Introduction to Electric Power Systems
An electric power system exists because energy is available in one place and wanted in another, and because it cannot be stored on the way — so everything in this book follows from the demand that generation match consumption instant by instant, over a network described by two numbers at every point: real power and reactive power.
- Why a power system is built at all — energy at the source, load at the city, and no bulk storage in between.
- How the sinusoidal steady state collapses a differential equation into algebra through the phasor, and why RMS is the natural magnitude.
- Why instantaneous power \(p(t)\) splits into a unidirectional part of mean \(P\) and a purely oscillating part of amplitude \(Q\) — the entire physical basis of reactive power.
- How \(S = \vec V\,\vec I^{\,*} = P + jQ\) packages both, and why \(P\) and \(Q\) are each conserved across a network.
- Why power factor is an economic quantity: conductors are sized by \(|S|\), losses scale as \(1/\cos^2\phi\), and plant is rated in MVA.
- How to size a compensating capacitor from \(Q_c = P(\tan\phi_1 - \tan\phi_2)\), and what it buys in current, loss and released capacity.
- The standard voltage levels, units and magnitudes you will meet in every remaining chapter.
The Problem a Power System Solves
Coal sits where the geology left it. A river falls where the terrain makes it fall. Wind is strongest offshore and along ridges, and sunlight is most reliable over deserts. None of these places is where people live. The first fact about electric power engineering is this mismatch of geography: the primary energy is in one place, and the demand for it is somewhere else entirely, usually hundreds of kilometres away.
In the early years of the industry this was not so. A generating plant stood in the middle of the district it lit, supplying a few streets at a few hundred volts, and if the district grew a second plant was built beside the first. That arrangement failed as soon as the loads became large. Fuel had to be carried to the city rather than the energy carried out of the coalfield; the plants were too small to be efficient; and a fault in one meant darkness for everyone it served. Building large stations at the fuel — and accepting the burden of moving the energy — turned out to be cheaper on every count. What has to be built to move it is the electric power system.
The second fact is harder. Electrical energy, in the quantities a country uses, cannot be stored. A grid-scale battery installation holds minutes of a nation's demand, and pumped storage holds hours at best; neither changes the essential situation. At every instant the mechanical power delivered by the turbines must equal the electrical power drawn by every lamp, motor and furnace on the network, plus the losses in the wires that connect them. Nobody schedules this. It happens automatically, and it happens through the physics of the synchronous machine: if consumption exceeds generation, the generators are decelerated by the extra torque demanded of them, the system frequency falls, and the governors respond by admitting more steam. Chapter 33 turns that reflex into a control loop. For now it is enough to see the consequence — the whole system is one machine, and a disturbance anywhere is felt everywhere.
Four requirements sit behind that sentence, and every later chapter serves one of them: the energy must arrive (Parts 2–4), the system must survive disturbances (Parts 5–6), it must be run cheaply (Part 7), and it must be protected (Part 8).
Two questions therefore run through the entire subject. What flows, and where? — a question about currents, voltages and losses in a network of impedances, answered by the load flow of Part 4. Does it hold together? — a question about whether the machines stay in step and whether protection clears a fault before the system breaks apart, answered by Parts 5 and 6. Everything in this first chapter is the vocabulary those two questions are asked in.
Sinusoidal Steady State and the Phasor
Generation, transmission and distribution are carried out with alternating quantities at a single frequency — 50 Hz across India, Europe and most of Asia and Africa, 60 Hz in the Americas. Chapter 2 gives the reasons; here we take the choice as made and ask what it allows us to assume.
A network of resistors, inductors and capacitors is linear and time-invariant. Drive it with a sinusoid of frequency \(f\) and wait long enough for the transients to die away, and every voltage and every current in it is a sinusoid of the same frequency. Only two things distinguish one from another: how large it is and when it peaks. Two real numbers therefore describe any waveform in the network completely, and the phasor is the device that carries them.
Which measure of "how large" should we use? The peak value is the obvious candidate, but power is what the system exists to deliver, so the magnitude ought to be the one that predicts power correctly. Take a resistor \(R\) carrying \(i(t)=I_m\cos\omega t\). The energy dissipated in one period \(T\) is
So the RMS value is precisely the direct current that would heat the same resistor at the same rate — which is why every meter, every nameplate and every specification in power engineering is in RMS, and why unqualified symbols \(V\) and \(I\) in this book always mean RMS values. Writing the waveforms with that convention built in,
The phasor is not the waveform; it is a complex number that stands for it, on the understanding that the frequency is known and common to all. The reason for going to the trouble is that differentiation and integration become multiplication. Since \(\frac{d}{dt}\,\mathrm{Re}\{\sqrt2\,\vec V e^{j\omega t}\} = \mathrm{Re}\{\sqrt2\,(j\omega \vec V)e^{j\omega t}\}\), the operator \(d/dt\) is replaced by the factor \(j\omega\), and \(\int dt\) by \(1/(j\omega)\). A series \(R\)–\(L\)–\(C\) loop obeys
What was a differential equation is now a division of two complex numbers. That single substitution is what makes networks of thousands of branches tractable at all, and it is assumed silently in every chapter that follows.
The method is valid only in the steady state and only at one frequency. Switching surges, lightning and fault transients violate both conditions — which is why Chapter 21 must return to the time domain, and why harmonics need the separate treatment of Chapter 39.
Impedance and the Three Similar Triangles
Collecting the bracket of the last equation gives the impedance of the branch,
\(Z\) is a complex number, but it is not a phasor: it does not stand for any sinusoid and has no time-domain counterpart. It is a ratio of two phasors, and the distinction matters when the two are mixed carelessly in a calculation.
The angle \(\phi\) is the amount by which the current lags the voltage, since \(\vec I = \vec V/Z\) subtracts \(\angle Z\) from \(\angle \vec V\). An inductive branch (\(X>0\)) draws a lagging current; a capacitive branch (\(X<0\)) draws a leading current. Almost every load a utility serves is a motor, a transformer or a fluorescent ballast, so almost every load is inductive and almost every current in the system lags. That one-sidedness is the reason reactive compensation is always about supplying \(Q\), never absorbing it — a theme Chapter 34 develops in full.
Now scale the impedance triangle. Multiply every side by the current magnitude \(I\) and it becomes the triangle of voltage drops \(IR\), \(IX\), \(I|Z|\). Multiply again by \(I\) and it becomes the triangle of powers that Section 1-5 will name. The three are similar, and \(\phi\) is common to all of them — which is why the same angle appears as an impedance angle, as a phase displacement between waveforms, and as the argument of complex power.
Instantaneous Power: Where Two Quantities Come From
Reactive power is usually introduced by definition, which leaves students with a formula and no picture. It is better to obtain it, as it was historically obtained, by multiplying out the instantaneous power and looking at what the product contains.
Connect two subcircuits, calling one the source and the other the load, and let \(p(t) = v(t)\,i(t)\) be the power delivered from the first to the second. When \(p(t)\) is positive, energy is flowing that way; when it is negative, the roles reverse for that instant. Take the simplest case first, a purely resistive load, with \(v\) and \(i\) in phase:
Two features stand out. The power pulsates at twice the supply frequency, and it never becomes negative: energy flows from source to load throughout the cycle and never returns. Its average over a cycle is \(VI\), the second term contributing nothing.
Now let the current lag the voltage by \(\phi\), as it does for every real load, and repeat:
The expansion \(\cos(2\omega t-\phi)=\cos 2\omega t\cos\phi+\sin 2\omega t\sin\phi\) is all that separates the two lines, but the second line is the important one. The instantaneous power has split, uniquely, into two pieces of entirely different character.
The first piece is \(VI\cos\phi\,(1+\cos2\omega t)\). Because \(1+\cos2\omega t\) is never negative, this component carries energy in one direction only, from source to load, and it does so at the average rate \(VI\cos\phi\). It is real, useful, irreversible transfer: shaft work, heat, light.
The second piece is \(VI\sin\phi\,\sin2\omega t\). It is a pure sinusoid at \(2\omega\) with no offset, so its average over a cycle is exactly zero. Energy travels out to the load for a quarter of the doubled period and comes back in the next quarter, over and over, transferring nothing on balance. This is the energy that is alternately stored in the magnetic field of every motor and transformer and returned to the network — real energy, in real motion, doing no net work. Its amplitude, \(VI\sin\phi\), is what we shall call reactive power.
Active, Reactive and Complex Power
The decomposition of Section 1-4 produced two numbers. It is convenient to carry them as one complex quantity, and there is only one product of the two phasors that does the job. Take \(\vec V = V\angle\phi_v\) and \(\vec I = I\angle\phi_i\), and form
The conjugate is not a convention chosen for tidiness. Without it the product would carry the angle \(\phi_v+\phi_i\), which depends on where the reference was placed and therefore cannot mean anything physical. With it, the angle is the difference \(\phi_v-\phi_i=\phi\) — the same \(\phi\) as the impedance angle — and the real and imaginary parts land exactly on the two components extracted from \(p(t)\).
Read the components off the current instead of the angle and the picture of Section 1-3 returns: \(I\cos\phi\) is the part of the current in phase with \(\vec V\), and it produces \(P\); \(I\sin\phi\) is the part in quadrature, and it produces \(Q\). Only the first delivers energy. The second is present in the conductor, heats it exactly as the first does, and delivers nothing.
\(P\) is active (or real, or average) power — the rate of irreversible energy conversion. \(Q\) is reactive power — the amplitude of the energy sloshing back and forth, zero on average. \(|S|\) is apparent power — what the conductors and the plant actually have to be built for.
Substituting \(\vec V = Z\vec I\) gives the branch forms that will be used constantly from Chapter 10 onward:
Both sums run over every branch of the network, and both are exact: complex power is conserved, so real and reactive power are separately conserved. The generators must supply the sum of all the loads' \(P\) plus every \(I^2R\) in between, and the sum of all the loads' \(Q\) plus every \(I^2X\) in between. Load flow (Chapter 18) is nothing more than the problem of enforcing those two balances at every bus at once.
Signs follow from \(X\). An inductor has \(X=+\omega L\) and therefore \(Q=+I^2\omega L\): it absorbs reactive power. A capacitor has \(X=-1/\omega C\) and \(Q=-I^2/\omega C\): it absorbs a negative amount, which is the same as saying it supplies reactive power. That antisymmetry is the entire mechanism of compensation.
A generator delivering into the network shows negative \(P\) under this convention, which is why generator quantities are almost always written with the opposite sign — current out of the machine. Fix a convention at the start of a problem and never change it mid-way.
Power Factor and What It Costs
Dividing the useful power by the total that has to be handled gives the ratio that names the whole issue:
It is dimensionless and lies between 0 and 1. The last equality — \(R/|Z|\) — is the impedance triangle again, and it holds only for a single branch; the ratio \(P/|S|\) is the general definition. A power factor is always quoted with the word lagging or leading, because \(\cos\phi\) alone cannot distinguish \(+\phi\) from \(-\phi\), and the two need opposite corrective action.
Why does a number that describes phase have an economic meaning? Because the physical size of every piece of equipment in a power system is set not by \(P\) but by \(V\) and \(I\) separately:
| Quantity | What it fixes | Why |
|---|---|---|
| Voltage \(V\) | Insulation thickness, clearances, bushing and transformer core size | Dielectric stress and, for a fixed frequency, the flux the core must carry |
| Current \(I\) | Conductor cross-section, busbar size, breaker rating | \(I^2R\) heating and the permissible temperature rise |
| Product \(VI = |S|\) | The nameplate rating of generators, transformers and cables | Both limits act at once, and neither knows about \(\cos\phi\) |
A transformer rated 1000 kVA is limited by its insulation and its winding heating. If it is asked to supply a load at unity power factor it delivers 1000 kW; at 0.7 lagging it delivers only 700 kW, from exactly the same iron and copper. The remaining 300 kVA of capacity has been consumed by reactive power that does no work. Generation, transmission and distribution equipment is therefore rated in MVA throughout this book, and the reader should treat any rating given in MW with suspicion until the power factor is stated.
Losses behave worse still. At a fixed load power and fixed voltage the current is
Falling from unity to 0.7 lagging doubles the line losses for the same delivered kilowatts, and it increases the voltage drop along the feeder in the same proportion — with the consequence for regulation that Chapter 13 quantifies. This is why tariffs penalise a low power factor (Chapter 30), and why the correction of Section 1-7 is one of the few investments in a power system that pays for itself out of loss savings alone.
Reactive Power Compensation
Connect a capacitor in shunt with an inductive load. The load continues to draw exactly the reactive power it needs — nothing about it has changed — but that reactive power is now furnished locally by the capacitor rather than dragged along the feeder from the generator. Upstream of the connection point, only the difference is seen.
Let the load take \(P\) at an angle \(\phi_1\), so \(Q_1 = P\tan\phi_1\). We want the combination to present an angle \(\phi_2\), with \(Q_2 = P\tan\phi_2\). Real power is untouched by a capacitor (it is lossless to first order), so \(P\) is common to both, and the capacitor must supply the shortfall:
The gains follow immediately. The apparent power falls from \(P/\cos\phi_1\) to \(P/\cos\phi_2\), so the current falls in the ratio \(\cos\phi_1/\cos\phi_2\) and the losses upstream fall as the square of that ratio:
Three cautions belong with the formula. Correcting all the way to unity is rarely economic: the capacitor cost rises steadily while the loss saving flattens out, since the last few kilovars remove very little current. Over-correction turns the current leading, which raises the receiving-end voltage instead of supporting it and can be worse than the original condition — the same mechanism as the Ferranti effect of Chapter 14. And a shunt capacitor forms a resonant circuit with the inductance of the supply, so a harmonic current near that resonance can be amplified badly; Chapter 39 returns to this.
The capacitor is only the cheapest of several sources of reactive power. Synchronous condensers, static VAR compensators and the reactive capability of the generators themselves all do the same job with more control, and Chapters 34 and 38 take up the comparison.
Scale, Units and Standard Voltages
Power system arithmetic runs across nine orders of magnitude, and a habit of carrying units carefully is the cheapest error-checking available. The three powers take three different unit names deliberately, so that a stray \(j\) is caught by inspection.
Energy, what the consumer actually buys, is the time integral of \(P\) and is metered in kilowatt-hours; a large consumer is additionally charged on the maximum kVA demanded, which is where power factor reaches the bill. Chapter 30 sets out the tariff structures.
Voltage levels are standardised, and the ladder below recurs in every single-line diagram in this book. The Indian values are given; other systems differ in detail but not in pattern.
| Stage | Typical voltage | Arrangement | Reason for the level |
|---|---|---|---|
| Generation | 11 kV (6.6, 15.75, 21 kV) | 3-phase, star, neutral earthed through impedance | Slot insulation inside a rotating machine limits how high the stator can be wound |
| Primary transmission | 132, 220, 400, 765 kV | 3-phase, 3-wire overhead | Conductor volume falls as \(1/V^2\) — Chapter 2 |
| Secondary transmission | 33, 66 kV | 3-phase, 3-wire | Bulk supply into and around a city |
| Primary distribution | 11 kV | 3-phase, 3-wire | Feeds distribution substations and large consumers directly |
| Secondary distribution | 400 V line, 230 V phase | 3-phase, 4-wire with neutral | Safe at the point of use; single-phase loads tap phase to neutral |
For scale: a domestic connection is 1–5 kW, a large industrial plant tens of megawatts, a modern thermal unit 500–800 MW, and a national system a few hundred gigawatts. A 400 kV double-circuit line carries of the order of 1000 MW. These numbers are worth memorising, because they make an implausible answer visible before its algebra is checked.
One more convention, used from Chapter 3 onward: a balanced three-phase system is analysed one phase at a time, using the phase (line-to-neutral) voltage and the line current, and the three-phase power is three times the per-phase result. Chapter 3 justifies this; Chapter 4 goes further and removes the transformer turns ratios as well by expressing everything in per unit, after which the voltage ladder of the table above disappears from the arithmetic entirely.
The Road Through This Book
The subject is built in one direction: first the components, then the network they form, then what happens to that network when something goes wrong, and finally how it is operated. Knowing the plan makes it clear why the early chapters spend so long on the geometry of conductors.
| Part | Chapters | The question it answers |
|---|---|---|
| 1 · The Power System | 1–4 | What is the system, how is three-phase power handled, and how do we draw and normalise it? |
| 2 · Transmission Line Parameters | 5–9 | What are \(R\), \(L\), \(C\) of a line, and how do conductor geometry and bundling set them? |
| 3 · Line Performance | 10–15 | Given those parameters, what voltage and power arrive at the far end? |
| 4 · Network Modelling and Load Flow | 16–20 | How do we solve a whole interconnected network for its operating point? |
| 5 · Fault Analysis | 21–25 | How large is the current when the insulation fails, and what must the breaker interrupt? |
| 6 · Stability | 26–29 | After a disturbance, do the machines stay in synchronism? |
| 7 · Operation, Control and Economics | 30–34 | Which units run, at what output, and how are frequency and voltage held? |
| 8 · Protection and the Modern Grid | 35–39 | How is the faulted part removed, and what changes with HVDC, FACTS and renewables? |
Two threads bind the parts together, and both begin in this chapter. The first is the split of every flow into \(P\) and \(Q\): load flow computes both, dispatch schedules \(P\), voltage control schedules \(Q\), and the near-decoupling of the two is what makes the fast decoupled method of Chapter 20 possible. The second is the phasor itself, which survives untouched until the fault transients of Chapter 21 and the swing dynamics of Chapter 27 force a return to real time.
Worked Examples
Problem. A series circuit of \(R=8\ \Omega\), \(L=31.83\ \mathrm{mH}\) and \(C=795.8\ \mu\mathrm{F}\) is supplied from \(v(t)=\sqrt2\,(230)\cos\omega t\) at 50 Hz. Find the current, the three powers and the power factor.
Solution. With \(\omega = 2\pi(50)=314.16\ \mathrm{rad/s}\):
Taking \(\vec V = 230\angle0^\circ\ \mathrm{V}\), the current phasor and its waveform are
The powers follow either from the phasors or from \(I^2Z\), and agreeing by both routes is the check:
The individual reactive contributions are worth separating: \(Q_L = I^2X_L = 529(10) = 5290\) var absorbed by the inductor, \(Q_C = -I^2X_C = -529(4) = -2116\) var supplied by the capacitor, and the net \(5290-2116=3174\) var drawn from the supply. The capacitor is already compensating part of the inductor's demand — Section 1-7 in miniature.
Problem. For the circuit of Example 1, write \(p(t)\), find its maximum and minimum values, and determine the fraction of each cycle during which power flows from the load back to the source.
Solution. Using the decomposition of Section 1-4 with \(VI = 5290\) VA and \(\phi=36.87^\circ\):
The second form shows at once that \(p(t)\) oscillates at \(2\omega\) about the mean \(P = 4232\) W with amplitude \(|S| = 5290\) VA. Hence
Power is negative when \(\cos(2\omega t - 36.87^\circ) < -\cos 36.87^\circ = -0.8\), that is when the argument lies between \(180^\circ-36.87^\circ = 143.13^\circ\) and \(180^\circ+36.87^\circ = 216.87^\circ\). The width of that window is \(2\phi = 73.74^\circ\) out of the \(360^\circ\) period of the doubled angle:
At unity power factor the window closes to zero; at \(\cos\phi=0\) it would occupy half the cycle and the mean power would be zero. The lower the power factor, the larger the share of the time the conductors spend carrying energy that is about to come straight back.
Problem. A single-phase busbar is held at \(\vec V = 415\angle0^\circ\) V. A load draws \(\vec I_L = 50\angle{-25^\circ}\) A and a capacitor bank on the same bus draws \(\vec I_C = 12\angle 90^\circ\) A. Find the complex power of each, and the current and power factor seen by the supply.
Solution. For the load,
so the load absorbs 18.81 kW and 8.77 kvar at \(\cos 25^\circ = 0.906\) lagging. For the capacitor, whose current leads by \(90^\circ\),
Zero real power, as expected of an ideal capacitor, and \(Q_C = -4.98\) kvar — it supplies 4.98 kvar. The supply sees the phasor sum of the currents:
The real part is unchanged at 18.81 kW, exactly as the conservation statement of Section 1-5 requires, while the reactive part has fallen from 8769 to \(8769-4980 = 3789\) var. The supply current drops from 50 A to 46.23 A and the power factor improves from 0.906 to \(\cos11.39^\circ = 0.980\) lagging. The load has not been altered in any way; only the place where its reactive power originates has moved.
Problem. A 500 kW load operates at 0.7 power factor lagging on an 11 kV single-phase supply at 50 Hz. Find the capacitor rating and capacitance needed to raise the power factor to 0.95 lagging, and the resulting reduction in supply current and in feeder loss.
Solution. Convert both power factors to tangents:
The capacitance that develops this at 11 kV is
The apparent power falls from \(500/0.7 = 714.3\) kVA to \(500/0.95 = 526.3\) kVA, so at constant voltage
Going the last step to unity would need a further 164.3 kvar — almost half as much capacitance again — and would remove only the remaining 5% of current. That flattening return is why utilities set a target near 0.95 rather than 1.0.
Problem. A single-phase feeder of total loop resistance \(0.2\ \Omega\) supplies a 12 kW load at 240 V. Compare the feeder loss at 0.8 lagging power factor with the loss at unity, and value the difference over a year at ₹8 per kWh.
Solution. The current required to deliver a fixed 12 kW depends on the power factor:
The loss ratio is \(781.25/500 = 1.5625 = 1/0.8^2\), confirming the \(1/\cos^2\phi\) law of Section 1-6 without any further work. The extra dissipation is 281.25 W, and over a year of continuous operation
On one small feeder this is a modest sum; multiplied across a distribution network with tens of thousands of such feeders it is the reason power factor appears in tariffs at all. The loss is also dissipated as heat in conductors that must then be sized larger, so the capital cost moves in the same direction as the running cost.
Problem. A 1000 kVA distribution transformer is fully loaded supplying 800 kW at 0.8 power factor lagging. The plant wishes to add load, all of it at 0.8 lagging. How much extra real load can the same transformer carry if capacitors are installed to hold the transformer power factor at 0.95, and what capacitor rating is needed?
Solution. At present \(|S| = 800/0.8 = 1000\) kVA and \(Q = 800\tan(36.87^\circ) = 800(0.75) = 600\) kvar; the transformer is exactly at rating. The limit that matters is the 1000 kVA, so after compensation
The load itself, being 950 kW at 0.8 lagging, still demands its own reactive power:
A 400 kvar capacitor bank therefore releases \(950-800 = 150\) kW of extra capacity, an 18.75% increase, from a transformer that is not touched. Set against the cost of a larger transformer, new cabling and an outage to install them, the capacitor is almost always the cheaper answer — and it improves the voltage at the busbar as a by-product, for the reason Chapter 13 makes precise.
Chapter Summary
Energy at the source, load at the city, and no bulk storage between: generation must match demand instant by instant.
\(v(t)=\sqrt2 V\cos(\omega t+\phi_v)\Leftrightarrow \vec V=V\angle\phi_v\); \(d/dt\to j\omega\) turns differential equations into algebra.
\(V=V_m/\sqrt2\) is the value that gives equal heating — hence every rating and every meter reading.
\(p=VI\cos\phi(1+\cos2\omega t)+VI\sin\phi\sin2\omega t\): one never negative, one of zero mean.
\(S=\vec V\vec I^{\,*}=P+jQ\), with \(P=VI\cos\phi\), \(Q=VI\sin\phi\), \(|S|=VI\).
\(P=\sum I_k^2R_k\) and \(Q=\sum I_k^2X_k\) — real and reactive power balance separately at every node.
\(\cos\phi = P/|S| = \mathrm{kW/kVA}\); plant is rated in MVA and losses scale as \(1/\cos^2\phi\).
\(Q_c=P(\tan\phi_1-\tan\phi_2)\) supplies \(Q\) locally, cutting current, loss and released capacity all at once.
Problems
Unless stated otherwise, take 50 Hz, single-phase quantities, RMS magnitudes and the load sign convention of Section 1-5. Carry units through every line; a stray factor of \(\sqrt3\) or \(1000\) is the commonest error in this subject.
- A load takes \(\vec I = 40\angle{-40^\circ}\) A from a 415 V supply taken as reference. Find \(P\), \(Q\), \(|S|\) and the power factor, and state whether the load is inductive or capacitive.
- A coil of resistance \(6\ \Omega\) and inductance \(25.5\ \mathrm{mH}\) is connected across 230 V, 50 Hz. Find the current, the power factor, and the value of the shunt capacitor that raises the overall power factor to unity.
- Show from \(p(t)=VI\cos\phi(1+\cos2\omega t)+VI\sin\phi\sin2\omega t\) that the energy returned to the source over one cycle is zero if and only if \(\cos\phi=1\), and find the peak instantaneous power in terms of \(P\) and \(|S|\).
- Two loads are supplied from the same 400 V bus: 15 kW at 0.85 lagging, and 20 kVA at 0.6 leading. Find the total \(P\), \(Q\) and \(|S|\), the resulting power factor, and the supply current.
- A factory draws 300 kW at 0.75 lagging from a feeder of resistance \(0.35\ \Omega\) at 3.3 kV. Determine the feeder loss before and after correction to 0.92 lagging, and the kvar rating required.
- A 750 kVA transformer supplies 600 kW at 0.8 lagging. Capacitors of 250 kvar are added. Find the new apparent power, the new power factor, and how much additional real load at 0.8 lagging the transformer can then accept before reaching its rating.
- The current in a branch of impedance \(Z=4+j3\ \Omega\) is 20 A. Compute \(P\) and \(Q\) from \(I^2R\) and \(I^2X\), then verify the same values from \(S=\vec V\vec I^{\,*}\) with \(\vec V=Z\vec I\).
- A consumer's monthly bill is charged at ₹7.50 per kWh of energy plus ₹250 per kVA of maximum demand. The plant runs 500 kW for 400 hours in the month at 0.72 lagging. Compute the bill, then recompute it after correction to 0.95 lagging, and find the monthly saving attributable solely to the reduced kVA demand.