Alternating Current (AC): Peak, RMS, and average values of AC voltage and current

📌 1. Question Description

⚡ Alternating Current

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

For a sinusoidal AC voltage:

v = V₀ sin(ωt)

For sinusoidal AC current:

i = I₀ sin(ωt)

Here V₀ and I₀ are the peak values of voltage and current.

🔵 2. Important Terms

Peak Value

The maximum instantaneous value attained by AC voltage or current.

V₀ = Peak Voltage

I₀ = Peak Current

RMS Value

The RMS value of AC is the equivalent DC value producing the same heating effect in a resistor.

V_rms = V₀/√2

I_rms = I₀/√2

Average Value

For a complete sinusoidal cycle, positive and negative halves cancel. Therefore:

V_avg = 0

I_avg = 0

📈 3. Animated AC Waveform

Peak Voltage V₀ = 220 V

Frequency f = 50 Hz

🔵 Instantaneous AC voltage | 🟡 Peak value | 🟢 RMS level

📐 4. Peak Value of AC

For sinusoidal alternating voltage:

v = V₀ sin(ωt)

The maximum value occurs when:

sin(ωt) = 1

Therefore:

V_max = V₀

Similarly, for current:

I_max = I₀

🧮 5. RMS Value — Derivation

Consider a sinusoidal current:

i = I₀ sin(ωt)

The mean square current over one complete cycle is:

I²_rms = Average(i²)

Therefore:

I²_rms = I₀² Average(sin²ωt)

For a complete cycle:

Average(sin²ωt) = 1/2

Hence:

I²_rms = I₀²/2

I_rms = I₀/√2

Similarly:

V_rms = V₀/√2

⭐ For sinusoidal AC:

RMS value = 0.707 × Peak value

🟠 6. Average Value of AC

For a sinusoidal AC, the average value over one complete cycle is zero because the positive and negative halves cancel each other.

I_{avg, complete cycle} = 0

V_{avg, complete cycle} = 0

For the positive half cycle:

I_avg = 2I₀/π

Similarly:

V_avg = 2V₀/π

✔ Half-cycle average = 0.637 × Peak value

📊 7. Peak, RMS & Average — Comparison

Quantity	Formula	For 220 V AC
Peak Voltage	V₀	≈ 311 V
RMS Voltage	V₀/√2	220 V
Half-cycle Average	2V₀/π	≈ 198 V
Complete-cycle Average	0	0 V

🧪 8. Practical Demonstration — AC Measurement

Experiment: Measuring AC Voltage

Apparatus:

• AC source/function generator
• AC voltmeter or multimeter
• Oscilloscope
• Connecting wires

Procedure:

1. Connect the AC source to the measuring instrument.
2. Observe the sinusoidal waveform on an oscilloscope.
3. Measure the RMS voltage using an AC voltmeter.
4. Determine the peak voltage from the waveform.

If the measured RMS voltage is 220 V:

V₀ = √2 × 220 ≈ 311 V

📚 9. Important Formulae

v = V₀ sin(ωt)

i = I₀ sin(ωt)

ω = 2πf

T = 1/f

V_rms = V₀/√2

I_rms = I₀/√2

V_{avg, half} = 2V₀/π

I_{avg, half} = 2I₀/π

V_{avg, complete} = 0

I_{avg, complete} = 0

📝 10. MCQ Practice — 15 Questions

1. The maximum value of an AC voltage is called:

A. RMS value

B. Average value

C. Peak value

D. Effective value

✔ Answer: C

2. RMS value of sinusoidal AC voltage is:

A. V₀

B. V₀/2

C. V₀/√2

D. 2V₀

✔ Answer: C

3. RMS value of 100 V peak AC is:

A. 50 V

B. 70.7 V

C. 100 V

D. 141.4 V

✔ Answer: B

4. Average value of sinusoidal AC over a complete cycle is:

A. V₀

B. V₀/√2

C. 2V₀/π

D. 0

✔ Answer: D

5. Average value over a positive half-cycle is:

A. V₀/2

B. 2V₀/π

C. V₀/√2

D. Zero

✔ Answer: B

6. For a sinusoidal AC, V₀/V_rms is:

A. 1

B. √2

C. 1/√2

D. 2

✔ Answer: B

7. The frequency of AC having time period 0.02 s is:

A. 20 Hz

B. 25 Hz

C. 50 Hz

D. 100 Hz

✔ Answer: C

8. The angular frequency of AC is:

A. ω = f/2π

B. ω = 2πf

C. ω = 1/f

D. ω = π/f

✔ Answer: B

9. 220 V AC refers to its:

A. Peak value

B. RMS value

C. Average value

D. Instantaneous value

✔ Answer: B

10. Peak value corresponding to 220 V RMS is approximately:

A. 155.6 V

B. 220 V

C. 311 V

D. 440 V

✔ Answer: C

11. RMS value is also known as:

A. Effective value

B. Peak value

C. Maximum value

D. Instantaneous value

✔ Answer: A

12. The average value of AC over a complete cycle becomes zero because:

A. Current is constant

B. Positive and negative halves cancel

C. Frequency becomes zero

D. Voltage becomes infinite

✔ Answer: B

13. For sinusoidal AC, I_rms equals approximately:

A. 0.5I₀

B. 0.637I₀

C. 0.707I₀

D. 1.414I₀

✔ Answer: C

14. Which instrument normally indicates the RMS value of AC?

A. AC voltmeter

B. Compass

C. Galvanometer only

D. Potentiometer

✔ Answer: A

15. If peak current is 10 A, RMS current is:

A. 5 A

B. 7.07 A

C. 10 A

D. 14.14 A

✔ Answer: B

🟣 11. Assertion–Reason — 5 Questions

Options:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is not the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.

Assertion: The average value of sinusoidal AC over a complete cycle is zero.

Reason: The positive and negative half-cycles cancel each other.

Answer: A

Assertion: RMS value of sinusoidal AC is less than its peak value.

Reason: V_rms = V₀/√2.

Answer: A

Assertion: 220 V AC has a peak voltage of approximately 311 V.

Reason: V₀ = √2 V_rms.

Answer: A

Assertion: The average value of AC is not always zero.

Reason: Average value over a half-cycle is 2V₀/π.

Answer: A

Assertion: RMS value is called the effective value of AC.

Reason: It gives the equivalent DC value for the same heating effect in a resistor.

Answer: A

🟢 12. 2 Marks — 6 Questions

Q1 Define alternating current. Write its mathematical expression.

Q2 Define peak value of AC voltage and current.

Q3 Define RMS value of alternating current.

Q4 Write the RMS value of sinusoidal AC voltage in terms of its peak value.

Q5 Why is the average value of sinusoidal AC over a complete cycle zero?

Q6 Write the relation between RMS value and peak value of AC current.

🟡 13. 3 Marks — 6 Questions

Q1 Explain peak value and RMS value of AC.

Q2 Derive the relation I_rms = I₀/√2.

Q3 Explain the average value of sinusoidal AC over a half-cycle.

Q4 Distinguish between peak value, RMS value and average value.

Q5 An AC current has a peak value of 10 A. Calculate its RMS value.

Q6 An AC voltage is represented by v = 311 sin(100πt). Find peak voltage, RMS voltage and frequency.

🟠 14. 4 Marks — 6 Questions

Q1 Derive the expression for RMS value of a sinusoidal AC current.

Q2 Derive the average value of sinusoidal AC over a positive half-cycle.

Q3 Explain peak, RMS and average values of AC with suitable equations.

Q4 For v = 200 sin(100πt), calculate peak voltage, RMS voltage, frequency and time period.

Q5 An AC current has RMS value 5 A. Find its peak current and average current over a positive half-cycle.

Q6 Explain why an AC voltmeter measures RMS value rather than the peak value.

🔴 15. 5 Marks — 6 Questions

Q1 Derive the RMS value of sinusoidal AC voltage from its instantaneous value.

Q2 Derive the average value of sinusoidal AC over a half-cycle.

Q3 Explain the physical significance of RMS value of AC.

Q4 Compare peak, RMS and average values of sinusoidal AC and derive their mutual relations.

Q5 For a 220 V AC supply, calculate peak voltage and average voltage over a positive half-cycle.

Q6 An alternating current is given by i = 20 sin(100πt). Determine peak current, RMS current, frequency, time period and half-cycle average current.

🔵 16. 6 Marks — 6 Questions

Q1 Starting from i = I₀ sin(ωt), derive the RMS value of sinusoidal alternating current in detail.

Q2 Derive the average value of sinusoidal AC over a positive half-cycle and explain why the complete-cycle average is zero.

Q3 Explain peak value, RMS value and average value of AC voltage and current with equations, graphs and physical significance.

Q4 For v = 311 sin(100πt), calculate peak voltage, RMS voltage, frequency, time period and positive half-cycle average voltage.

Q5 A 220 V, 50 Hz AC source is connected to a resistor. Explain the meaning of 220 V and calculate its peak voltage.

Q6 Derive and compare the relations V_rms = V₀/√2 and V_avg = 2V₀/π for sinusoidal AC.

🧮 17. Numerical Practice — 6 Questions

Q1. An AC voltage has peak value 311 V. Calculate its RMS value.

Q2. An AC current has RMS value 5 A. Calculate its peak value.

Q3. The instantaneous voltage is v = 100 sin(200πt). Find peak voltage, frequency and time period.

Q4. Calculate the average voltage over a positive half-cycle for an AC having peak voltage 200 V.

Q5. A 220 V, 50 Hz AC supply is given to a circuit. Find its peak voltage and angular frequency.

Q6. A sinusoidal AC current has peak value 14.14 A. Calculate its RMS value and positive half-cycle average value.

🚀 18. Quick Revision

Instantaneous Voltage

v = V₀ sin(ωt)

Instantaneous Current

i = I₀ sin(ωt)

Peak → RMS

V_rms = 0.707V₀

Peak → Half-Cycle Average

V_avg = 0.637V₀

Complete Cycle

V_avg = 0

Angular Frequency

ω = 2πf