NCERT CLASS 12 PHYSICS

CHAPTER 7 – ALTERNATING CURRENT

PART 1: IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is alternating current?

Alternating current (AC) is an electric current whose magnitude and direction change periodically with time.

In an alternating current circuit, the current flows in one direction during one half-cycle and in the opposite direction during the next half-cycle.

2. What is direct current?

Direct current (DC) is an electric current that flows in only one direction. Its magnitude may remain constant or vary with time, but its direction does not change.

3. Write the mathematical expression for alternating voltage.

The instantaneous value of alternating voltage is given by:

v = Vₘ sin ωt

Where:

v = Instantaneous voltage
Vₘ = Peak or maximum voltage
ω = Angular frequency
t = Time

4. What is the peak value of alternating current?

The peak value of alternating current is the maximum value attained by the current during a complete cycle.

It is represented by Iₘ.

5. What is the time period of alternating current?

The time period is the time taken by alternating current to complete one full cycle.

It is represented by T.

T = 1/f

Where f is the frequency of alternating current.

6. What is the frequency of alternating current?

Frequency is the number of complete cycles of alternating current produced in one second.

Its SI unit is hertz (Hz).

f = 1/T

7. What is angular frequency?

Angular frequency is the rate of change of phase of an alternating quantity.

It is given by:

ω = 2πf

Its SI unit is radian per second (rad s⁻¹).

8. What is the RMS value of alternating current?

The root mean square (RMS) value of alternating current is the value of steady direct current that produces the same amount of heat in a given resistance in the same time.

For sinusoidal alternating current:

Iᵣₘₛ = Iₘ/√2

Iᵣₘₛ = 0.707 Iₘ

9. What is the RMS value of alternating voltage?

The RMS value of alternating voltage is the effective value of AC voltage that produces the same heating effect as an equivalent DC voltage.

For sinusoidal alternating voltage:

Vᵣₘₛ = Vₘ/√2

Vᵣₘₛ = 0.707 Vₘ

10. Why is the average value of alternating current zero over one complete cycle?

The average value of sinusoidal alternating current over one complete cycle is zero because the positive and negative half-cycles are equal in magnitude and opposite in direction.

Therefore, they cancel each other.

11. Why is the RMS value used instead of the average value of alternating current?

The average value of alternating current over a complete cycle is zero. However, AC produces heating in a resistance.

Therefore, the RMS value is used because it represents the effective value of AC that produces the same heating effect as DC.

12. What is the phase difference between voltage and current in a pure resistive AC circuit?

In a pure resistive circuit, voltage and current are in the same phase.

The phase difference is:

φ = 0°

13. What happens when an AC voltage is applied to a pure resistor?

When an AC voltage is applied to a pure resistor, an alternating current flows through it.

According to Ohm's law:

i = v/R

Both voltage and current reach their maximum, minimum and zero values at the same instant.

Therefore, they are in phase.

14. What is inductive reactance?

Inductive reactance is the opposition offered by an inductor to the flow of alternating current.

It is represented by Xₗ.

Xₗ = ωL = 2πfL

Its SI unit is ohm (Ω).

15. What is the phase relationship between voltage and current in a pure inductive circuit?

In a pure inductive circuit, current lags behind voltage by 90° or π/2 radians.

In other words, voltage leads current by 90°.

16. Why does an inductor oppose a change in current?

An inductor opposes a change in current because a changing current produces a changing magnetic field, which induces an opposing emf in the coil.

This phenomenon is explained by Lenz's law.

17. How does inductive reactance depend on frequency?

Inductive reactance is directly proportional to the frequency of alternating current.

Xₗ = 2πfL

Therefore, if frequency increases, inductive reactance also increases.

18. What is capacitive reactance?

Capacitive reactance is the opposition offered by a capacitor to the flow of alternating current.

It is represented by X꜀.

X꜀ = 1/ωC = 1/2πfC

Its SI unit is ohm (Ω).

19. What is the phase relationship between voltage and current in a pure capacitive circuit?

In a pure capacitive circuit, current leads voltage by 90° or π/2 radians.

20. How does capacitive reactance depend on frequency?

Capacitive reactance is inversely proportional to the frequency of alternating current.

X꜀ = 1/2πfC

Therefore, when frequency increases, capacitive reactance decreases.

21. Why does a capacitor block direct current but allow alternating current to flow?

When a capacitor is connected to a DC source, it charges up and eventually stops the flow of current.

When connected to an AC source, the voltage continuously changes direction and magnitude. The capacitor repeatedly charges and discharges, allowing alternating current to flow in the circuit.

22. What is a phasor?

A phasor is a rotating vector used to represent a sinusoidally varying quantity.

Its projection on a reference axis represents the instantaneous value of the alternating quantity.

Phasors help us understand the phase relationship between voltage and current.

23. What is a series LCR circuit?

A series LCR circuit is an AC circuit in which a resistor (R), inductor (L) and capacitor (C) are connected in series with an alternating voltage source.

The same current flows through all three components.

24. What is impedance?

Impedance is the total opposition offered by an AC circuit to the flow of alternating current.

It is represented by Z.

For a series LCR circuit:

Z = √[R² + (Xₗ − X꜀)²]

Its SI unit is ohm (Ω).

25. What is the phase angle in a series LCR circuit?

The phase angle is the angle between the applied voltage and the current in an AC circuit.

For a series LCR circuit:

tan φ = (Xₗ − X꜀)/R

The sign of the phase angle indicates whether the circuit behaves predominantly as an inductive or capacitive circuit.

26. What is electrical resonance in a series LCR circuit?

Electrical resonance is the condition in a series LCR circuit when inductive reactance becomes equal to capacitive reactance.

Xₗ = X꜀

At resonance:

Impedance is minimum.
Impedance becomes equal to resistance.
Current becomes maximum.
Voltage and current are in phase.
Power factor becomes unity.

27. What is resonant frequency?

The frequency at which a series LCR circuit becomes resonant is called its resonant frequency.

It is given by:

f₀ = 1/(2π√LC)

28. What is the power factor of an AC circuit?

Power factor is the cosine of the phase angle between alternating voltage and current.

Power factor = cos φ

For a series LCR circuit:

cos φ = R/Z

29. What is wattless current?

Wattless current is the alternating current that flows in a circuit without consuming any average power over a complete cycle.

It occurs in a purely inductive or purely capacitive circuit because the phase difference between voltage and current is 90°.

P = VI cos 90° = 0

30. Why is the average power consumed in a pure inductor zero?

In a pure inductive circuit, current lags behind voltage by 90°.

Average power is:

P = VI cos φ

P = VI cos 90°

P = 0

The inductor stores energy in its magnetic field during one part of the cycle and returns it to the source during another part.

31. Why is the average power consumed in a pure capacitor zero?

In a pure capacitive circuit, current leads voltage by 90°.

Average power is:

P = VI cos 90° = 0

The capacitor stores energy in its electric field and returns it to the source during the cycle.

32. What is a transformer?

A transformer is an electrical device used to increase or decrease alternating voltage without changing its frequency.

It works on the principle of mutual induction.

33. What is mutual induction?

Mutual induction is the phenomenon in which a change in current in one coil produces a changing magnetic flux that induces an emf in another nearby coil.

34. What is a step-up transformer?

A step-up transformer is a transformer that increases the alternating voltage from the primary coil to the secondary coil.

In a step-up transformer:

Nₛ > Nₚ

Vₛ > Vₚ

35. What is a step-down transformer?

A step-down transformer is a transformer that decreases the alternating voltage from the primary coil to the secondary coil.

In a step-down transformer:

Nₛ < Nₚ

Vₛ < Vₚ

36. Why are transformers used in the transmission of electrical energy?

Transformers are used to increase voltage at power stations for long-distance transmission.

High voltage allows the same power to be transmitted at a lower current, reducing energy loss due to heating in transmission wires.

At the receiving end, step-down transformers reduce the voltage to suitable levels.

37. Why does a transformer not work on direct current?

A transformer works on the principle of mutual induction, which requires changing magnetic flux.

Direct current produces a constant magnetic field after the initial switching period. Therefore, it does not produce continuous induced emf in the secondary coil.

38. What are the main energy losses in a practical transformer?

The main energy losses in a practical transformer are:

Copper loss: Due to resistance of the windings.
Eddy current loss: Due to induced currents in the iron core.
Hysteresis loss: Due to repeated magnetisation and demagnetisation of the core.
Flux leakage: Due to some magnetic flux not linking both coils.

These losses are reduced through suitable design and materials.

PART 2: NCERT EXERCISE QUESTIONS AND ANSWERS

QUESTION 7.1

A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply.

(a) What is the RMS value of current in the circuit?

(b) What is the net power consumed over a full cycle?

Solution:

Given:

Resistance, R = 100 Ω

RMS voltage, V = 220 V

Frequency, f = 50 Hz

(a) RMS value of current

According to Ohm's law:

I = V/R

I = 220/100

I = 2.2 A

(b) Net power consumed

For a pure resistor:

P = VI

P = 220 × 2.2

P = 484 W

Final Answer:

(a) RMS current = 2.2 A

(b) Net power consumed = 484 W

QUESTION 7.2

(a) The peak voltage of an AC supply is 300 V. What is the RMS voltage?

(b) The RMS value of current in an AC circuit is 10 A. What is the peak current?

Solution:

(a) RMS voltage

Given:

Peak voltage, Vₘ = 300 V

Formula:

Vᵣₘₛ = Vₘ/√2

Substituting:

Vᵣₘₛ = 300/1.414

Vᵣₘₛ = 212.1 V

(b) Peak current

Given:

RMS current, Iᵣₘₛ = 10 A

Formula:

Iₘ = √2 Iᵣₘₛ

Substituting:

Iₘ = 1.414 × 10

Iₘ = 14.14 A

Final Answer:

(a) RMS voltage = 212.1 V

(b) Peak current = 14.14 A

QUESTION 7.3

A 44 mH inductor is connected to 220 V, 50 Hz AC supply. Determine the RMS value of the current in the circuit.

Solution:

Given:

Inductance, L = 44 mH

L = 44 × 10⁻³ H

RMS voltage, V = 220 V

Frequency, f = 50 Hz

Step 1: Calculate inductive reactance.

Formula:

Xₗ = 2πfL

Substituting:

Xₗ = 2 × 3.14 × 50 × 44 × 10⁻³

Xₗ ≈ 13.82 Ω

Step 2: Calculate RMS current.

I = V/Xₗ

I = 220/13.82

I ≈ 15.9 A

Final Answer:

RMS current = 15.9 A

QUESTION 7.4

A 60 μF capacitor is connected to a 110 V, 60 Hz AC supply. Determine the RMS value of the current in the circuit.

Solution:

Given:

Capacitance, C = 60 μF

C = 60 × 10⁻⁶ F

RMS voltage, V = 110 V

Frequency, f = 60 Hz

Step 1: Calculate capacitive reactance.

Formula:

X꜀ = 1/2πfC

Substituting:

X꜀ = 1/[2 × 3.14 × 60 × 60 × 10⁻⁶]

X꜀ ≈ 44.2 Ω

Step 2: Calculate RMS current.

I = V/X꜀

I = 110/44.2

I ≈ 2.49 A

Final Answer:

RMS current = 2.49 A

QUESTION 7.5

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Solution:

The average power consumed in an AC circuit is given by:

P = VI cos φ

Case 1: Pure Inductive Circuit

In a pure inductive circuit, current lags voltage by 90°.

Therefore:

φ = 90°

P = VI cos 90°

P = 0 W

Case 2: Pure Capacitive Circuit

In a pure capacitive circuit, current leads voltage by 90°.

Therefore:

φ = 90° in magnitude

P = VI cos 90°

P = 0 W

Explanation:

In both circuits, energy is stored temporarily in the magnetic field of the inductor or the electric field of the capacitor and is returned to the source.

Therefore, the net power absorbed over one complete cycle is zero.

Final Answer:

Net power absorbed by the inductor = 0 W

Net power absorbed by the capacitor = 0 W

QUESTION 7.6

A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Solution:

Given:

Capacitance, C = 30 μF

C = 30 × 10⁻⁶ F

Inductance, L = 27 mH

L = 27 × 10⁻³ H

Formula:

Angular frequency of free oscillations:

ω₀ = 1/√LC

Substituting:

ω₀ = 1/√[(27 × 10⁻³) × (30 × 10⁻⁶)]

ω₀ = 1/√(810 × 10⁻⁹)

ω₀ = 1/√(8.1 × 10⁻⁷)

ω₀ ≈ 1.11 × 10³ rad/s

Final Answer:

Angular frequency = 1.11 × 10³ rad/s

QUESTION 7.7

A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V AC supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Solution:

Given:

Resistance, R = 20 Ω

Inductance, L = 1.5 H

Capacitance, C = 35 μF

RMS voltage, V = 200 V

At resonance:

Xₗ = X꜀

Therefore, impedance:

Z = R

Z = 20 Ω

Step 1: Calculate RMS current.

I = V/Z

I = 200/20

I = 10 A

Step 2: Calculate average power.

At resonance, voltage and current are in phase.

Therefore:

cos φ = 1

P = VI cos φ

P = 200 × 10 × 1

P = 2000 W

Final Answer:

Average power transferred = 2000 W or 2 kW

QUESTION 7.8

A series LCR circuit is connected to a variable-frequency 230 V source with L = 5.0 H, C = 80 μF and R = 40 Ω.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

(c) Determine the RMS potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Solution:

Given:

R = 40 Ω

L = 5.0 H

C = 80 μF = 80 × 10⁻⁶ F

RMS voltage, V = 230 V

(a) Calculate resonant frequency.

Formula:

ω₀ = 1/√LC

Substituting:

ω₀ = 1/√[5 × 80 × 10⁻⁶]

ω₀ = 1/√(400 × 10⁻⁶)

ω₀ = 1/0.02

ω₀ = 50 rad/s

Now,

f₀ = ω₀/2π

f₀ = 50/(2 × 3.14)

f₀ ≈ 7.96 Hz

(b) Calculate impedance and amplitude of current.

At resonance:

Xₗ = X꜀

Therefore:

Z = R

Z = 40 Ω

RMS current:

I = V/Z

I = 230/40

I = 5.75 A

Peak or amplitude of current:

Iₘ = √2 I

Iₘ = 1.414 × 5.75

Iₘ ≈ 8.13 A

(c) Calculate RMS potential drops.

Potential drop across resistor:

Vᵣ = IR

Vᵣ = 5.75 × 40

Vᵣ = 230 V

Potential drop across inductor:

First, calculate inductive reactance.

Xₗ = ω₀L

Xₗ = 50 × 5

Xₗ = 250 Ω

Therefore:

Vₗ = IXₗ

Vₗ = 5.75 × 250

Vₗ = 1437.5 V

Potential drop across capacitor:

X꜀ = 1/ω₀C

X꜀ = 1/[50 × 80 × 10⁻⁶]

X꜀ = 250 Ω

Therefore:

V꜀ = IX꜀

V꜀ = 5.75 × 250

V꜀ = 1437.5 V

Show that potential drop across LC combination is zero.

At resonance, the voltage across the inductor and capacitor are equal in magnitude but opposite in phase.

Therefore:

Vₗ꜀ = Vₗ − V꜀

Vₗ꜀ = 1437.5 − 1437.5

Vₗ꜀ = 0 V

Final Answer:

(a) Resonant frequency = 7.96 Hz

(b) Impedance = 40 Ω; peak current = 8.13 A

Net potential drop across LC combination = 0 V

PART 3: IMPORTANT DERIVATIONS

1. Derive the expression for RMS value of alternating current.

Let the instantaneous alternating current be:

i = Iₘ sin ωt

The RMS value is defined as:

Iᵣₘₛ = √[mean of i² over one complete cycle]

Therefore:

Iᵣₘₛ = √[(1/T) ∫₀ᵀ i² dt]

Substituting i = Iₘ sin ωt:

Iᵣₘₛ = √[(Iₘ²/T) ∫₀ᵀ sin²ωt dt]

Using the average value of sin²ωt over one complete cycle:

Average of sin²ωt = 1/2

Therefore:

Iᵣₘₛ = √(Iₘ²/2)

Iᵣₘₛ = Iₘ/√2

Hence proved.

2. Derive the expression for RMS voltage.

Let the instantaneous voltage be:

v = Vₘ sin ωt

The RMS voltage is:

Vᵣₘₛ = √[(1/T) ∫₀ᵀ v² dt]

Substituting:

Vᵣₘₛ = √[(Vₘ²/T) ∫₀ᵀ sin²ωt dt]

Since the average value of sin²ωt over one complete cycle is 1/2:

Vᵣₘₛ = √(Vₘ²/2)

Vᵣₘₛ = Vₘ/√2

Hence proved.

3. Derive the expression for inductive reactance.

For a pure inductor connected to an AC source:

v = Vₘ sin ωt

The voltage across an inductor is:

v = L(di/dt)

Therefore:

L(di/dt) = Vₘ sin ωt

di/dt = (Vₘ/L) sin ωt

Integrating:

i = −Vₘ/(ωL) cos ωt

Using:

−cos ωt = sin(ωt − π/2)

Therefore:

i = [Vₘ/(ωL)] sin(ωt − π/2)

Comparing with:

i = Iₘ sin(ωt − π/2)

We get:

Iₘ = Vₘ/(ωL)

Therefore:

Vₘ/Iₘ = ωL

The inductive reactance is:

Xₗ = ωL = 2πfL

Hence proved.

4. Derive the expression for capacitive reactance.

For a capacitor connected to an AC source:

v = Vₘ sin ωt

Charge on the capacitor:

q = Cv

q = CVₘ sin ωt

Current is:

i = dq/dt

Differentiating:

i = ωCVₘ cos ωt

Using:

cos ωt = sin(ωt + π/2)

Therefore:

i = ωCVₘ sin(ωt + π/2)

Comparing with:

i = Iₘ sin(ωt + π/2)

We get:

Iₘ = ωCVₘ

Therefore:

Vₘ/Iₘ = 1/ωC

The capacitive reactance is:

X꜀ = 1/ωC = 1/2πfC

Hence proved.

5. Derive the expression for impedance in a series LCR circuit.

In a series LCR circuit, the same current flows through the resistor, inductor and capacitor.

Let:

Vᵣ = IR

Vₗ = IXₗ

V꜀ = IX꜀

The voltage across the resistor is in phase with current.

The voltage across the inductor leads current by 90°, while the voltage across the capacitor lags current by 90°.

Using phasor addition:

V² = Vᵣ² + (Vₗ − V꜀)²

Substituting:

V² = (IR)² + (IXₗ − IX꜀)²

V² = I²[R² + (Xₗ − X꜀)²]

Dividing by I²:

(V/I)² = R² + (Xₗ − X꜀)²

Since impedance Z = V/I:

Z = √[R² + (Xₗ − X꜀)²]

Hence proved.

6. Derive the expression for resonant frequency in a series LCR circuit.

At resonance:

Xₗ = X꜀

We know:

Xₗ = ωL

X꜀ = 1/ωC

Therefore:

ωL = 1/ωC

ω²LC = 1

ω² = 1/LC

Therefore:

ω₀ = 1/√LC

Since:

ω₀ = 2πf₀

Therefore:

f₀ = 1/(2π√LC)

Hence proved.

7. Derive the expression for average power in an AC circuit.

Let the alternating voltage be:

v = Vₘ sin ωt

Let the alternating current be:

i = Iₘ sin(ωt − φ)

Instantaneous power:

p = vi

p = VₘIₘ sin ωt sin(ωt − φ)

Using the trigonometric identity:

2 sin A sin B = cos(A − B) − cos(A + B)

We get:

p = (VₘIₘ/2)[cos φ − cos(2ωt − φ)]

The average value of cos(2ωt − φ) over a complete cycle is zero.

Therefore:

P = (VₘIₘ/2) cos φ

Since:

V = Vₘ/√2 and I = Iₘ/√2

Therefore:

P = VI cos φ

Hence proved.

8. Derive the transformer equation.

A transformer works on the principle of mutual induction.

Let:

Nₚ = Number of turns in primary coil

Nₛ = Number of turns in secondary coil

Φ = Magnetic flux linked with each turn

According to Faraday's law:

Induced emf in primary:

Eₚ = −Nₚ(dΦ/dt)

Induced emf in secondary:

Eₛ = −Nₛ(dΦ/dt)

Dividing:

Eₛ/Eₚ = Nₛ/Nₚ

For an ideal transformer:

Vₛ/Vₚ = Nₛ/Nₚ

Therefore:

Vₛ/Vₚ = Nₛ/Nₚ

For an ideal transformer, input power equals output power:

VₚIₚ = VₛIₛ

Therefore:

Iₛ/Iₚ = Nₚ/Nₛ

Hence proved.

PART 4: IMPORTANT NUMERICAL QUESTIONS

1. An alternating current has a peak value of 20 A. Calculate its RMS value.

Given:

Iₘ = 20 A

Formula:

Iᵣₘₛ = Iₘ/√2

Iᵣₘₛ = 20/1.414

Iᵣₘₛ = 14.14 A

2. An AC voltage has an RMS value of 220 V. Calculate its peak value.

Given:

Vᵣₘₛ = 220 V

Formula:

Vₘ = √2 Vᵣₘₛ

Vₘ = 1.414 × 220

Vₘ = 311.1 V

3. Calculate the inductive reactance of a 0.5 H inductor connected to a 50 Hz AC supply.

Given:

L = 0.5 H

f = 50 Hz

Formula:

Xₗ = 2πfL

Xₗ = 2 × 3.14 × 50 × 0.5

Xₗ = 157 Ω

4. Calculate the capacitive reactance of a 100 μF capacitor connected to a 50 Hz AC supply.

Given:

C = 100 × 10⁻⁶ F

f = 50 Hz

Formula:

X꜀ = 1/2πfC

X꜀ = 1/[2 × 3.14 × 50 × 100 × 10⁻⁶]

X꜀ ≈ 31.85 Ω

5. A series LCR circuit has resistance 30 Ω, inductive reactance 40 Ω and capacitive reactance 20 Ω. Calculate its impedance.

Given:

R = 30 Ω

Xₗ = 40 Ω

X꜀ = 20 Ω

Formula:

Z = √[R² + (Xₗ − X꜀)²]

Z = √[30² + (40 − 20)²]

Z = √[900 + 400]

Z = √1300

Z ≈ 36.06 Ω

6. A transformer has 500 turns in its primary coil and 1000 turns in its secondary coil. If the primary voltage is 220 V, calculate the secondary voltage.

Given:

Nₚ = 500

Nₛ = 1000

Vₚ = 220 V

Formula:

Vₛ/Vₚ = Nₛ/Nₚ

Vₛ/220 = 1000/500

Vₛ = 220 × 2

Vₛ = 440 V

The transformer is a step-up transformer.

PART 5: IMPORTANT FORMULA SHEET

Alternating Current

Instantaneous voltage:
v = Vₘ sin ωt
Instantaneous current:
i = Iₘ sin ωt
Angular frequency:
ω = 2πf
Frequency:
f = 1/T
Time period:
T = 1/f

RMS Values

RMS current:
Iᵣₘₛ = Iₘ/√2
RMS voltage:
Vᵣₘₛ = Vₘ/√2

Pure Resistive Circuit

Ohm's law:
I = V/R
Average power:
P = VI = I²R = V²/R

Pure Inductive Circuit

Inductive reactance:
Xₗ = ωL = 2πfL
RMS current:
I = V/Xₗ
Phase difference:
φ = 90° in magnitude

Pure Capacitive Circuit

Capacitive reactance:
X꜀ = 1/ωC = 1/2πfC
RMS current:
I = V/X꜀
Phase difference:
φ = 90° in magnitude

Series LCR Circuit

Impedance:
Z = √[R² + (Xₗ − X꜀)²]
Current:
I = V/Z
Phase angle:
tan φ = (Xₗ − X꜀)/R
Power factor:
cos φ = R/Z
Resonant angular frequency:
ω₀ = 1/√LC
Resonant frequency:
f₀ = 1/(2π√LC)
Impedance at resonance:
Z = R
Current at resonance:
I = V/R

AC Power

Average power:
P = VI cos φ
Wattless current condition:
cos φ = 0

Transformer

Voltage transformation ratio:
Vₛ/Vₚ = Nₛ/Nₚ
Current transformation ratio:
Iₛ/Iₚ = Nₚ/Nₛ
Ideal transformer power:
VₚIₚ = VₛIₛ

PART 6: IMPORTANT SI UNITS

Physical Quantity	SI Unit
Current	Ampere (A)
Voltage	Volt (V)
Resistance	Ohm (Ω)
Impedance	Ohm (Ω)
Inductive reactance	Ohm (Ω)
Capacitive reactance	Ohm (Ω)
Inductance	Henry (H)
Capacitance	Farad (F)
Frequency	Hertz (Hz)
Angular frequency	radian per second (rad/s)
Time period	Second (s)
Power	Watt (W)
Magnetic flux	Weber (Wb)
Phase angle	Radian (rad) or degree (°)

PART 7: QUICK REVISION POINTS

Alternating current changes its magnitude and direction periodically.
The RMS value of sinusoidal current is 0.707 times its peak value.
In a pure resistor, voltage and current are in phase.
In a pure inductor, current lags voltage by 90°.
In a pure capacitor, current leads voltage by 90°.
Inductive reactance increases with frequency.
Capacitive reactance decreases with frequency.
Impedance is the total opposition offered by an AC circuit.
At resonance, inductive reactance equals capacitive reactance.
At resonance, impedance is minimum and current is maximum in a series LCR circuit.
The power factor of a pure resistive circuit is one.
The average power consumed by a pure inductor or capacitor is zero.
A transformer works on the principle of mutual induction.
A step-up transformer increases voltage and decreases current in an ideal transformer.
A step-down transformer decreases voltage and increases current in an ideal transformer.
A transformer cannot continuously operate on direct current because it requires changing magnetic flux.