average and rms value of ac waveform

Alternating Current | CBSE | NEET | JEE Main | JEE Advanced

Average Value and RMS Value of AC Waveforms

This section explains average value, RMS value, peak value, peak-to-peak value and heating effect of AC waveforms in a clean exam-oriented style. Students will learn sinusoidal AC, rectified waves, square waves and triangular waves with formulas, diagrams and solved questions.

Average Value RMS Value Waveforms Result Table Questions Guidance

1. Average Value of AC

Definition

Average value of an AC waveform is the value of DC which would produce the same average effect over a given interval.

For complete cycle: Vavg = (1/T) ∫0T v(t) dt Angular form: Vavg = (1/2π) ∫02π v(θ) dθ

For Pure Sine Wave

For a sinusoidal AC waveform:

v = V0 sin θ Vavg = (1/2π) ∫02π V0 sin θ dθ = 0

Average value over a full cycle of pure sinusoidal AC is zero because the positive half cycle and negative half cycle cancel each other.

Average Value Over Positive Half Cycle

Vavg = (1/π) ∫0π V0 sin θ dθ Vavg = 2V0/π Iavg = 2I0/π

2. RMS Value of AC in Terms of Heating Effect

Definition

RMS value of AC is the value of steady DC which produces the same heating effect in a resistor as the given AC over the same time.

Vrms = √[(1/T) ∫0T v2(t) dt] Vrms = √[(1/2π) ∫02π (V0 sin θ)2 dθ]

Final RMS Result for Sine Wave

Vrms = V0/√2 Irms = I0/√2

V0 is the peak value of voltage and I0 is the peak value of current. RMS value is also called the effective value of AC.

3. SVG Waveform Diagrams

Full Sine Wave

+V0 -V0 t v Vavg = 0, Vrms = V0/√2

Positive Half Sine Wave

V0 theta Vavg = 2V0/π, Vrms = V0/√2

Full Wave Rectified Sine

V0 Vavg = 2V0/π, Vrms = V0/√2

Half Wave Rectified Sine

V0 Vavg = V0/π, Vrms = V0/2

Square Wave +V0 and -V0

+V0 -V0 Vavg = 0, Vrms = V0

Unipolar Square Wave 0 to V0

V0 Vavg = V0/2, Vrms = V0/√2

Triangular Wave +V0 and -V0

+V0 -V0 Vavg = 0, Vrms = V0/√3

4. Result Table for Important AC Waveforms

Waveform Average Value RMS Value Exam Note
Full sine wave Vavg = 0 Vrms = V0/√2 Positive and negative halves cancel.
Positive half sine wave Vavg = 2V0/π Vrms = V0/√2 Used for half-cycle mean value.
Full wave rectified sine Vavg = 2V0/π Vrms = V0/√2 All negative halves are converted positive.
Half wave rectified sine Vavg = V0/π Vrms = V0/2 Current flows only for half cycle.
Square wave +V0 and -V0 Vavg = 0 Vrms = V0 Magnitude is always V0.
Unipolar square wave 0 to V0 Vavg = V0/2 Vrms = V0/√2 ON for half cycle, OFF for half cycle.
Triangular wave +V0 and -V0 Vavg = 0 Vrms = V0/√3 Linear rise and fall waveform.

5. Screenshot Question Section

Extracted Question

The instantaneous emf of an AC source is given by E = 300 sin 314t. What is the RMS value of the emf?

Step-by-Step Solution

Compare the given equation with the standard equation:

E = E0 sin ωt

Given peak value:

E0 = 300 V

For sinusoidal AC:

Erms = E0/√2

Substitution:

Erms = 300 / 1.414 = 212 V Final Answer: 212 V

6. Practice Questions with Answers

CBSE Class 12 Conceptual Questions

1. Why is average value of sinusoidal AC over a full cycle zero?
AnswerPositive and negative half cycles are equal and opposite, so their algebraic sum is zero.
2. Why is RMS value also called effective value?
AnswerBecause it gives the same heating effect as an equivalent DC value.
3. Does a household 230 V AC supply mean peak voltage?
AnswerNo. It is RMS voltage. Peak voltage is about 230√2 = 325 V.
4. Which value is measured by an AC voltmeter?
AnswerAn AC voltmeter generally reads RMS value.
5. What is peak-to-peak voltage?
AnswerPeak-to-peak voltage is the difference between maximum positive and maximum negative values. For sine wave it is 2V0.

CBSE Class 12 Numerical Questions

1. If V = 200 sin 100πt, find Vrms.
AnswerV0 = 200 V, so Vrms = 200/√2 = 141.4 V.
2. For I = 10 sin 314t, find Irms.
AnswerIrms = 10/√2 = 7.07 A.
3. Find peak voltage of 230 V AC supply.
AnswerV0 = √2 Vrms = 1.414 x 230 = 325 V.
4. Peak value of AC current is 14.14 A. Find RMS current.
AnswerIrms = 14.14/√2 = 10 A.
5. A half wave rectified sine has V0 = 100 V. Find Vavg and Vrms.
AnswerVavg = V0/π = 31.8 V, Vrms = V0/2 = 50 V.

NEET MCQs

1. RMS value of V = V0 sin ωt is:
A) V0 B) V0/2 C) V0/√2 D) 2V0/π
AnswerC) V0/√2.
2. Average value of full sinusoidal AC cycle is:
A) V0 B) 0 C) V0/√2 D) 2V0/π
AnswerB) 0.
3. Peak value of 220 V AC is nearly:
A) 110 V B) 220 V C) 311 V D) 440 V
AnswerC) 311 V.
4. An AC ammeter reads:
A) Peak value B) RMS value C) Average full-cycle value D) Peak-to-peak value
AnswerB) RMS value.
5. For square wave ±V0, RMS value is:
A) 0 B) V0/2 C) V0/√2 D) V0
AnswerD) V0.
6. For full-wave rectified sine, average value is:
A) 0 B) V0/π C) 2V0/π D) V0/√2
AnswerC) 2V0/π.
7. RMS value is linked with:
A) Magnetic effect only B) Heating effect C) Chemical effect only D) Capacitance only
AnswerB) Heating effect.
8. V = 100 sin 314t. Frequency is:
A) 25 Hz B) 50 Hz C) 100 Hz D) 314 Hz
AnswerB) 50 Hz because ω = 314 = 2πf.
9. Peak-to-peak value for V0 = 50 V is:
A) 25 V B) 50 V C) 70.7 V D) 100 V
AnswerD) 100 V.
10. For triangular wave ±V0, RMS value is:
A) V0 B) V0/2 C) V0/√3 D) V0/√2
AnswerC) V0/√3.

JEE Main MCQs

1. A sine voltage has peak-to-peak value 400 V. Its RMS value is:
A) 100 V B) 141 V C) 200 V D) 283 V
AnswerB) V0 = 200 V, Vrms = 200/√2 = 141 V.
2. For half-wave rectified sine with V0 = 20 V, Vrms is:
A) 5 V B) 10 V C) 14.14 V D) 20 V
AnswerB) Vrms = V0/2 = 10 V.
3. For full-wave rectified sine, which is correct?
A) Vavg=0 B) Vrms=V0 C) Vavg=2V0/pi D) Vrms=V0/2
AnswerC) Vavg = 2V0/π.
4. A 230 V AC is applied to 46 ohm resistor. RMS current is:
A) 2 A B) 3 A C) 5 A D) 7 A
AnswerC) Irms = Vrms/R = 230/46 = 5 A.
5. For unipolar square wave 0 to V0, average value is:
A) 0 B) V0/2 C) V0/radic2 D) V0
AnswerB) V0/2.
6. If Vavg over positive half cycle is 100 V for sine wave, peak voltage is:
A) 100 V B) 157 V C) 200 V D) 314 V
AnswerB) Vavg = 2V0/π, so V0 = 100π/2 = 157 V.
7. If RMS current is 4 A, peak current is:
A) 2 A B) 4 A C) 5.66 A D) 8 A
AnswerC) I0 = √2 Irms = 5.66 A.
8. Heating in resistor depends on:
A) Iavg B) Irms C) Ipeak-to-peak D) Frequency only
AnswerB) Irms.
9. V = 150 sin(200πt). Frequency is:
A) 50 Hz B) 75 Hz C) 100 Hz D) 200 Hz
AnswerC) 200π = 2πf, so f = 100 Hz.
10. RMS of bipolar square wave is greater than RMS of sine wave with same peak because:
A) Square wave has constant magnitude V0 B) Frequency is higher C) Average is zero D) Phase changes
AnswerA) Square wave remains at magnitude V0, so Vrms = V0.

JEE Advanced Challenging Questions

1. A waveform is V0 for half cycle and 0 for half cycle. Find Vavg and Vrms.
AnswerVavg = V0/2, Vrms = V0/√2.
2. A sine wave and square wave have same peak value V0. Which produces more heat in same resistor?
AnswerSquare wave. Sine RMS = V0/√2, square RMS = V0.
3. For triangular wave ±60 V, find RMS voltage.
AnswerVrms = V0/√3 = 60/1.732 = 34.6 V.
4. A full-wave rectified sine and original sine have same peak V0. Compare RMS values.
AnswerBoth have same RMS value V0/√2 because squaring removes sign.
5. A 325 V peak sine wave is connected to 100 ohm resistor. Find average power.
AnswerVrms = 325/√2 = 230 V. P = Vrms2/R = 2302/100 = 529 W.

7. Important Notes

Average value of a full sine wave is zero.
RMS value is related to heating effect.
Household AC voltage is RMS value, not peak value.
Peak voltage = √2 x Vrms.
For India, 230 V AC means RMS value.
Peak value of 230 V AC is approximately 325 V.
Average Value RMS Value Peak Value Peak-to-Peak Value Heating Effect Rectifier Waveforms

Need Help in AC Waveforms?

If average value, RMS value, heating effect, sine wave, square wave, triangular wave, half wave rectifier or full wave rectifier is not clear, you can contact Kumar Sir for one-to-one Physics classes.

Contact: +91-9958461445 Website: KumarPhysicsClasses.com
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Alternating Current | CBSE | NEET | JEE | IB | IGCSE | A-Level

Average Value and RMS Value of Alternating Current / Alternating Voltage

A complete classroom-style AC waveform guide by Kumar Sir covering average value, RMS value, peak value, peak-to-peak value, heating effect, waveform formulas, solved screenshot questions, exam questions and revision tables.

Screenshot Questions Theory Derivations Waveforms Exam Questions Revision Table

1. Questions With Complete Solutions

Question 1

The instantaneous emf of an AC source is given by E = 300 sin 314t. What is the RMS value of the emf?

Solution

Formula used: Erms = E0/√2

Given equation: E = 300 sin 314t. Comparing with E = E0 sin ωt, peak value E0 = 300 V.

Erms = 300/√2 = 300/1.414 = 212 V.

Final answer: 212 V

Common mistake: Students often write 300 V as RMS value, but 300 V is the peak value.

Question 2

The emf of an AC source is given by E = 300 sin 314t. Write the value of peak voltage and frequency of the source.

Solution

Formula used: E = E0 sin ωt and ω = 2πf

Peak voltage E0 = 300 V.

ω = 314 rad s-1. Therefore f = ω/2π = 314/(2 × 3.14) = 50 Hz.

Final answer: Peak voltage = 300 V, frequency = 50 Hz

Common mistake: Do not confuse angular frequency ω with frequency f.

Question 3

The instantaneous current from an AC source is I = 5 sin 314t. What is the RMS value of current?

Solution

Formula used: Irms = I0/√2

I0 = 5 A.

Irms = 5/√2 = 3.54 A.

Final answer: 3.54 A

Common mistake: Writing 5 A as meter reading. AC meters read RMS value.

Question 4

An alternating voltage V = 140 sin 314t is connected across a pure resistor of 50 Ω. Find (i) frequency of the source, (ii) RMS current through the resistor.

Solution

Formula used: ω = 2πf, Vrms = V0/√2, Irms = Vrms/R

ω = 314 rad s-1, so f = 314/(2 × 3.14) = 50 Hz.

V0 = 140 V, so Vrms = 140/√2 = 99 V.

Irms = 99/50 = 1.98 A.

Final answer: (i) 50 Hz, (ii) 1.98 A

Common mistake: Using peak voltage directly in Ohm's law for RMS current.

Question 5

An alternating emf of peak value 350 V is applied across an AC ammeter of resistance 100 Ω. What is the reading of the ammeter?

Solution

Formula used: Vrms = V0/√2, Irms = Vrms/R

V0 = 350 V.

Vrms = 350/√2 = 247.5 V.

Irms = 247.5/100 = 2.47 A.

Final answer: 2.47 A

Common mistake: AC ammeter does not show peak current; it shows RMS current.

Question 6

The effective value of current in a 50 cycle AC circuit is 5 A. What is the value of current 1/300 second after it was zero?

Solution

Formula used: I0 = √2 Irms, i = I0 sin ωt, ω = 2πf

Irms = 5 A, so I0 = 5√2 = 7.07 A.

f = 50 Hz, t = 1/300 s.

ωt = 2πft = 2π × 50 × 1/300 = π/3.

i = 7.07 sin(π/3) = 7.07 × 0.866 = 6.123 A.

Final answer: 6.123 A

Common mistake: Forgetting to convert RMS current into peak current before using instantaneous equation.

Question 7

The peak value of an alternating current of frequency 50 Hz is 14.14 A. Find its RMS value. How much time will the current take in reaching from zero to maximum?

Solution

Formula used: Irms = I0/√2 and time from zero to maximum = T/4

I0 = 14.14 A.

Irms = 14.14/√2 = 10 A.

Frequency f = 50 Hz, so T = 1/f = 1/50 = 0.02 s.

Time from zero to maximum = T/4 = 0.02/4 = 0.005 s = 5 ms.

Final answer: 10 A, 5 ms

Common mistake: Taking time from zero to maximum as half cycle instead of quarter cycle.

Question 8

A 100 Ω iron is connected to a 220 V, 50 cycle wall plug. Find (i) peak potential difference, (ii) average potential difference, and (iii) RMS current.

Solution

Formula used: V0 = √2 Vrms, average over full cycle = 0, Irms = Vrms/R

Given wall plug voltage 220 V is RMS value.

Peak voltage V0 = √2 × 220 = 311 V.

Average potential difference over a complete cycle = 0.

Irms = 220/100 = 2.2 A.

Final answer: (i) 311 V, (ii) 0 V, (iii) 2.2 A

Common mistake: Taking 220 V as peak voltage. Domestic supply value is RMS value.

Question 9

The equation of AC in a circuit is I = 50 sin 100πt. Find (i) frequency of AC, (ii) mean value of AC over positive half cycle, (iii) RMS value of current, and (iv) value of current 1/300 second after it was zero.

Solution

Formula used: ω = 2πf, Iavg = 2I0/π, Irms = I0/√2, i = I0 sin ωt

I0 = 50 A and ω = 100π rad s-1.

100π = 2πf, so f = 50 Hz.

Mean value over positive half cycle = 2I0/π = 100/π = 31.8 A.

RMS value = 50/√2 = 35.35 A.

At t = 1/300 s, ωt = 100π × 1/300 = π/3.

i = 50 sin(π/3) = 50 × 0.866 = 43.3 A.

Final answer: 50 Hz, 31.8 A, 35.35 A, 43.3 A

Common mistake: Using average value over full cycle instead of positive half cycle.

2. Theory: Average Value and RMS Value

Average Value of AC

Average value of AC is the DC value that gives the same average effect over a selected interval. For a complete cycle of pure sinusoidal AC, the average value is zero because the positive half cycle and negative half cycle cancel each other.

Vavg = (1/T) ∫0T v(t) dt Vavg over positive half cycle = 2V0/π

RMS Value of AC

RMS value is the value of steady DC that produces the same heat in a resistor as the given AC during the same time. RMS value is more useful than average value because electrical power and heat depend on the square of current.

Vrms = √[(1/T) ∫0T v²(t) dt] Vrms = V0/√2
RMS value is connected with heating effect.
Average value over full sine cycle is zero.
AC meters usually show RMS values.

3. Formula and Derivation Section

Average Value of Sine Wave Over Positive Half Cycle

v = V0 sin θ Vavg = (1/π) ∫0π V0 sin θ dθ Vavg = (V0/π) [-cos θ]0π = 2V0/π Iavg = 2I0/π Vavg = 2V0/π

Average Value Over Full Cycle

Vavg = (1/2π) ∫02π V0 sin θ dθ = 0

Positive and negative half cycles have equal areas with opposite signs. Hence the net average over a full cycle is zero.

RMS Value of Sine Wave

Vrms = √[(1/2π) ∫02π V0² sin²θ dθ] Since average value of sin²θ over one cycle is 1/2 Vrms = V0/√2 Irms = I0/√2

Square Wave RMS

For a bipolar square wave, voltage magnitude remains V0 throughout the cycle.

Vrms = V0

Triangular Wave RMS

For a symmetrical triangular wave varying from -V0 to +V0:

Vrms = V0/√3

Half-Wave Rectified Sine

Vavg = V0/π Vrms = V0/2

Full-Wave Rectified Sine

Vavg = 2V0/π Vrms = V0/√2

Form Factor, Peak Factor and Power

Form factor = RMS value / Average value Peak factor = Peak value / RMS value Average power = VrmsIrms cosφ Heat produced = Irms²Rt

4. Classroom Waveform Diagrams

Sinusoidal AC

+V0 -V0 Vavg = 0, Vrms = V0/√2

Full-Wave Rectified Sine

V0 Vavg = 2V0/π, Vrms = V0/√2

Half-Wave Rectified Sine

V0 Vavg = V0/π, Vrms = V0/2

Square Wave and Triangular Wave

Square: Vrms = V0 Triangle: Vrms = V0/√3

5. Exam-Level Question Bank with Answers

IGCSE / GCSE / British Curriculum

Q1

Why is RMS voltage used instead of average voltage for AC supply?

Answer

Formula: H = Irms²Rt. RMS value is used because heating and power depend on square of current or voltage. Average voltage over full AC cycle may be zero, but heating is not zero.

Mistake: Thinking zero average means no heating.

Q2

A 12 V RMS AC source is connected to a lamp. What peak voltage is applied?

Answer

Formula: V0 = √2Vrms. V0 = 1.414 × 12 = 17.0 V.

Mistake: Using 12 V as peak.

IB Physics

Q3

A sinusoidal AC voltage has peak value 20 V. Calculate RMS value and explain its physical meaning.

Answer

Formula: Vrms = V0/√2 = 20/1.414 = 14.1 V. It means this AC produces the same heating effect as 14.1 V DC in the same resistor.

Mistake: Defining RMS as simple average.

Q4

A square wave alternates between +8 V and -8 V. Find average and RMS values.

Answer

Formula: For bipolar square wave, Vavg = 0 and Vrms = V0. Therefore Vavg = 0, Vrms = 8 V.

Mistake: Using sine wave RMS formula.

A-Level Physics

Q5

A 240 V RMS supply is connected to a 60 Ω heater. Find RMS current and average power.

Answer

Formula: Irms = Vrms/R, P = Vrms²/R. I = 240/60 = 4 A. P = 240²/60 = 960 W.

Mistake: Using peak voltage in power formula.

Q6

A triangular voltage waveform has peak value 30 V. Find RMS value.

Answer

Formula: Vrms = V0/√3 = 30/1.732 = 17.3 V.

Mistake: Applying sine wave formula.

CBSE Class 12

Q7

Derive the RMS value of sinusoidal current I = I0 sinωt.

Answer

Formula: Irms = √[(1/T)∫I²dt]. Substitute I = I0sinωt. Since average of sin²ωt over one cycle is 1/2, Irms = I0/√2.

Mistake: Averaging sinωt instead of sin²ωt.

Q8

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

Answer

For pure sine AC, positive and negative half cycles are symmetrical and equal in magnitude. Their algebraic sum over a complete cycle is zero.

Mistake: Confusing full-cycle average with half-cycle average.

NEET

Q9

An AC current is I = 10 sin 100πt. The RMS value is: A) 10 A B) 7.07 A C) 5 A D) 14.14 A

Answer

Correct answer: B. Formula: Irms = I0/√2 = 10/√2 = 7.07 A.

Mistake: Choosing peak current as RMS current.

Q10

The average value of a sinusoidal AC current over complete cycle is: A) I0 B) I0/√2 C) 2I0/π D) 0

Answer

Correct answer: D. Full-cycle average is zero.

Mistake: Using half-cycle average formula.

JEE Main

Q11

A sinusoidal voltage has peak-to-peak value 400 V. Find RMS voltage.

Answer

Formula: Vpp = 2V0. Therefore V0 = 200 V. Vrms = 200/√2 = 141.4 V.

Mistake: Taking 400 V as peak value.

Q12

A half-wave rectified sine wave has peak current 8 A. Find RMS current.

Answer

Formula: Irms = I0/2. Therefore Irms = 8/2 = 4 A.

Mistake: Using I0/√2 for half-wave rectified sine.

JEE Advanced

Q13

A waveform is +V0 for T/4, 0 for T/2, and -V0 for T/4. Find average and RMS value over one cycle.

Answer

Formula: Vavg = algebraic time average, Vrms = √(mean of V²). Average = [V0(T/4) - V0(T/4)]/T = 0. RMS = √[(V0²T/4 + V0²T/4)/T] = V0/√2.

Mistake: Ignoring zero interval in RMS calculation.

Q14

A sine wave and a square wave have the same peak voltage. Which produces more heat in the same resistor?

Answer

Formula: Heating ∝ Vrms². Sine wave RMS = V0/√2. Square wave RMS = V0. Square wave produces more heat.

Mistake: Comparing peak values only.

Assertion-Reason Questions

Q15

Assertion: RMS value of AC is important in heating calculations. Reason: Heat produced depends on square of current.

Answer

Both assertion and reason are true, and reason correctly explains assertion. Formula: H = Irms²Rt.

Mistake: Using average current for heat.

Q16

Assertion: Average value of sinusoidal AC over full cycle is zero. Reason: Positive and negative halves are equal and opposite.

Answer

Both assertion and reason are true, and reason correctly explains assertion.

Mistake: Thinking RMS is also zero.

6. Case Study Section

Case Study 1: Household AC Supply

In India, domestic AC supply is nearly 230 V and 50 Hz. This value is RMS value, not peak value.

Questions and Answers

Q1: What is peak voltage? Answer: V0 = √2 × 230 = 325 V.

Q2: What is average voltage over complete cycle? Answer: Zero.

Formula used: V0 = √2Vrms.

Common mistake: Writing 230 V as peak voltage.

Case Study 2: Heating Effect of AC Current

A heater of resistance 100 Ω is connected to 200 V RMS AC supply.

Questions and Answers

Q1: Find RMS current. Answer: I = 200/100 = 2 A.

Q2: Find power. Answer: P = V²/R = 200²/100 = 400 W.

Formula used: P = Vrms²/R.

Common mistake: Using peak voltage to calculate average power.

Case Study 3: Sine Wave, Square Wave and Rectified Wave

Three waveforms have the same peak value V0: sine wave, bipolar square wave and full-wave rectified sine wave.

Questions and Answers

Q1: Which has maximum RMS value? Answer: Bipolar square wave, because Vrms = V0.

Q2: Which sine-based wave has non-zero average value? Answer: Full-wave rectified sine wave.

Formula used: Sine RMS = V0/√2, square RMS = V0, full-wave rectified average = 2V0/π.

Common mistake: Assuming rectification changes RMS value of sine wave.

7. Kumar Sir Exam Tips

In AC, meter reading is RMS value.
220 V AC or 230 V AC means RMS value.
Peak value = √2 × RMS value.
Average over full cycle of pure sine AC is zero.
Heating depends on RMS value, not simple average value.
For positive half cycle, average sine value = 2V0/π.

8. Final Revision Table

Waveform Average Value RMS Value Form Factor Important Exam Point
Full sine wave 0 V0/√2 Not defined for full-cycle average Full-cycle average is zero.
Sine wave positive half cycle 2V0/π V0/√2 1.11 Used in rectifier and AC theory.
Full-wave rectified sine 2V0/π V0/√2 1.11 Average is non-zero.
Half-wave rectified sine V0/π V0/2 π/2 Current flows only in half cycle.
Bipolar square wave 0 V0 Not defined for full-cycle average Maximum RMS for same peak value.
Unipolar square wave V0/2 V0/√2 √2 ON for half cycle, OFF for half cycle.
Triangular wave 0 V0/√3 Not defined for full-cycle average RMS is less than sine for same peak.

Kumar Physics Classes

If average value, RMS value, peak value, peak-to-peak value, rectifier waveforms, AC heating effect or numerical questions are not clear, students can contact Kumar Sir for one-to-one Physics guidance.

Contact: +91-9958461445 Email: kumarsirphysics@gmail.com Website: https://kumarphysicsclasses.com
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