In short

Multiply instantaneous voltage v(t) = V_0\sin(\omega t) by instantaneous current i(t) = I_0\sin(\omega t - \phi) and average over one cycle. The oscillating \sin\cdot\sin product collapses to a single clean number — the average power:

\boxed{\;P \;=\; V_\text{rms}\,I_\text{rms}\,\cos\phi\;}

The factor \cos\phi is the power factor. It is 1 for a pure resistor (voltage and current in phase), 0 for a pure inductor or capacitor (90° out of phase), and somewhere between for a realistic load like a ceiling fan or an induction motor.

Three powers, three names.

  • Apparent power S = V_\text{rms}I_\text{rms}, in volt-amperes (VA). What the wires and switchgear have to handle.
  • Real power P = V_\text{rms}I_\text{rms}\cos\phi, in watts (W). What the load actually consumes and the meter bills you for.
  • Reactive power Q = V_\text{rms}I_\text{rms}\sin\phi, in volt-amperes reactive (VAR). What sloshes back and forth between the source and the load's magnetic and electric fields without doing net work.

They form a right triangle: S^2 = P^2 + Q^2.

Wattless current is the component of I_\text{rms} that is 90° out of phase with V. It carries zero average power but still flows through the wires, heating them up and requiring bigger copper. This is why the Maharashtra State Electricity Distribution Company levies a power-factor penalty on commercial and industrial consumers whose \cos\phi drops below about 0.9.

Transformers. Two coils sharing a single iron core. Faraday's law on each:

\boxed{\;\dfrac{V_s}{V_p} \;=\; \dfrac{N_s}{N_p}\;}

(secondary-to-primary voltage ratio equals secondary-to-primary turns ratio). For an ideal transformer, P_\text{in} = P_\text{out} forces I_s/I_p = N_p/N_s — voltage steps up, current steps down, in exact inverse proportion. Every step of the Indian grid — from the 11 kV feeder to the 132 kV subtransmission network to the 400 kV interstate backbone and back down to your 230 V socket — is one transformer doing exactly this.

At Kalol GIDC outside Ahmedabad, a large pharmaceutical plant draws its electricity from a 22 kV feeder off the Gujarat grid. The bill for April arrives: 4,80,000 units (kWh) consumed, at the commercial tariff, would have been ₹38 lakh. But the bill is ₹41 lakh. The extra ₹3 lakh is a single line item: "Power factor penalty, PF recorded 0.78, billing PF 0.90, penalty 4.5%." Somewhere in the plant, the current in the wires is not doing as much useful work as the ammeter reading suggests. The voltage says one thing; the current says another; and the angle between them is costing the company ₹1,000 per hour, continuously, 24 hours a day.

This chapter is about where that angle comes from, why it costs money, and how the engineers at the NTPC Mouda thermal plant push 25 kV up to 400 kV and send it across a thousand kilometres of transmission line at 50 Hz and lose only a few percent in the wires. Everything flows from one equation — P = V_\text{rms}I_\text{rms}\cos\phi — which we will derive in three lines of calculus and then unpack for the rest of the article.

Instantaneous power — why it oscillates

Take the AC voltage v(t) = V_0\sin(\omega t) across a load, and the current it drives, i(t) = I_0\sin(\omega t - \phi). The \phi is whatever phase angle the load imposes — zero for a pure resistor, +\pi/2 for a pure inductor, -\pi/2 for a pure capacitor, something in between for a realistic LCR circuit.

Instantaneous power is

p(t) \;=\; v(t)\,i(t) \;=\; V_0 I_0 \sin(\omega t)\sin(\omega t - \phi)

The product of two sines is not enlightening as written. The product-to-sum identity

\sin A\sin B \;=\; \tfrac{1}{2}\bigl[\cos(A - B) - \cos(A + B)\bigr]

turns it into a sum of two pieces, each of which is easy to think about.

Step 1. Apply the identity with A = \omega t and B = \omega t - \phi:

\sin(\omega t)\sin(\omega t - \phi) \;=\; \tfrac{1}{2}\bigl[\cos\phi - \cos(2\omega t - \phi)\bigr]

Why: A - B = \omega t - (\omega t - \phi) = \phi (constant — independent of t). A + B = 2\omega t - \phi (oscillates at twice the mains frequency). The product of two sinusoids at \omega is a constant plus a sinusoid at 2\omega.

Step 2. Substitute back:

p(t) \;=\; \tfrac{1}{2}V_0 I_0\cos\phi \;-\; \tfrac{1}{2}V_0 I_0\cos(2\omega t - \phi)

The first term is a constant. The second term oscillates sinusoidally at 2\omega — twice the mains frequency. Every cycle of AC, power pulses twice.

Step 3. Average over a full cycle. The time average of any cosine is zero over its period, so the second term vanishes:

\boxed{\;\langle p(t)\rangle \;=\; \tfrac{1}{2}V_0 I_0\cos\phi\;}

Step 4. Convert to RMS. Since V_\text{rms} = V_0/\sqrt{2} and I_\text{rms} = I_0/\sqrt{2}, we have V_0 I_0 = 2\,V_\text{rms}I_\text{rms}:

\boxed{\;P \;\equiv\; \langle p(t)\rangle \;=\; V_\text{rms}\,I_\text{rms}\,\cos\phi\;}

Why: the 1/2 and the \sqrt{2}\sqrt{2} = 2 cancel perfectly, leaving the clean RMS formula. This is the whole purpose of defining RMS — to make the average-power formula look like its DC counterpart P = VI, just with an extra factor of \cos\phi.

Result. For any linear AC load, the average power consumed is the product of RMS voltage, RMS current, and the cosine of the phase angle between them. A 1500 W kettle with \cos\phi = 1 on a 230 V supply draws 6.5 A RMS. A 1500 W-real ceiling fan with \cos\phi = 0.8 draws 1500/(230 × 0.8) = 8.15 A RMS — more current for the same real power, because 20% of the apparent power is just sloshing.

Animated voltage, current, and power for a load with phase angle Over two full cycles at 50 Hz: voltage v(t) (red) oscillates between plus and minus one. Current i(t) (dark) oscillates similarly but lags voltage by about 40 degrees. Instantaneous power p(t) = v(t) times i(t) (grey) oscillates at twice the frequency around a constant positive average value of about 0.38. +1 0 −0.5 time (two cycles at 50 Hz) P = ½V₀I₀cos φ
Voltage (red), current (dark) — lagging by 40° — and instantaneous power (grey) for a realistic load. Power oscillates at 100 Hz (twice the mains frequency), dipping briefly below zero whenever current and voltage have opposite signs. The dashed red line is the time-average $P = \tfrac{1}{2}V_0 I_0\cos\phi$, which is what the energy meter integrates. Click replay.

Power factor — the physical meaning of cos φ

Look at the formula P = V_\text{rms}I_\text{rms}\cos\phi and ask: what does \cos\phi actually mean, physically?

Decompose the current into two components. Split the total current phasor into a component parallel to the voltage and a component perpendicular to it.

I_\text{rms} \;=\; I_\text{rms}\cos\phi \;\hat{v} \;+\; I_\text{rms}\sin\phi \;\hat{\perp}

The parallel component I_p = I_\text{rms}\cos\phi is the active or watt-producing current — it is in phase with the voltage, and the two together produce the \int v\cdot i integral that gives real power. The perpendicular component I_w = I_\text{rms}\sin\phi is the wattless current — it is 90° out of phase with the voltage, and the integral \int v\cdot i over a full cycle for this component is zero.

Power triangle: real power, reactive power, and apparent power A right triangle with horizontal leg P (real power, in watts), vertical leg Q (reactive power, in VAR), and hypotenuse S (apparent power, in VA). The angle between S and P is phi, with cos phi labeled as the power factor. The power triangle P = V·I·cos φ (watts) Q = V·I·sin φ (VAR) S = V·I (VA) φ S² = P² + Q², cos φ = P/S
The power triangle. Real power $P$ (horizontal, heats water) and reactive power $Q$ (vertical, sloshes in and out of reactances) are the two legs. Apparent power $S = V_\text{rms}I_\text{rms}$ is the hypotenuse — the "size of the current" that the wires and switchgear must be rated for. The angle $\phi$ is the same phase shift that appears in $\cos\phi$.

The three "powers" form a right triangle — the power triangle of any AC load:

P \;=\; S\cos\phi,\qquad Q \;=\; S\sin\phi,\qquad S^2 \;=\; P^2 + Q^2

This is more than a mnemonic. It is the practical accounting tool every utility uses. A factory's monthly bill is computed from three totals: kWh (integral of P), kVARh (integral of Q), and billing kVA (peak demand S). The penalty in the Ahmedabad pharma plant comes from having too much Q relative to P — too much current flowing without doing work.

Why wattless current isn't free

The instantaneous version of the wattless current does zero average work, but it is still current. It flows through the feeder, the distribution transformer, the generator's windings, and it dissipates I^2 R in every resistance it passes through. The resistance of a long transmission line might be 0.1\,\Omega per kilometre; a factory drawing 800 A RMS through 10 km of feeder at \cos\phi = 1 wastes (800)^2 \times 1 = 640\,\text{kW} in line heating. The same factory drawing the same real power but at \cos\phi = 0.6 pulls 800/0.6 \approx 1333 A — and loses 1333^2 \times 1 \approx 1.78\,\text{MW} in the line. Nearly three times the loss, for the same real-power delivery.

That is why the grid cares about \cos\phi and why industrial tariffs reward consumers who keep it close to 1. The cheapest fix is a bank of power-factor-correction capacitors in parallel with the inductive load. The capacitor's current leads the voltage by 90°; the motor's current lags by some angle; the two combine so that the net current is nearly in phase with the voltage. The capacitor supplies the motor's reactive power locally, without having to drag it all the way back to the generator.

Explore: watch the power factor shift

Drag the phase angle \phi and see the real and reactive powers change.

Interactive: real and reactive power vs phase angle Two curves on a plot with phase angle phi on the x-axis, from 0 to 90 degrees, and fractional power on the y-axis, from 0 to 1. Red curve: real-power fraction cos phi, going from 1 at phi=0 down to 0 at phi=90. Dark curve: reactive-power fraction sin phi, going from 0 at phi=0 up to 1 at phi=90. Drag the red point to see both at any phase. phase angle φ (degrees) fraction of S = V·I 0 0.5 1 30° 60° 90° cos φ (real) sin φ (reactive) drag φ; the curves cross at 45°
Drag the phase angle $\phi$ between 0 and 90°. At $\phi = 0$ (resistive load), all apparent power is real. At $\phi = 90°$ (pure reactance), all apparent power is wattless. The two curves cross at $\phi = 45°$ — the point where real and reactive powers are equal, corresponding to $\cos\phi = \sin\phi = 1/\sqrt{2} \approx 0.707$.

Special cases

Load \phi \cos\phi Comment
Pure resistor (heater, kettle, incandescent bulb) 0 1.00 All power is real
Pure inductor (ideal choke) -\pi/2 0 No real power; only reactive
Pure capacitor (ideal) +\pi/2 0 No real power; only reactive, opposite sign
Ceiling fan (induction motor) \approx -35° 0.8 Lagging, typical residential
Fluorescent tube with magnetic ballast \approx -50° 0.65 Heavily lagging
Modern LED driver \approx -20° 0.95 Near-unity, by design
Induction furnace at steel plant \approx -50° 0.65 Heavily lagging, needs correction

In the general AC circuit the phase angle comes from the LCR combination: \tan\phi = (X_L - X_C)/R. Pure resistance gives \phi = 0; excess inductance gives lagging (negative) \phi; excess capacitance gives leading (positive) \phi.

Transformers — two coils, one core, four equations

All of the above concerns a single load. The other half of this chapter concerns how AC power is transported. The clever device that makes long-distance transmission possible is the transformer — two coils wrapped around a common ferromagnetic core, with no electrical connection between them. Apply AC to one coil (the primary) and AC appears on the other (the secondary), typically at a different voltage and current.

How a transformer works — in four steps

Step 1. Apply an AC voltage v_p(t) = V_{p,0}\sin(\omega t) to the primary, a coil of N_p turns wrapped around an iron core. The magnetising current produces an alternating flux \Phi(t) in the core.

Step 2. For an ideal transformer (no leakage, no core losses, no winding resistance), Faraday's law applied to the primary gives

v_p(t) \;=\; -N_p\,\dfrac{d\Phi}{dt}

Why: each turn of the primary encloses the same core flux \Phi(t); the total flux linkage is N_p\Phi; Faraday's law says the induced EMF equals the negative time derivative of flux linkage. In steady state the source voltage equals this back-EMF (since there is no other impedance to drop the voltage across).

Step 3. The same flux \Phi(t) threads every turn of the secondary as well, because the core guides essentially all of it:

v_s(t) \;=\; -N_s\,\dfrac{d\Phi}{dt}

Step 4. Divide the two equations.

\dfrac{v_s(t)}{v_p(t)} \;=\; \dfrac{N_s}{N_p}

Why: the d\Phi/dt term is common to both and cancels. This is the voltage transformation ratio: secondary voltage is to primary voltage as secondary turns are to primary turns. Wind 10 turns of secondary for every 1 turn of primary and you get 10× the voltage on the secondary side.

Writing this for RMS values (since both sinusoids are the same sine function):

\boxed{\;\dfrac{V_s}{V_p} \;=\; \dfrac{N_s}{N_p}\;}

Current transformation

An ideal transformer also conserves power: whatever goes in on the primary side comes out on the secondary side, with no loss.

P_\text{in} \;=\; P_\text{out}
V_p I_p \cos\phi_p \;=\; V_s I_s \cos\phi_s

For an ideal transformer on a linear load the phase angles match, so \cos\phi_p = \cos\phi_s and

V_p I_p \;=\; V_s I_s

Substituting the voltage ratio:

\dfrac{I_s}{I_p} \;=\; \dfrac{V_p}{V_s} \;=\; \dfrac{N_p}{N_s}
\boxed{\;\dfrac{I_s}{I_p} \;=\; \dfrac{N_p}{N_s}\;}

Voltage goes up by the turns ratio; current goes down by the inverse turns ratio. This is the entire commercial value of the transformer. The product VI (apparent power in VA) is preserved; what changes is how that VA is divided between a "high-V, low-I" form (efficient for transmission) and a "low-V, high-I" form (safe for household use).

Cross-section of a simple two-winding transformer An iron core shaped like a rectangle with a central vertical strut. The left limb carries the primary winding with Np turns, connected to an AC source labelled Vp. The right limb carries the secondary winding with Ns turns, connected to a load labelled Vs. Magnetic flux arrows show the flux circulating around the core loop. Ideal two-winding transformer on a laminated iron core primary, N_p turns secondary, N_s turns Φ(t) V_p ~ V_s → load
A simple two-winding transformer. Both coils share the same iron-core flux $\Phi(t)$. Faraday's law on each coil gives an induced EMF proportional to $N\,d\Phi/dt$ — the voltage ratio is the turns ratio.

Step-up, step-down, and the Indian transmission backbone

A step-up transformer has N_s > N_p: voltage goes up, current goes down. Every generating station in India uses a step-up transformer right at the generator terminals to lift the 11–25 kV generated voltage up to the transmission level. Mouda thermal plant's 500 MW units generate at 21 kV; a large step-up transformer (typically 400 MVA or larger) raises this to 400 kV before sending the power out on the Maharashtra interstate transmission network.

A step-down transformer does the opposite: N_s < N_p, voltage down, current up. On the Indian grid the power is stepped down through several stages:

Stage Voltage level Where it appears
Generator terminals 11–25 kV NTPC Mouda, Koradi
Interstate transmission 400 kV, 765 kV Mouda → Sholapur HVDC backbone
Subtransmission 132 kV, 220 kV Regional grid substations
Primary distribution 33 kV, 22 kV City feeder lines
Secondary distribution 11 kV Neighbourhood distribution transformer
Consumer 230 V / 400 V Your house and workshop

Each of those arrows is a transformer doing its simple voltage-ratio job. From generator to socket, the power typically passes through at least four transformers — each introducing a small loss (around 0.5–1% per stage for modern units) that accumulates to a full-chain efficiency of around 94–96% before any transmission-line resistance is considered.

Why transmit at high voltage?

Transmission-line loss is I^2 R per unit length. For a fixed real power P = VI\cos\phi and fixed \cos\phi, doubling V halves I — and halves I squared to one-quarter. Line loss drops by a factor of 4. Going from 11 kV to 400 kV (36×) drops the line loss by 36² ≈ 1300×. That is why even though the voltage has to be inconveniently high for long-distance transmission, you pay for the transformer and then save thirteen hundred times more in the wires. The break-even distance for HVAC (high-voltage AC) transmission to justify the extra transformer cost is around 40–50 km; above that, it's always worth stepping up.

Losses — why real transformers are not ideal

Real transformers have three loss mechanisms:

  1. Copper loss (I^2 R in the windings) — scales as current squared. Dominates at full load.
  2. Iron loss (hysteresis + eddy currents in the core) — scales with the square of the peak flux, roughly independent of load. Dominates at light load.
  3. Stray loss (leakage flux coupling into nearby steel, skin effect) — small, geometry-dependent.

Efficiency of a modern 400 MVA grid transformer runs to 99.7% or better at full load. Lower-voltage distribution transformers are somewhat less efficient — around 97–98% — but still far better than any other large electrical machine ever built.

Worked examples

Example 1: The Ahmedabad factory's power factor penalty

A textile factory in Naroda GIDC draws 500 kW of real power from the 11 kV, 50 Hz feeder. The load's power factor is measured as \cos\phi = 0.72 (lagging, because the induction motors dominate).

(a) What is the RMS line current? (b) What is the apparent power in kVA? (c) What is the reactive power in kVAR? (d) If the tariff penalises power factor below 0.90 at 1.5% of the monthly bill per 0.01 below threshold, and the monthly energy consumption is 300,000 kWh at ₹10 per kWh, what is the penalty?

Power triangle for the Naroda textile factory Right triangle showing P = 500 kW horizontal, S = 694 kVA hypotenuse, and Q = 482 kVAR vertical. Angle between S and P is arc-cos 0.72 ≈ 44 degrees. P = 500 kW Q ≈ 482 kVAR S ≈ 694 kVA φ ≈ 44° cos φ = 0.72, lagging
Power triangle for the Naroda factory at 500 kW real and $\cos\phi = 0.72$.

Step 1. Apparent power from real power and power factor.

S \;=\; \dfrac{P}{\cos\phi} \;=\; \dfrac{500\,\text{kW}}{0.72} \;\approx\; 694.4\,\text{kVA}

Why: P = S\cos\phi rearranged. S is in kVA because "real power over a fraction" is still the product V_\text{rms}I_\text{rms} in thousands of volt-amperes.

Step 2. RMS line current (for a single-phase equivalent or one phase of three-phase).

I_\text{rms} \;=\; \dfrac{S}{V_\text{rms}} \;=\; \dfrac{694\,400\,\text{VA}}{11\,000\,\text{V}} \;\approx\; 63.1\,\text{A}

Why: apparent power is just the product of RMS volts and RMS amps, without any cos-phi factor — the full current flows through the wire regardless of phase.

Step 3. Reactive power.

Q \;=\; S\sin\phi \;=\; S\sqrt{1 - \cos^2\phi} \;=\; 694.4\,\times\,\sqrt{1 - 0.72^2}
= 694.4\,\times\,\sqrt{0.4816} \;=\; 694.4\,\times\,0.694 \;\approx\; 481.8\,\text{kVAR}

Why: the power triangle says P^2 + Q^2 = S^2, so Q = \sqrt{S^2 - P^2} = S\sqrt{1-\cos^2\phi} = S\sin\phi. Using \sin\phi = \sqrt{1-\cos^2\phi} avoids computing \phi itself.

Step 4. Monthly penalty.

Bill before penalty: 300{,}000 \times 10 = ₹30{,}00{,}000 (thirty lakh).

\cos\phi deficit below threshold: 0.90 - 0.72 = 0.18 = 18\times 0.01.

Penalty rate: 18 \times 1.5\% = 27\%.

Penalty amount: 30{,}00{,}000 \times 0.27 = ₹8{,}10{,}000 (eight lakh ten thousand).

Why: real Indian industrial tariffs from MSEDCL, GUVNL, and TSSPDCL follow this kind of rule, though the exact slab and rate vary. The point is that a 0.18 shortfall in \cos\phi can translate to a 27% bill increase — enough to pay for a capacitor bank in a month or two.

Result. S \approx 694 kVA; I \approx 63 A at 11 kV; Q \approx 482 kVAR; penalty ≈ ₹8.1 lakh per month.

What this shows. Poor power factor is not about wasted energy (the meter only counts real kWh) — it is about current the utility had to deliver that did no paying work. The fix is a ~480 kVAR capacitor bank in parallel with the factory's main busbar, which supplies the reactive power locally and corrects \cos\phi to near 1. That capacitor bank typically pays for itself in 3–6 months from the penalty savings alone.

Example 2: Neighbourhood distribution transformer

A 100 kVA, 11 kV / 400 V three-phase distribution transformer serves a residential cluster in Koramangala, Bengaluru. The transformer's primary has N_p = 3300 turns per phase.

(a) How many turns are on each secondary? (b) At full load, what is the primary RMS current per phase and the secondary RMS current per phase? (c) If the copper loss at full load is 2.1 kW and the iron loss is 0.6 kW, what is the efficiency at full load with \cos\phi = 0.95?

Step 1. Turns ratio from voltage ratio.

For a three-phase transformer in star-star configuration, the phase voltages are V_{p,\text{phase}} = 11{,}000/\sqrt{3} \approx 6351 V and V_{s,\text{phase}} = 400/\sqrt{3} \approx 231 V.

\dfrac{N_s}{N_p} \;=\; \dfrac{V_{s,\text{phase}}}{V_{p,\text{phase}}} \;=\; \dfrac{231}{6351} \;\approx\; 0.0364
N_s \;=\; N_p \times 0.0364 \;=\; 3300 \times 0.0364 \;\approx\; 120 \text{ turns per phase}

Why: the voltage-ratio equation applies to the actual voltage across each coil — in a star-star three-phase transformer, this is the phase voltage, which is 1/\sqrt{3} times the line-to-line voltage.

Step 2. Full-load currents per phase.

Total apparent power is 100 kVA, divided over three phases: 33.33 kVA per phase.

Primary phase current: I_p = (33{,}333)/6351 \approx 5.25 A.

Secondary phase current: I_s = (33{,}333)/231 \approx 144.3 A.

Check the ratio: I_s/I_p = 144.3/5.25 \approx 27.5 \approx N_p/N_s.

Step 3. Efficiency.

Output real power: P_\text{out} = S\cos\phi = 100 \times 0.95 = 95 kW.

Total losses: 2.1 + 0.6 = 2.7 kW.

Input power: P_\text{in} = P_\text{out} + P_\text{loss} = 95 + 2.7 = 97.7 kW.

\eta \;=\; \dfrac{P_\text{out}}{P_\text{in}} \;=\; \dfrac{95}{97.7} \;\approx\; 0.972 \;=\; 97.2\%

Why: efficiency is output-over-input, not output-over-rated. The iron loss is roughly constant (from magnetising the core); the copper loss is load-dependent. At lighter loads the copper loss drops, and efficiency can actually improve slightly — until the constant iron loss starts to dominate at very light load.

Result. N_s = 120 turns; I_p = 5.25 A, I_s = 144 A; \eta = 97.2\% at full load with \cos\phi = 0.95.

What this shows. A real distribution transformer routinely pushes 97–98% efficiency — a remarkable achievement given that it is just iron and copper in a can. The 2.7 kW lost in each 100 kVA transformer, multiplied over the roughly 12 lakh distribution transformers on the Indian grid, is about 3 GW of continuous loss — small as a fraction (≈1.5% of national demand) but huge in absolute terms. This is why Bureau of Energy Efficiency stars on distribution transformers are a real national priority.

Example 3: Power factor correction with a capacitor bank

Return to the Naroda textile factory. It currently draws 500 kW at \cos\phi_1 = 0.72 from an 11 kV line at 50 Hz. The owner wants to improve the power factor to \cos\phi_2 = 0.95 by adding a parallel capacitor bank.

(a) What reactive power must the capacitor bank supply? (b) What is the total capacitance required (treating it as a single-phase equivalent across 11 kV)?

Correcting the power triangle by adding parallel capacitance Two overlapping power triangles. Taller triangle: S1 = 694 kVA, Q1 = 482 kVAR, P = 500 kW, phi1 = 44 degrees. Shorter triangle: S2 = 526 kVA, Q2 = 164 kVAR, same P, phi2 = 18 degrees. The capacitor supplies Q1 - Q2 = 318 kVAR of leading reactive power. P = 500 kW Q₁ = 482 kVAR (before) Q₂ = 164 kVAR (after) S₁ = 694 kVA S₂ = 526 kVA Q_cap = Q₁ − Q₂ = 318 kVAR
Before correction (dashed outer triangle): $S_1 = 694$ kVA, $Q_1 = 482$ kVAR, $\cos\phi_1 = 0.72$. After a 318 kVAR capacitor bank is added (solid shorter triangle): $S_2 = 526$ kVA, $Q_2 = 164$ kVAR, $\cos\phi_2 = 0.95$. $P$ is unchanged.

Step 1. Reactive power before correction (from Example 1).

Q_1 \;=\; 500\,\tan\phi_1 \;=\; 500\,\times\,\tan(\arccos 0.72) \;=\; 500\,\times\,0.964 \;=\; 482\,\text{kVAR}

Step 2. Reactive power after correction.

Q_2 \;=\; 500\,\tan\phi_2 \;=\; 500\,\times\,\tan(\arccos 0.95) \;=\; 500\,\times\,0.329 \;=\; 164.3\,\text{kVAR}

Why: real power hasn't changed (the factory's motors still do the same mechanical work); only the reactive power has dropped because the capacitor supplies part of it locally.

Step 3. Reactive power the capacitor must supply.

Q_\text{cap} \;=\; Q_1 - Q_2 \;=\; 482 - 164 \;=\; 318\,\text{kVAR}

Step 4. Capacitance required.

A capacitor rated V_\text{rms} = 11 kV produces reactive power Q_\text{cap} = V_\text{rms}^2/X_C = V_\text{rms}^2\omega C. Solve for C:

C \;=\; \dfrac{Q_\text{cap}}{V_\text{rms}^2\,\omega} \;=\; \dfrac{318{,}000}{(11{,}000)^2 \times 314.16}
C \;=\; \dfrac{318{,}000}{3.8\times 10^{10}} \;\approx\; 8.4\times 10^{-6}\,\text{F} \;=\; 8.4\,\mu\text{F}

Result. The factory needs a 318 kVAR, 8.4 μF capacitor bank at 11 kV.

What this shows. A modest amount of capacitance — less than 10 μF at 11 kV — corrects a 318 kVAR deficit and saves the factory ₹8 lakh a month. Note that the real power stays at 500 kW: the capacitor does not reduce the work the motors do. It only stops reactive current from being dragged all the way from the generator.

Common confusions

If you can compute real/reactive/apparent power and use the transformer voltage-turns ratio, you have the operational machinery. The deeper treatment connects power factor to the full complex-impedance picture, explains transformer behaviour under non-ideal conditions, and derives the peak-flux density that determines transformer core sizing.

Complex power — the cleanest single quantity

In the phasor formulation, let \tilde{V} and \tilde{I} be the complex voltage and current phasors (using peak amplitudes or RMS by convention — we'll use RMS). Define the complex power as

\tilde{S} \;\equiv\; \tilde{V}\,\tilde{I}^{*}

where \tilde{I}^{*} is the complex conjugate of the current phasor. Expanding, if \tilde{V} = V_\text{rms}e^{j\theta_v} and \tilde{I} = I_\text{rms}e^{j\theta_i},

\tilde{S} \;=\; V_\text{rms}I_\text{rms}\,e^{j(\theta_v - \theta_i)} \;=\; V_\text{rms}I_\text{rms}(\cos\phi + j\sin\phi)

where \phi = \theta_v - \theta_i. The real part is P; the imaginary part is Q; the magnitude is S. Three powers in one number.

This form is why engineers use the conjugate: it produces a complex number whose real part is the real power. Without the conjugate, the phase angle would come out with a sign that is harder to interpret. The convention is universal.

Maximum power transfer and impedance matching

Drive a source of internal impedance Z_s = R_s + jX_s with a load Z_L = R_L + jX_L. The load current is

\tilde{I} \;=\; \dfrac{\tilde{V}_s}{Z_s + Z_L}

The real power delivered to the load is P_L = |\tilde{I}|^2 R_L. Maximising this with respect to R_L and X_L gives the maximum-power-transfer condition:

R_L \;=\; R_s,\qquad X_L \;=\; -X_s

That is, Z_L = Z_s^{*} — the load impedance must be the complex conjugate of the source impedance. The load's reactance must cancel the source's reactance, so all the voltage is dropped across the resistive parts. This is why RF engineers agonise over impedance matching at antennas.

In the power-grid context, you do not want maximum power transfer — you want maximum efficiency, which requires R_L \gg R_s (so that very little of the real power is lost in the source). The grid's transformers, lines, and generators are designed to present low internal impedance so that the load dominates.

Transformer core: why the peak flux matters

The peak flux in the core is set by the primary voltage:

V_{p,\text{rms}} \;=\; \dfrac{N_p\,\omega\,\Phi_\text{peak}}{\sqrt{2}}

Why: from v_p = N_p\,d\Phi/dt with \Phi = \Phi_\text{peak}\sin\omega t, the peak EMF is N_p\omega\Phi_\text{peak}, and RMS is 1/\sqrt{2} of that.

Rearrange:

\Phi_\text{peak} \;=\; \dfrac{\sqrt{2}\,V_{p,\text{rms}}}{N_p\,\omega}

For a 230 V / 12 V transformer on 50 Hz with N_p = 1000 turns:

\Phi_\text{peak} \;=\; \dfrac{1.414 \times 230}{1000 \times 314.16} \;\approx\; 1.04\,\text{mWb}

The peak flux density B_\text{peak} = \Phi_\text{peak}/A, where A is the core cross-section. For silicon steel the saturation flux density is around 1.5 T — so for a 1.2 T operating point you need A \geq 1.04\,\text{mWb}/1.2\,\text{T} \approx 8.7 cm². This is where transformer design lives: the number of primary turns, the core cross-section, and the operating frequency together set the flux density, which in turn sets the iron loss. Double the frequency and you can halve the core cross-section for the same flux density and the same primary voltage — which is why modern aerospace and server-rack power supplies run at tens or hundreds of kilohertz (with tiny ferrite cores) rather than at 50 Hz.

Real transformer model and the short-circuit test

A realistic transformer has four non-ideal elements in series with each winding and one parallel magnetising branch. The equivalent circuit (referred to the primary side):

  • R_p and R_s': copper resistance of primary and secondary (secondary referred to primary by the turns-ratio squared).
  • X_p and X_s': leakage reactance of primary and secondary (representing flux that doesn't couple between the two coils).
  • R_c in parallel: core-loss resistance (representing hysteresis + eddy-current losses).
  • X_m in parallel: magnetising reactance (the primary's self-inductance seen through the core).

Measuring these is a standard engineering exercise. The short-circuit test shorts the secondary and applies reduced primary voltage until rated current flows; the voltage and power readings give R_p + R_s' and X_p + X_s'. The open-circuit test leaves the secondary open and applies rated primary voltage; the small primary current and power readings give R_c and X_m. Together these four numbers characterise the transformer for all operating conditions.

Why three-phase?

India's grid, like most of the world's, uses three-phase AC rather than single-phase. Three conductors carry three sinusoidal voltages that are 120° out of phase with each other. The total instantaneous power is

p_\text{3ph}(t) \;=\; V_{p,0}I_{p,0}\bigl[\sin(\omega t)\sin(\omega t - \phi) + \sin(\omega t - 2\pi/3)\sin(\omega t - 2\pi/3 - \phi) + \sin(\omega t - 4\pi/3)\sin(\omega t - 4\pi/3 - \phi)\bigr]

Using the product-to-sum identity on each term and summing, all the 2\omega cosines cancel (three cosines at 120° phase shifts sum to zero). What's left is three times the single-phase average power, with no oscillation:

p_\text{3ph}(t) \;=\; 3\,V_\text{rms,phase}\,I_\text{rms,phase}\,\cos\phi \;=\; \text{constant}

This is why three-phase motors run smoothly. The power is constant in time (not pulsing at 2\omega as in single phase), so the motor produces a constant torque. Single-phase motors, in contrast, have a pulsing torque at 100 Hz (twice the 50 Hz mains) and need clever tricks (a capacitor for the auxiliary winding, as in a ceiling fan) to produce a rotating magnetic field at all. Three-phase is self-starting, smoother, and three times more efficient per unit of copper — which is why every large industrial motor in India is three-phase.

Where this leads next