dependent source appears.
In the sinusoidal steady state every derivative is a jω, so resistance becomes impedance and every DC theorem survives verbatim. Keep Z rectangular to add and polar to divide, and settle peak-versus-RMS before you start.
S = V I* splits into the P that heats and the Q that only sloshes. Correcting the 4232 W, 0.8 pf load to unity takes 191 μF and cuts line loss by 36 per cent — but 68 μF of that reaches 0.9, and the last 123 μF buys almost nothing.
Maximum power needs Z_L = Z_th*, the conjugate. Copying Z_th instead delivers only 64 per cent; a load restricted to pure resistance should be |Z_th|, not R_th. And since the fraction 4k/(1+k)² is invariant under k → 1/k, mismatching by 2.5 times either way costs exactly the same.

Practice questions (8)

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.

A connected resistive network has 6 nodes and 9 branches. Three of its ideal independent voltage sources each have one terminal on the reference node, and one ideal independent current source lies wholly inside a single mesh. After those sources have been used to delete unknowns, which method is left with fewer unknowns, and by how many?
1. Mesh analysis, by one unknown
2. Mesh analysis, by two unknowns
3. Node analysis, by one unknown
4. Both methods are left with the same number of unknowns
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Answer: C — Node analysis, by one unknown
Node analysis starts from N - 1 = 6 - 1 = 5 equations, and each ideal voltage source with a terminal on the reference node makes that node voltage known outright, so three of them remove three unknowns and leave 2. Mesh analysis starts from B - N + 1 = 9 - 6 + 1 = 4 equations, and a current source confined to one mesh fixes that mesh current, removing one and leaving 3. Node analysis therefore wins by one unknown. The tempting answer that mesh analysis is cheaper comes from comparing the raw counts 5 and 4 and stopping there, before the sources are allowed to delete anything; the deletions are exactly what reverses the verdict, and they are worth the ten seconds it takes to count them.
A linear resistive network is driven by one independent voltage source V_s and one independent current source I_s, and the voltage v across a particular branch is observed. With V_s = 10 V and I_s switched off, v = 4 V. With V_s switched off and I_s = 2 A, v = 3 V. What is v when V_s = 25 V and I_s = 5 A act together?
1. 17.5 V
2. 10.0 V
3. 7.5 V
4. 14.0 V
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Answer: A — 17.5 V
Linearity forces v = a·V_s + b·I_s for two constants fixed by the network alone. The first measurement gives 10a = 4, so a = 0.4 volts per volt. The second gives 2b = 3, so b = 1.5 volts per ampere. Then v = 0.4(25) + 1.5(5) = 10 + 7.5 = 17.5 V. The answer 10.0 V is what you get by carrying only the voltage-source contribution and forgetting that the current source scales too, and 7.5 V is the mirror error of keeping only the current-source term; each is a correct partial response, and superposition guarantees nothing about a part on its own — only the sum of both parts is the measured voltage.
In a small-signal model, an ideal 9 V battery sits in series with 3 kilo-ohm and the pair drives terminal a. A 6 kilo-ohm resistance is tied from terminal a down to ground. A controlled current source injects 0.5i into terminal a out of ground, i being the current the battery pushes through the 3 kilo-ohm towards a. A resistive load placed between terminal a and ground is picked to take the greatest power it can. How much power does it take?
1. 5.70 mW
2. 7.59 mW
3. 13.5 mW
4. 3.80 mW
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Answer: B — 7.59 mW
Work in volts, milliamps and kilohms. With the port open, i = (9 - Va)/3 and KCL at node a reads i + 0.5i = Va/6, so 1.5(9 - Va)/3 = Va/6, giving Va = 6.75 V, which is V_oc. Shorting the port puts Va = 0, so i = 3 mA and the short carries 1.5i = 4.5 mA, which is I_sc. Hence R_th = 6.75/4.5 = 1.5 kilo-ohm and the matched load takes V_oc squared over 4R_th = 45.5625/6000 W = 7.59 mW. The value 5.70 mW is what a candidate gets by deactivating the dependent source along with the 9 V source and reducing 3 in parallel with 6 to 2 kilo-ohm; a dependent source is part of the network description and switching it off describes a different circuit.
A two-port network is built from resistors, capacitors, an ideal transformer and one voltage-controlled current source. An experimenter drives port 1 with a voltage source and measures the short-circuit current at port 2, then interchanges the source and the ammeter and finds a different reading. What accounts for the change?
1. The voltage-controlled current source: a dependent source of any kind destroys reciprocity
2. The capacitors, because reciprocity holds only in a purely resistive network
3. The ideal transformer, which passes power in one direction only
4. The two ports presenting different driving-point impedances, so the network is unsymmetric
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Answer: A — The voltage-controlled current source: a dependent source of any kind destroys reciprocity
Reciprocity is guaranteed for a linear, passive, time-invariant network containing no dependent sources, and every element listed here except the controlled source satisfies that. Resistors, capacitors, inductors and ideal transformers are all reciprocal, so the capacitors and the transformer are innocent. The idea that an unsymmetric network cannot be reciprocal is the most tempting wrong answer and confuses two different conditions: symmetry is z11 equal to z22 and means the ports are interchangeable, while reciprocity is z12 equal to z21 and says only that the transfer impedance is the same both ways. A thoroughly lopsided ladder is reciprocal; a network with one controlled source is not.
A voltage v(t) = 100 cos(500t + 30 degrees) volts is applied across a series connection of a 60 ohm resistor and a 0.16 H inductor. What is the amplitude of the steady-state current?
1. 0.714 A
2. 1.67 A
3. 1.25 A
4. 1.00 A
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Answer: D — 1.00 A
The inductive reactance is omega·L = 500 times 0.16 = 80 ohm, so the impedance is 60 + j80 ohm whose magnitude is the square root of 3600 + 6400 = 100 ohm. The current amplitude is therefore 100/100 = 1.00 A, lagging the voltage by arctan(80/60) = 53.13 degrees. The value 0.714 A comes from adding the resistance and the reactance arithmetically to get 140 ohm, which treats two quantities ninety degrees apart as if they were in line; resistance and reactance combine in quadrature, never by simple addition. Using 60 ohm alone gives 1.67 A and using 80 ohm alone gives 1.25 A, each ignoring one element entirely.
A single-phase load draws 12 A RMS from a 240 V RMS supply at a power factor of 0.6 lagging. What reactive power does the load absorb?
1. 1728 VAr
2. 2304 VAr
3. 2880 VAr
4. 1152 VAr
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Answer: B — 2304 VAr
The apparent power is the product of the RMS voltage and the RMS current, 240 times 12 = 2880 VA. With a power factor of 0.6 the phase angle satisfies cos(phi) = 0.6 and therefore sin(phi) = 0.8, so the reactive power is 2880 times 0.8 = 2304 VAr, and it is positive because the current lags. The value 1728 VAr is the active power 2880 times 0.6, which heats the load rather than sloshing back and forth, and quoting it as reactive power is the commonest slip here. The 2880 value is the apparent power itself, in volt-amperes, and is the magnitude of the complex power rather than either of its two components.
A linear source has a Thevenin equivalent of 40 V behind 10 ohm. A purely resistive load of 40 ohm is attached to it. What fraction of the greatest power this source could deliver to any resistive load does the 40 ohm load actually receive?
1. 0.80
2. 0.16
3. 0.50
4. 0.64
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Answer: D — 0.64
The greatest power is delivered at a matched 10 ohm load and equals 40 squared over 4 times 10 = 40 W. With a 40 ohm load the current is 40/(40 + 10) = 0.8 A, so the load receives 0.8 squared times 40 = 25.6 W, a fraction 25.6/40 = 0.64. Writing the load as k times the source resistance, the fraction is 4k/(1 + k) squared, which at k = 4 is 16/25 = 0.64 and is unchanged if k is replaced by 1/4 — mismatching by four times either way costs exactly the same. The value 0.80 is the current-divider ratio 40/(40 + 10) and is not a power ratio at all; powers go as the square of that ratio, not as the ratio itself.
A network is made of linear resistors, several independent sources and one current-controlled voltage source. Which of the following procedures give correct results for this network?
1. Obtaining the Thevenin resistance at a port as the open-circuit voltage divided by the short-circuit current
2. Adding the branch voltages produced by each independent source acting alone, with the controlled source left active throughout
3. Obtaining the Thevenin resistance by shorting every voltage source, opening every current source and reducing the remaining resistors
4. Adding the powers dissipated in a branch by each independent source acting alone to obtain the total power in that branch
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Answer: A — Obtaining the Thevenin resistance at a port as the open-circuit voltage divided by the short-circuit current; B — Adding the branch voltages produced by each independent source acting alone, with the controlled source left active throughout
The open-circuit over short-circuit route leaves every source exactly as it is, so it survives a controlled source untouched and is the safe method here. Superposition of a voltage or a current is a consequence of linearity and remains valid provided only the independent sources are switched one at a time while the controlled source stays active, since it describes the network rather than exciting it. Shorting and opening sources before combining resistors is legal only with no dependent source present, and it is precisely the recipe that returns a wrong resistance here. Adding powers is wrong for any network whatever, because power is quadratic in the response and the cross term 2ab/R is thrown away; that cross term can be either sign, so the error is not even a safe bound.