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Latest RMS Value of Time Varying Waveforms MCQ Objective Questions

Top RMS Value of Time Varying Waveforms MCQ Objective Questions

RMS Value of Time Varying Waveforms Question 1:

For a sinusoidal waveform, the RMS value of current will be _______ times the maximum value of current.

  1. 1.1
  2. 1.414
  3. 0.637
  4. 0.707

Answer (Detailed Solution Below)

Option 4 : 0.707

RMS Value of Time Varying Waveforms Question 1 Detailed Solution

Crest Factor ‘or’ Peak Factor is defined as the ratio of the maximum value to the R.M.S value of an alternating quantity.

C.F. ‘or’ P.F. = MaximumValueR.M.SValue

For a sinusoidal waveform, the value of the crest factor is 1.41.

∴ RMS Value = MaximumValue1.41=0.707×Maximum value

Hence, For a sinusoidal waveform, the RMS value of current will be 0.707 times the maximum value of current.

Form Factor:

The form factor is defined as the ratio of the RMS value to the average value of an alternating quantity.

F.F. (Form factor) = R.M.SValueAverageValue

For a sinusoidal waveform, the value of the form factor is 1.11.

:

WAVEFORM

 

SHAPE

 

MAX.

VALUE

AVERAGE VALUE

RMS VALUE

FORM FACTOR

CREST FACTOR

SINUSOIDAL WAVE

DMRC JE Official Paper Shift 2 images shubham D 1

Am

2Amπ

Am2

Am22Amπ=1.11

AmAm2=2

SQUARE WAVE

DMRC JE Official Paper Shift 2 images shubham D 2

Am

 

Am

 

 

Am

 

AmAm=1

AmAm=1

TRIANGULAR WAVE

DMRC JE Official Paper Shift 2 images shubham D 3

Am

Am2

Am3

Am3Am2=23

AmAm3=3

HALF-WAVE RECTIFIED WAVE

 

DMRC JE Official Paper Shift 2 images shubham D 4

Am

Amπ

Am2

Am2Amπ=π2

2

RMS Value of Time Varying Waveforms Question 2:

The rms value of a sinusoidal ac current is numerically equal to its value at an angle of _______ degrees

  1. 60
  2. 30
  3. 45
  4. 90

Answer (Detailed Solution Below)

Option 3 : 45

RMS Value of Time Varying Waveforms Question 2 Detailed Solution

RMS (Root mean square) value:

  • RMS value is based on the heating effect of wave-forms.
  • The value at which the heat dissipated in the AC circuit is the same as the heat dissipated in the DC circuit is called the RMS value provided, both the AC and DC circuits have equal value of resistance and are operated at the same time.
  • RMS value 'or' the effective value of an alternating quantity is calculated as:

    Vrms=1T0Tv2(t)dt

    T = Time period

F1 U.B. Nita 11.11.2019 D 2

RMS value = Vm sin θ = Vm sin 45° 

= 0.707 Vm 

Note:

  • Average or mean value of alternating current is that value of steady current which sends the same amount of charge through the circuit in a certain interval of time as is sent by alternating current through the same circuit in the same interval of time
  • The ratio of the maximum value (peak value) to RMS value is known as the peak factor or crest factor.
  • Peakfactor=maximumvaluermsvalue
  • The ratio of RMS value to the average value is known as the form factor.
  • Formfactor=rmsvalueaveragevalue
 

RMS Value of Time Varying Waveforms Question 3:

An alternating voltage has the equation V(t) = 200 sin 377t V. What is the value of r.m.s. voltage and frequency?

  1. 200 V, 50 Hz
  2. 200 V, 60 Hz
  3. 2002 V, 60 Hz
  4. 2002 V, 50 Hz

Answer (Detailed Solution Below)

Option 3 : 2002 V, 60 Hz

RMS Value of Time Varying Waveforms Question 3 Detailed Solution

Concept:

A general sinusoidal voltage is expressed as:

v(t) = Vm sin (ωt + ϕ)   ---(1)

Vm = Maximum amplitude of the wave

ϕ = Phase angle

ω = angular frequency, given by:

ω=2πf=2πT

T = Time period of the wave

The rms value of any general expression is calculated as:

Vrms=1T0Tv2(t)dt

For the general sinusoidal wave of equation (1), the RMS value is:

Vrms=Vm2

Calculation:

V(t) = 200 sin 377t V

Comparing this with the general expression of Equation (1), we get:

Vm = 200

ω = 377

2πf = 377

f=3772π

= 60 Hz

The time period will be:

T=1f=160

T = 0.0167 s

The RMS value will be:

Vrms=Vm2=2002

The correct answer is RMS voltage = 200/√2 V and frequency = 60 Hz

RMS Value of Time Varying Waveforms Question 4:

What is the peak to peak voltage of 500 V AC RMS voltage?

  1. 500V
  2. 1414 V
  3. 1000 V
  4. 707 V

Answer (Detailed Solution Below)

Option 2 : 1414 V

RMS Value of Time Varying Waveforms Question 4 Detailed Solution

Concept:

For a sinusoidal alternating waveform

  • Peak to peak value of voltage (Vp-p) = 2 (Peak value)
  • The peak value of voltage (Vp) = √2 Vrms
  • RMS value of voltage (Vrms)=Vp2
  • The average value of voltage (Vavg)=2Vpπ

 

Calculation:

Vrms = 500 V

Vp = √2 × 500 V

Vp-p = 2 × √2 × 500

Vp-p = 1414 V

RMS Value of Time Varying Waveforms Question 5:

An ac current is given as:

i = 10 + 10 sin 314t

The average and rms values of the current, respectively, are.

  1. 10 A, 12.2 A
  2. 10 A, 17.07 A
  3. 16.36 A, 12.2 A
  4. 16.36 A, 17.7 A

Answer (Detailed Solution Below)

Option 1 : 10 A, 12.2 A

RMS Value of Time Varying Waveforms Question 5 Detailed Solution

The Correct Answer is Option "1".
 

Concept:

To find the average and RMS (root mean square) values of the given AC current, we'll use the following formulas,

Iavg=1T0Ti(t)dt

Irms=1T0T[i(t)]2dt

The integral of  sin(wt) over one complete cycle is zero, resulting in Iavg=0 for a DC signal, the average value is equal to the amplitude of the signal.

Calculation:
​​i=10+10sin314t

 Iavg=10+0=10A

For a sinusoidal signal, : RMS=A2=102=7.07

 Irms=102+7.072=12.24 A.

RMS Value of Time Varying Waveforms Question 6:

The AC current flowing through a 10 Ω resistance in a closed power circuit is denoted by i (t) = 3 + 4 sin (ωt) + 4 sin (2 ωt) A. Find the rms value of the current.

  1. 3 A
  2. 10 A
  3. 8 A
  4. 5 A

Answer (Detailed Solution Below)

Option 4 : 5 A

RMS Value of Time Varying Waveforms Question 6 Detailed Solution

Concept:

RMS value of current for different frequency functions can be determined as

Irms=Irms12+Irms22+Irms32+.

Calculation:

Given-

i (t) = 3 + 4 sin (ωt) + 4 sin (2 ωt) A

Irms1=3 A

irms2=42A

irms2=42A

Irms=(3)2+(42)2+(42)2

Irms=25=5 A

RMS Value of Time Varying Waveforms Question 7:

The RMS value of the voltage u(t) = 3 + 4 cos (3t) is

  1. 17V
  2. 5 V
  3. 7 V
  4. (3+22)V

Answer (Detailed Solution Below)

Option 1 : 17V

RMS Value of Time Varying Waveforms Question 7 Detailed Solution

RMS (Root mean square) value:

  • RMS value is based on the heating effect of wave-forms.
  • The value at which the heat dissipated in AC circuit is the same as the heat dissipated in DC circuit is called RMS value provided, both the AC and DC circuits have equal value of resistance and are operated at the same time.
  • RMS value 'or' the effective value of an alternating quantity is calculated as:

V = a0 + a1 sin (ω1t + θ1) + a2 sin (ω2t + θ2) + a3 sin (ω3t + θ3) +...........

Here a0 = DC value = average value of current

RMS value Vrms a02+12(a12+a22+a32+)

Calculation:

For the given 

u(t) = 3 + 4 cos (3t) 

urms a02+12(a12+a22+a32+)

Rms value of given voltage is,

=9+(42)2=9+8=17V

RMS Value of Time Varying Waveforms Question 8:

Which of the following statement is CORRECT?

  1. The maximum value of alternating voltage is given by the co-efficient of sine of the time angle
  2. The maximum value of alternating voltage is given by only frequency
  3. The maximum value of alternating voltage is given by only Alternation
  4. The maximum value of alternating voltage is given by negative value of amplitude

Answer (Detailed Solution Below)

Option 1 : The maximum value of alternating voltage is given by the co-efficient of sine of the time angle

RMS Value of Time Varying Waveforms Question 8 Detailed Solution

Different Forms of Alternating Voltage:

The standard form of an alternating voltage is given by:

v = Vsin θ = Vsin ωt  = Vsin (2πft) = Vsin 2πTt

Where,

v is the instantaneous value of voltage
Vm is the maximum value of voltage
(θ = ωt = 2πft = 2πTt) is Co-efficient of sine of time angle

From the above equation following point can be noted:

  • The maximum value of alternating voltage is given by the co-efficient of sine of the time angle i.e. Maximum value of voltage = Co-efficient of sine of time angle
  • The frequency f of alternating voltage is given by dividing the co-efficient of time in the angle by 2π i.e.,
    f = Co-efficient of time in the angle/

RMS Value of Time Varying Waveforms Question 9:

What will be the average value of given waveform:

F1 Nakshtra 16-11-21 Savita D1

  1. 1 V
  2. 2 V
  3. 3 V
  4. 4 V
  5. 5 V

Answer (Detailed Solution Below)

Option 3 : 3 V

RMS Value of Time Varying Waveforms Question 9 Detailed Solution

Average Value: The average value of any waveform is the ratio of the area under the curve to the time or length of the base (X-axis).

Av=Areat

Method 1: Calculating by using Conventional Method:

F1 Nakshtra 16-11-21 Savita D2 

Here, total area under one cycle = Area 1 + Area 2

Total Time = 0.6 sec

Area 1 = (15 × 0.2) = 3 Vsec

Area 2 = (-3 × 0.4) = -1.2 Vsec

Total Area = (3 - 1.2) = 1.8 Vsec

F1 Nakshtra 16-11-21 Savita D3

Hence, Average value = Areat=1.80.6=3 V

Method 2: Calculating by using Method of integration:

Area = 00.215dt+0.20.63dt = 1.8 Vsed

Hence, Average value = ​Areat=1.80.6=3 V

RMS Value of Time Varying Waveforms Question 10:

The rms value of the resultant current in a wire which carries a dc current of 20 A and a sinusoidal alternating current of peak value 20 A is 

  1. 34.1 A
  2. 17.3 A
  3. 24.5 A
  4. 40.0 A

Answer (Detailed Solution Below)

Option 3 : 24.5 A

RMS Value of Time Varying Waveforms Question 10 Detailed Solution

Given 

DC current / Average current = 20 A.

Peak value of sinusoidal current = 20 A

Calcualation:

RMS value of resultant current =  I02+I1rms2+I2rms2+I3rms2.....

I1rms = 202 A.

Irmsresultant=202+(202)2=400+200=24.5A

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