Power Factor & Power Factor Correction: Causes, Disadvantages, Capacitor Sizing (Solved Problems + MCQs)

Two factories may draw exactly the same real power, yet one needs a bigger transformer, thicker cables and pays a penalty to the utility. The difference is power factor (PF). Power factor and its correction are core topics for PGCB, BPDB, BREB/PBS, DPDC, DESCO, WZPDCL and BCS (EEE) exams and for every power-sector viva. This post explains the concept visually, then covers causes, disadvantages, correction methods and capacitor sizing with fully solved problems.

⚡ Key takeaways

• PF = cos φ = P / S = R / Z, where φ is the angle between voltage and current.
• P (kW) does useful work, Q (kVAR) only builds magnetic/electric fields, S (kVA) is what the system must carry: S² = P² + Q².
• Low PF ⇒ more current for the same power ⇒ bigger equipment, higher I²R loss, poorer voltage regulation and penalty charges.
• Main cause: inductive loads, especially lightly loaded induction motors and transformers.
• Correction: shunt capacitors, synchronous condensers, phase advancers. Capacitor size Q_C = P (tan φ₁ − tan φ₂).

📑 Contents

1. Active, reactive and apparent power
2. What is power factor?
3. Seeing power factor (animation)
4. Causes of low power factor
5. Disadvantages of low power factor
6. Methods of power factor correction
7. Capacitor sizing formula
8. Solved problems
9. Most economical power factor
10. One-line exam facts
11. Practice MCQs
12. বাংলায় দ্রুত রিভিশন

1. Active, Reactive and Apparent Power

Quantity	Symbol & unit	Formula (single-phase)	Meaning
Active (real/true) power	P — W, kW, MW	P = V I cos φ	Power actually converted into work, heat, light
Reactive power	Q — VAR, kVAR	Q = V I sin φ	Power that flows back and forth to build magnetic fields (inductors) or electric fields (capacitors); average value zero
Apparent power	S — VA, kVA	S = V I = √(P² + Q²)	Total power the source, cables and transformers must carry

For a balanced three-phase system multiply by √3 with line values: P = √3 V_L I_L cos φ, Q = √3 V_L I_L sin φ, S = √3 V_L I_L. These three form the power triangle.

💡 Beer analogyIn a glass of beer, the liquid is P (what you actually want), the foam is Q (takes up space but doesn’t quench thirst) and the full glass is S. Power factor = liquid ÷ full glass. Less foam = better PF.

2. What Is Power Factor?

Power factor is the ratio of active power to apparent power. For sinusoidal voltage and current it equals the cosine of the phase angle between them:

PF = cos φ = P / S = kW / kVA = R / Z

• Lagging PF: current lags voltage — inductive loads (motors, transformers, chokes).
• Leading PF: current leads voltage — capacitive loads (capacitor banks, over-excited synchronous motors, lightly loaded long cables).
• Unity PF (= 1): current in phase with voltage — purely resistive loads (heaters, incandescent lamps), or an RLC circuit at resonance.

PF lies between 0 and 1; it is never expressed as negative — we always say “lagging” or “leading”. For non-sinusoidal (harmonic) currents, true PF = displacement PF (cos φ) × distortion factor.

3. Seeing Power Factor (Animation)

In the animation, the phase angle φ between voltage (blue) and current (orange) grows from 0° to 75°. The product p(t) = v·i is the instantaneous power. Green areas are energy delivered to the load; red areas are energy returned to the source. As φ increases, more energy sloshes back and forth, the average power (dashed line) falls, and the power triangle shows Q growing while P shrinks — even though the current magnitude stays the same.

Animation 1 – As φ increases, PF = cos φ drops: the same current delivers less real power.

4. Causes of Low Power Factor

• Induction motors — the largest cause in industry. They need magnetising current; PF is about 0.2–0.3 at no load and 0.8–0.9 at full load, so lightly loaded (oversized) motors are very poor.
• Transformers, especially at light load (magnetising current is a larger share of total current).
• Discharge lighting with magnetic ballasts (fluorescent tubes, mercury-vapour and sodium-vapour lamps).
• Arc furnaces, induction furnaces and welding sets.
• Variable industrial loads — at light load the reactive power stays almost constant while active power falls.
• Increased supply voltage above rated — raises magnetising current of motors and transformers.
• Harmonic-producing loads (rectifiers, drives, SMPS) reduce true PF through distortion.

5. Disadvantages of Low Power Factor

For a given power, current is inversely proportional to PF: I = P / (√3 V_L cos φ). Example — a 100 kW, 400 V three-phase load draws 144 A at PF 1.0 but 241 A at PF 0.6.

1. Larger kVA rating of equipment: generators, transformers and switchgear are rated in kVA = kW / cos φ, so they must be bigger and costlier.
2. Larger conductor size: more current needs thicker cables and busbars.
3. Higher copper losses: I²R loss ∝ 1 / cos²φ. At PF 0.6 the loss is 2.78 times that at unity PF, so efficiency drops.
4. Poor voltage regulation: larger current causes more voltage drop in lines, transformers and alternators, so extra voltage-regulating equipment is needed.
5. Reduced handling capacity: lines and transformers reach their current limit while carrying less useful kW.
6. Penalty in electricity bill: utilities charge industrial consumers extra (or bill on kVA demand) when PF falls below the limit set in the tariff.

6. Methods of Power Factor Correction

A lagging load draws reactive power from the supply. The cure is to supply that reactive power locally with a device that draws leading current, so that the supply only needs to provide P plus a small Q.

6.1 Static capacitors (shunt capacitor banks)

Capacitors connected in parallel with the load take a leading current that cancels part of the lagging reactive current. They can be fixed (connected permanently to a motor), or switched in steps by an Automatic Power Factor Correction (APFC) panel that monitors PF and switches capacitor stages through contactors or thyristors. In three-phase systems they are usually connected in delta, because the required capacitance is one-third of that in star.

Placement: individual (at each large motor), group (at a distribution board), or central (at the main incoming bus). The closer to the load, the more of the system is relieved of reactive current.

6.2 Synchronous condenser

A synchronous motor running at no mechanical load and over-excited draws leading current and behaves like a variable capacitor. By adjusting field current, its reactive output can be varied smoothly. Used at grid sub-stations for large-scale reactive power and voltage control.

6.3 Phase advancer

An AC exciter mounted on the shaft of a large induction motor and connected in its rotor circuit. It supplies the magnetising ampere-turns at slip frequency, so the motor no longer draws magnetising current from the supply and can even run at leading PF.

Comparison

Feature	Static capacitor	Synchronous condenser
Moving parts	None — little maintenance	Rotating machine — needs maintenance
Losses	Very low	Higher
Control of kVAR	In steps	Smooth, continuous
Can absorb kVAR (lagging)?	No	Yes (under-excited)
Economical range	Small to medium ratings	Large ratings (tens of MVAR and above)
Weak points	Damaged by over-voltage & harmonics; limited life	Noise, high initial cost

7. Capacitor Sizing Formula

Real power P stays the same; only Q is reduced. If the PF is to be improved from cos φ₁ to cos φ₂:

Q_C = P (tan φ₁ − tan φ₂) kVAR

Capacitance needed (f = supply frequency, ω = 2πf):

• Single-phase: C = Q_C / (ω V²)
• Three-phase, star-connected bank: C per phase = Q_C / (ω V_L²)
• Three-phase, delta-connected bank: C per phase = Q_C / (3 ω V_L²)

Animation 2 – A capacitor bank supplies Q_C locally, shrinking the power triangle: the same 100 kW now needs only 105 kVA instead of 143 kVA.

8. Solved Problems

Problem 1 — Three-phase plant: raise PF from 0.70 to 0.95

QuestionA 415 V, 50 Hz three-phase plant draws 100 kW at 0.7 PF lagging. Find the capacitor kVAR and the capacitance per phase (delta and star) needed to raise the PF to 0.95 lagging.

Given	To find
VL = 415 V, f = 50 Hz, P = 100 kW cos φ₁ = 0.70 (lag), cos φ₂ = 0.95 (lag)	QC (kVAR); C per phase for Δ and Y; current and loss saving

Fig. P1(a) — Single-line diagram of the plant with the shunt capacitor bank.

Step 1 — Find the power-factor angles and their tangents

φ₁ = cos⁻¹(0.70) = 45.57°  ⇒  tan φ₁ = 1.0202 φ₂ = cos⁻¹(0.95) = 18.19°  ⇒  tan φ₂ = 0.3287

Step 2 — Build the power triangle before correction

S₁ = P / cos φ₁ = 100 / 0.70 = 142.86 kVA Q₁ = P tan φ₁ = 100 × 1.0202 = 102.02 kVAR

Step 3 — Power triangle after correction (P does not change)

S₂ = P / cos φ₂ = 100 / 0.95 = 105.26 kVA Q₂ = P tan φ₂ = 100 × 0.3287 = 32.87 kVAR

Step 4 — Capacitor kVAR required

Q_C = Q₁ − Q₂ = P (tan φ₁ − tan φ₂) = 100 × (1.0202 − 0.3287) = 69.15 kVAR

Per phase (one capacitor of the three-unit bank):

Q_C,ph = 69.15 / 3 = 23.05 kVAR = 23 050 VAR

Fig. P1(b) — Power triangle: P stays 100 kW; the capacitor removes 69.2 kVAR of the 102 kVAR.

Step 5 — Angular frequency

ω = 2πf = 2 × 3.1416 × 50 = 314.16 rad/s

Step 6 — Capacitance per phase, delta (Δ) connection

In delta, each capacitor is connected between two lines, so it sees the full line voltage V_L = 415 V.

X_C = V_L² / Q_C,ph = 415² / 23 050 = 172 225 / 23 050 = 7.472 Ω C_Δ = 1 / (ω X_C) = 1 / (314.16 × 7.472) = 4.260 × 10⁻⁴ F ≈ 426 µF per phase

Check with the direct formula: C_Δ = Q_C / (3 ω V_L²) = 69 150 / (3 × 314.16 × 415²) = 426 µF ✔

Step 7 — Capacitance per phase, star (Y) connection

In star, each capacitor is connected from a line to neutral, so it sees only the phase voltage.

V_ph = V_L / √3 = 415 / 1.732 = 239.6 V X_C = V_ph² / Q_C,ph = 239.6² / 23 050 = 57 408 / 23 050 = 2.491 Ω C_Y = 1 / (ω X_C) = 1 / (314.16 × 2.491) = 1.278 × 10⁻³ F ≈ 1278 µF per phase

C_Y / C_Δ = 1278 / 426 = 3 — star needs three times the capacitance, which is why delta-connected banks are preferred on LV systems.

Fig. P1(c) — Same 23.05 kVAR per capacitor: in delta each unit sees √3 times more voltage, so it needs 1/3 of the capacitance.

Step 8 — Line current and loss saving

I₁ = S₁ / (√3 V_L) = 142 857 / (1.732 × 415) = 142 857 / 718.8 = 198.7 A I₂ = S₂ / (√3 V_L) = 105 263 / 718.8 = 146.4 A Current reduction = (198.7 − 146.4) / 198.7 = 26.3 % Loss ratio = (I₂ / I₁)² = (0.70 / 0.95)² = 0.543  ⇒  I²R loss falls by ≈ 45.7 %

Capacitor-bank line current: I_C = Q_C / (√3 V_L) = 69 150 / 718.8 = 96.2 A.

Answer — Problem 1 Q_C = 69.2 kVAR  |  C_Δ = 426 µF/phase  |  C_Y = 1278 µF/phase
kVA: 142.9 → 105.3 (−26 %)  |  Line current: 198.7 A → 146.4 A  |  Copper loss: −46 %

Problem 2 — Single-phase motor: raise PF from 0.6 to unity

QuestionA single-phase motor takes 10 A at 0.6 PF lagging from a 230 V, 50 Hz supply. Find the capacitance needed (in parallel) to raise the PF to unity, and the new supply current.

Given	To find
V = 230 V, f = 50 Hz, IM = 10 A cos φ₁ = 0.6 (lag), target cos φ₂ = 1	QC, IC, C and the new line current I

Fig. P2(a) — Capacitor connected in parallel with the motor terminals.

Step 1 — Power-factor angle

φ₁ = cos⁻¹(0.6) = 53.13°  ⇒  sin φ₁ = √(1 − 0.6²) = 0.8

Step 2 — Power drawn by the motor

S = V I = 230 × 10 = 2300 VA P = V I cos φ₁ = 230 × 10 × 0.6 = 1380 W Q = V I sin φ₁ = 230 × 10 × 0.8 = 1840 VAR (lagging)

Step 3 — Reactive power the capacitor must supply

For unity PF, φ₂ = 0 so tan φ₂ = 0 and Q₂ = 0 — all of the motor's reactive power must be cancelled:

Q_C = Q₁ − Q₂ = 1840 − 0 = 1840 VAR

Step 4 — Capacitor current and reactance

I_C = Q_C / V = 1840 / 230 = 8 A X_C = V / I_C = 230 / 8 = 28.75 Ω

Step 5 — Capacitance

ω = 2π × 50 = 314.16 rad/s C = 1 / (ω X_C) = 1 / (314.16 × 28.75) = 1 / 9032 = 1.107 × 10⁻⁴ F = 110.7 µF

Check with the direct formula: C = Q_C / (ω V²) = 1840 / (314.16 × 230²) = 1840 / 16 619 025 = 110.7 µF ✔

Fig. P2(b) — Phasor diagram: I = I_M + I_C. The capacitor current cancels the lagging (reactive) part of the motor current.

Step 6 — New supply current

After correction the supply only provides the active component (the capacitor provides the reactive one):

I = P / (V cos φ₂) = 1380 / (230 × 1) = 6 A

Phasor check: I = √[(I_M cos φ)² + (I_M sin φ − I_C)²] = √[6² + (8 − 8)²] = 6 A ✔  — a 40 % drop from 10 A.

Answer — Problem 2 Q_C = 1840 VAR  |  I_C = 8 A  |  C = 110.7 µF  |  Supply current: 10 A → 6 A

Exam tipUnity PF is rarely chosen in practice — the last few kVAR give very little saving and risk over-correction (leading PF) at light load. 0.95–0.98 lagging is the usual target.

9. Most Economical Power Factor

Raising PF all the way to unity needs a very large capacitor bank for a small extra benefit. If x = annual cost per kVA of maximum demand (tariff charge) and y = annual cost per kVAR of the correcting equipment (interest + depreciation), the most economical PF is:

cos φ₂ = √(1 − (y / x)²)

This is why industries usually correct to about 0.95–0.98 lagging, not to exactly 1. Over-correction to a leading PF can cause over-voltage at light load.

10. One-Line Exam Facts

• PF of a pure resistor = 1; pure inductor = 0 lagging; pure capacitor = 0 leading.
• Series RLC circuit at resonance: PF = 1.
• Over-excited synchronous motor → leading PF; under-excited → lagging PF (V-curves / inverted V-curves).
• Wattmeter reads P; varmeter reads Q; PF meter reads cos φ.
• In the two-wattmeter method: if W₁ = W₂, PF = 1; if one reading is zero, PF = 0.5; if one reading is negative, PF < 0.5.
• Capacitor banks also raise voltage and release system capacity (kVA).
• Reactive power is measured in VAR (volt-ampere reactive).

11. Practice MCQs (click to see answer)

1. Power factor is defined as: (a) kVA/kW (b) kW/kVA (c) kVAR/kW (d) kW/kVAR

✔ (b) kW/kVA = cos φ.

2. The main cause of low power factor in industry is: (a) Heaters (b) Incandescent lamps (c) Induction motors (d) Capacitors

✔ (c) Induction motors (especially at light load).

3. Power factor of a purely resistive circuit is: (a) 0 (b) 0.5 (c) 1 (d) 0.707

✔ (c) 1 (unity).

4. Static capacitors for PF correction are connected: (a) In series with load (b) In parallel with load (c) In series with neutral (d) Across the generator field

✔ (b) In parallel (shunt) with the load.

5. An over-excited synchronous motor at no load used for PF improvement is called: (a) Phase advancer (b) Synchronous condenser (c) Reactor (d) Booster

✔ (b) Synchronous condenser.

6. Copper loss in a line is inversely proportional to: (a) cos φ (b) cos²φ (c) sin φ (d) tan φ

✔ (b) cos²φ (for constant P and V).

7. kVAR rating of capacitor to improve PF from cos φ₁ to cos φ₂ is: (a) P(tan φ₂ − tan φ₁) (b) P(tan φ₁ − tan φ₂) (c) P(cos φ₁ − cos φ₂) (d) S(sin φ₁ − sin φ₂)

✔ (b) P(tan φ₁ − tan φ₂).

8. For the same kVAR, a delta-connected capacitor bank needs capacitance per phase equal to ______ of star connection. (a) 3 times (b) 1/3 (c) √3 times (d) Same

✔ (b) 1/3.

9. A load takes 80 kW at 100 kVA. Its power factor is: (a) 0.6 (b) 0.8 (c) 1.25 (d) 0.75

✔ (b) 0.8; reactive power = 60 kVAR.

10. Low power factor results in: (a) Lower line losses (b) Better voltage regulation (c) Larger kVA rating of equipment (d) Lower current

✔ (c) Larger kVA rating of equipment.

11. In the two-wattmeter method, one wattmeter reads zero. The PF is: (a) 1 (b) 0.866 (c) 0.5 (d) 0

✔ (c) 0.5.

12. Unit of reactive power is: (a) Watt (b) VA (c) VAR (d) Joule

✔ (c) VAR (volt-ampere reactive).

12. বাংলায় দ্রুত রিভিশন

📝 পরীক্ষার আগে এক নজরে

• পাওয়ার ফ্যাক্টর = সক্রিয় শক্তি ÷ আপাত শক্তি = kW/kVA = cos φ, যেখানে φ হলো ভোল্টেজ ও কারেন্টের মধ্যবর্তী কোণ।
• P (kW) কাজ করে, Q (kVAR) চৌম্বক ক্ষেত্র তৈরি করে, S (kVA) মোট শক্তি; S² = P² + Q²।
• কম PF-এর কারণ: আবেশ মোটর (হালকা লোডে), ট্রান্সফরমার, ফ্লুরোসেন্ট বাতির চোক, আর্ক ফার্নেস, ওয়েল্ডিং।
• অসুবিধা: একই শক্তির জন্য বেশি কারেন্ট — বড় যন্ত্র ও মোটা তার লাগে, I²R লস বাড়ে (1/cos²φ অনুপাতে), ভোল্টেজ রেগুলেশন খারাপ হয়, বিলে জরিমানা।
• সংশোধন: লোডের সমান্তরালে স্ট্যাটিক ক্যাপাসিটর, সিনক্রোনাস কনডেন্সার (অতি-উত্তেজিত সিনক্রোনাস মোটর), ফেজ অ্যাডভান্সার।
• ক্যাপাসিটরের মান: Qc = P (tan φ₁ − tan φ₂)। ডেল্টা সংযোগে স্টারের তুলনায় এক-তৃতীয়াংশ ক্যাপাসিট্যান্স লাগে।

Related Posts

• 220 Most Repeated Power Sector MCQs (PGCB, BPDB, BREB, DPDC)
• Rooftop Solar in Bangladesh: Net Metering Rules & Exam Questions
• EEE Job Viva Voce Preparation
• EEE MCQ Syllabus for Job Exams

References: V. K. Mehta & Rohit Mehta, Principles of Power System; B. L. Theraja, Electrical Technology Vol. I; IEEE Std 1459 (power definitions). Images and animations: original work by Totalecer.