UPDA-1 - Notes | FM Electricals

🔷 SET 1: BASIC CONCEPTS 🔷

📖 1. WHAT IS ALTERNATING CURRENT (AC)?

Alternating Current (AC) is an electric current that periodically reverses direction and continuously changes its magnitude with time.

Direct Current (DC) — Flows in one direction (like a battery)
Alternating Current (AC) — Alternates between positive and negative cycles

💡 Real life example: The 230V supply in your home is AC. The voltage oscillates between +325V and -325V, 50 times per second (50 Hz).

📐 2. SINUSOIDAL WAVEFORM

The sinusoidal waveform is the most common form of AC. It can be expressed mathematically as:

v(t) = Vm × sin(ωt + φ)

Symbol	Meaning	Typical Value
v(t)	Instantaneous voltage at time t	Changes with time
Vm	Maximum value / Peak value	325V for 230V supply
ω	Angular frequency (rad/s)	2πf = 314 rad/s (for 50Hz)
f	Frequency (Hertz)	50 Hz in Qatar
φ	Phase angle	0° to 360°

⚡ 3. IMPORTANT PARAMETERS

A. Frequency (f)

Number of cycles in 1 second • Unit: Hertz (Hz) • Qatar: 50 Hz
Formula: f = 1/T

B. Time Period (T)

Time for one cycle • Unit: Seconds (s)
Formula: T = 1/f
Example: For 50 Hz, T = 1/50 = 0.02 sec = 20 ms

C. Peak Value (Vm)

Maximum instantaneous value • Also called Amplitude or Crest value

D. Peak-to-Peak Value (Vp-p)

Difference between max positive and max negative
Formula: Vp-p = 2 × Vm

📊 4. AVERAGE VALUE (Vavg)

Definition: Average of all instantaneous values over one complete cycle.

⚠️ Important: For a symmetrical waveform (sine wave), positive and negative half cycles cancel → average over full cycle = ZERO.

Half-cycle average value (practical use):

Vavg (half cycle) = (2/π) × Vm = 0.636 × Vm

🔍 Example: If Vm = 100V, then Vavg = 63.6V
Applications: Rectifier circuits, battery charging, DC meters

🎯 5. RMS VALUE (Vrms) — MOST IMPORTANT!

Definition: Root Mean Square value — the DC value that produces the same heating effect in a resistive load.

Vrms = Vm / √2 = 0.707 × Vm

🔍 Example: If Vm = 311V, then Vrms = 311 × 0.707 = 220V

Why RMS is important?

"230V AC supply" means RMS voltage is 230V.
Peak voltage = 230 × √2 = 325V.
All standard AC voltmeters display RMS value.

Power in AC: P = Vrms × Irms × cosφ

📐 6. FORM FACTOR & PEAK FACTOR

A. Form Factor

Definition: Ratio of RMS value to Average value

Form Factor = Vrms / Vavg = 1.11 (for sine wave)

B. Peak Factor (Crest Factor)

Definition: Ratio of Maximum value to RMS value

Peak Factor = Vm / Vrms = √2 = 1.414 (for sine wave)

🔄 7. PHASE DIFFERENCE

Definition: Angular difference between two same-frequency waveforms.

In Phase: Same zero crossing → φ = 0°
Leading: Wave A crosses zero earlier → A leads B
Lagging: Wave A crosses zero later → A lags B

Phase Difference = θB - θA

🔍 Example: Wave A at 30°, Wave B at 50° → Difference = 20°, A leads B by 20°

📋 8. SUMMARY TABLE

Parameter	Formula	Sine Wave Value
Peak Value	Vm	Given
Peak-to-Peak	2 × Vm	2Vm
Average (half-cycle)	(2/π) × Vm	0.636 Vm
RMS	Vm / √2	0.707 Vm
Form Factor	Vrms / Vavg	1.11
Peak Factor	Vm / Vrms	1.414
Angular Frequency	2πf	314 (50Hz)
Time Period	1/f	0.02 sec

📝 9. IMPORTANT FORMULAS

ω = 2πf • T = 1/f • Vavg = 0.636Vm • Vrms = 0.707Vm • Vm = √2 × Vrms
Form Factor = 1.11 • Peak Factor = 1.414

🧮 10. NUMERICAL EXAMPLE

Question: v = 311 sin(314t). Find Vm, Vrms, Vavg, f, T.

Solution:
Vm = 311V • Vrms = 220V • Vavg = 198V • f = 50Hz • T = 0.02s

🔷 SET 2: AC THROUGH R, L & C 🔷

⚡ 11. AC THROUGH PURE RESISTOR (R)

Current is in phase with voltage.

v = Vm sin(ωt) • i = Im sin(ωt) • Im = Vm/R

Phase difference: 0°
Power factor: 1
Power: P = Vrms × Irms = I²R

🧲 12. AC THROUGH PURE INDUCTOR (L)

Current lags voltage by 90°.

XL = ωL = 2πfL • i = Im sin(ωt - 90°) • Im = Vm/XL

Power factor: 0 lagging
Real power: 0
Reactive power: QL = Vrms × Irms (VAR)

⚠️ XL ∝ f (higher frequency = higher reactance = less current)

⚛️ 13. AC THROUGH PURE CAPACITOR (C)

Current leads voltage by 90°.

XC = 1/(ωC) = 1/(2πfC) • i = Im sin(ωt + 90°) • Im = Vm/XC

Power factor: 0 leading
Real power: 0
Reactive power: QC = Vrms × Irms (VAR)

⚠️ XC ∝ 1/f (higher frequency = lower reactance = more current)

📊 14. COMPARISON TABLE: R, L, C

Parameter	R	L	C
Opposes	Current flow	Change in current	Change in voltage
Reactance	R (Ω)	XL = ωL	XC = 1/(ωC)
Phase	In phase	Current lags 90°	Current leads 90°
Power Factor	1	0 lagging	0 leading
Real Power	P = VI	0	0
Frequency Effect	No effect	XL ↑	XC ↓

🧮 15. NUMERICAL EXAMPLES

Example 1: L = 0.1H, 230V, 50Hz → XL = 31.4Ω, I = 7.32A

Example 2: C = 100µF, 230V, 50Hz → XC = 31.85Ω, I = 7.22A

Example 3: R = 100Ω, 230V, 50Hz → I = 2.3A, P = 529W

🔗 16. SERIES RL CIRCUIT

A series RL circuit consists of a resistor (R) and an inductor (L) connected in series with an AC supply.

Impedance: Z = √(R² + XL²)
Phase Angle: φ = tan⁻¹(XL / R)
Current: I = V / Z

Voltage across resistor (VR): In phase with current
Voltage across inductor (VL): Leads current by 90°
Supply voltage (V): Phasor sum of VR and VL → V = √(VR² + VL²)
Power factor: cos φ = R / Z (lagging)

📌 Key Point: In RL circuit, current lags the supply voltage. The lag angle depends on the ratio of XL to R.

Parameter	Formula
Impedance (Z)	Z = √(R² + XL²)
Phase Angle (φ)	φ = tan⁻¹(XL / R)
Power Factor	cos φ = R / Z (lagging)
Power Consumed	P = I²R = Vrms × Irms × cos φ

🔗 17. SERIES RC CIRCUIT

A series RC circuit consists of a resistor (R) and a capacitor (C) connected in series with an AC supply.

Impedance: Z = √(R² + XC²)
Phase Angle: φ = tan⁻¹(-XC / R)
Current: I = V / Z

Voltage across resistor (VR): In phase with current
Voltage across capacitor (VC): Lags current by 90°
Supply voltage (V): Phasor sum of VR and VC → V = √(VR² + VC²)
Power factor: cos φ = R / Z (leading)

📌 Key Point: In RC circuit, current leads the supply voltage. The lead angle depends on the ratio of XC to R.

Parameter	Formula
Impedance (Z)	Z = √(R² + XC²)
Phase Angle (φ)	φ = tan⁻¹(-XC / R)
Power Factor	cos φ = R / Z (leading)
Power Consumed	P = I²R = Vrms × Irms × cos φ

🔗 18. SERIES RLC CIRCUIT

A series RLC circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series with an AC supply.

Net Reactance: X = XL - XC (or XC - XL, whichever is larger)
Impedance: Z = √(R² + (XL - XC)²)
Phase Angle: φ = tan⁻¹((XL - XC) / R)
Current: I = V / Z

If XL > XC: Circuit is inductive (current lags voltage)
If XC > XL: Circuit is capacitive (current leads voltage)
If XL = XC: Circuit is resistive (current in phase with voltage) — This is RESONANCE

⚠️ Important: The voltages across L and C can be much larger than the supply voltage at resonance! This is called voltage magnification.

Parameter	Series RL	Series RC	Series RLC
Impedance (Z)	√(R²+XL²)	√(R²+XC²)	√(R²+(XL-XC)²)
Phase Angle (φ)	+ (lagging)	- (leading)	+ (inductive) / - (capacitive) / 0 (resistive)
Power Factor	Lagging	Leading	Depends on XL vs XC
At Resonance	Not applicable	Not applicable	Z = R (minimum), I maximum

🧮 22. NUMERICAL EXAMPLES

Example 1 (Series RL): R = 10Ω, XL = 10Ω, V = 100V
Z = √(10² + 10²) = √200 = 14.14Ω
I = 100 / 14.14 = 7.07A
φ = tan⁻¹(10/10) = 45° (current lags)
P = I²R = 7.07² × 10 = 500W

Example 2 (Resonance): R = 10Ω, L = 0.1H, C = 100µF, V = 100V
f₀ = 1 / (2π√(0.1 × 100×10⁻⁶)) = 1 / (2π√(0.00001)) = 1 / (2π × 0.00316) = 50.3 Hz
At resonance: Z = R = 10Ω, I = 100/10 = 10A (maximum)
VL = I × XL = 10 × (2π × 50.3 × 0.1) = 10 × 31.6 = 316V (voltage magnification!)

Example 3 (Series RC): R = 10Ω, XC = 10Ω, V = 100V
Z = √(10² + 10²) = 14.14Ω
I = 100 / 14.14 = 7.07A
φ = tan⁻¹(-10/10) = -45° (current leads)
P = I²R = 7.07² × 10 = 500W

📐 24. POWER TRIANGLE

The Power Triangle shows the relationship between real power (P), reactive power (Q), and apparent power (S).

S = √(P² + Q²)
Power Factor = cos φ = P / S
φ = tan⁻¹(Q / P)

Component	Direction	Meaning
Real Power (P)	Horizontal (adjacent)	Useful power / Consumed power
Reactive Power (Q)	Vertical (opposite)	Stored and returned power
Apparent Power (S)	Hypotenuse	Total power supplied by source

🔍 Example: If P = 300W, Q = 400VAR, then:
S = √(300² + 400²) = √(90000 + 160000) = √250000 = 500 VA
Power Factor = P/S = 300/500 = 0.6 lagging

Circuit Type	Power Factor (cos φ)	Real Power (P)	Reactive Power (Q)
Pure Resistor	1 (unity)	Maximum	0
Pure Inductor	0 (lagging)	0	Positive (VAR)
Pure Capacitor	0 (leading)	0	Negative (VAR)
RL Circuit	Between 0 and 1 (lagging)	I²R	I²XL
RC Circuit	Between 0 and 1 (leading)	I²R	I²XC
RLC at Resonance	1 (unity)	Maximum	0

🎯 26. POWER FACTOR (PF)

Power Factor is the ratio of Real Power to Apparent Power.

Power Factor = cos φ = P / S = R / Z

Unity PF (1): P = S, no reactive power (pure resistor)
Lagging PF (0 to 1): Current lags voltage (inductive load — motors, transformers)
Leading PF (0 to 1): Current leads voltage (capacitive load — capacitor banks, synchronous motors)

⚠️ Importance of High Power Factor:
• Reduces line losses (I²R)
• Improves voltage regulation
• Reduces kVA demand (lower electricity bills)
• Increases system capacity

🔧 27. POWER FACTOR IMPROVEMENT

Low power factor is usually caused by inductive loads (motors, transformers, induction furnaces).

Methods to improve power factor:

Capacitor Banks: Connected in parallel with inductive load (most common)
Synchronous Condensers: Over-excited synchronous motors
Phase Advancers: Used for induction motors

Required capacitor size: Qc = P × (tan φ₁ - tan φ₂)
Where φ₁ = initial angle, φ₂ = desired angle

🔍 Example: Load P = 100kW, PF = 0.6 lagging (φ₁ = 53.13°, tan φ₁ = 1.333)
Desired PF = 0.9 lagging (φ₂ = 25.84°, tan φ₂ = 0.484)
Qc = 100 × (1.333 - 0.484) = 100 × 0.849 = 84.9 kVAR

💰 28. POWER FACTOR TARIFF & PENALTIES

Electricity companies impose penalties for low power factor because:

More current required for same real power
Increased line losses
Reduced system capacity

Power Factor Range	Typical Penalty/Benefit
Above 0.95	Rebate / Incentive
0.90 to 0.95	No penalty / Standard rate
0.85 to 0.90	Small penalty
Below 0.85	High penalty / Surcharge

💡 Qatar Kahramaa Requirement: Power factor must be maintained between 0.9 lag to unity. PF correction mandatory for commercial loads above 210kW.

🧮 30. NUMERICAL EXAMPLES

Example 1: A 230V, 50Hz AC supply feeds a load taking 10A at 0.8 lagging PF.
Find P, Q, S.

Solution:
S = V × I = 230 × 10 = 2300 VA
P = S × cos φ = 2300 × 0.8 = 1840 W
Q = S × sin φ = 2300 × 0.6 = 1380 VAR

Example 2: A motor draws 5kW at 0.7 lagging PF. What kVAR capacitor is needed to improve PF to 0.95?

Solution:
cos φ₁ = 0.7 → φ₁ = 45.57°, tan φ₁ = 1.020
cos φ₂ = 0.95 → φ₂ = 18.19°, tan φ₂ = 0.329
Qc = P × (tan φ₁ - tan φ₂) = 5 × (1.020 - 0.329) = 5 × 0.691 = 3.455 kVAR

Example 3: A 3-phase motor, 415V, 20A, PF 0.85. Find total power.

Solution:
P = √3 × VL × IL × cos φ = 1.732 × 415 × 20 × 0.85
P = 1.732 × 415 = 718.78 × 20 = 14375.6 × 0.85 = 12.22 kW