Average and RMS Value of Alternating Current and Voltage

Average and RMS Value of Alternating Current and Voltage: In a dc system, the voltage and current are constant and, therefore, there is no problem in specifying their magnitude. But in case of ac system, an alternating voltage or current varies from instant to instant and so poses a problem how to specify the magnitude of an alternating voltage or current. An alternating voltage or current may possibly be expressed in terms of peak (maximum) value, average (mean) value or effective (rms) value.

In specifying an alternating voltage or current, its peak or maximum value is rarely used because it has that value only twice each cycle. Furthermore, the average or mean value cannot be used because it is positive as much as it is negative, so the average value is zero. Although the average value over half cycle might be used, it would not be as logical a choice as what we shall find effective (virtual or rms) value which is related to the power developed in a resistance by the alternating voltage or current.

Average Value of Alternating Current.

The average (or mean) value of an alternating current is equal to the value of direct current which transfers across any circuit the same charge as is transferred by that alternating current during a given time.

Since in the case of a symmetrical alternating current (i.e. one whose two half cycles are exactly similar, whether sinusoidal or non-sinusoidal) the average or mean value over a complete cycle is zero hence for such alternating quantities average or mean value means the value determined by taking the average of instantaneous values during half cycle or one alternation only. However, for unsymmetrical alternating current, as half wave rectified current, the average value means the value determined by taking the mean of instantaneous values over the whole cycle.

Average Value of Alternating Current

The average value is determined by measuring the lengths of a number of equidistant ordinates and then taking their mean i.e. of i1, i2, i3….in. etc. which are mid-ordinates.

Average value of alternating current,

I_{av} = \dfrac{i_1+i_2+i_3+i_4+i_n}{n} =\dfrac{Area\ of\ one\ alternating\ (or\ half\ cycle)}{Length\ of\ base\ over\ one\ alternation\ (or\ half\ cycle)}

Using the integral calculus, the average (or mean) value of a function f(t) over a specific interval of time between t1 and t2 is given by

F_{av} = \dfrac{1}{t_2-t_1}\int_{t1}^{t2}f(t)dt        …..(1)

Any function whose cycle is repeated continuously, irrespective of its wave shape, is termed as periodic function, such as sinusoidal function, and its average value is given by

F_{av} = \dfrac{1}{T}\int_0^Tf(t)dt      ……(2)

where T, is time period of periodic function.

In case of a symmetrical alternating current, whether sinusoidal or non-sinusoidal the average value is determined by taking average of one half cycle or one alternation only.

i.e., for symmetrical waveforms,

F_{av} = \dfrac{1}{T}\int_0^{\dfrac{T}{2}}f(t)dt      ……(3)

RMS Value or Effective Value of Alternating Current.

The rms or effective value of an alternating current or voltage is given by that steady current or voltage which when flows or applied to a given resistance for a given time produces the same amount of heat as when the alternating current or voltage is flowing or applied to the same resistance for the same time.

RMS Value or Effective Value of Alternating Current

Consider an alternating current of waveform shown in Fig. 2 flowing through a resistor of R ohms. Divide the base of one alternation into n equal parts and let the mid-ordinates be i1, i2 i3…in, etc.

Heat produced during 1st interval = i_1^2R\times\dfrac{T}{n}joules

Heat produced during 2nd interval = i_2^2R\times\dfrac{T}{n}joules

Heat produced during 3rd interval = i_3^2R\times\dfrac{T}{n}joules

Heat produced during nth interval = i_n^2R\times\dfrac{T}{n}joules

Total heat produced in time T = RT(\dfrac{i_1^2+ i_2^2+ i_3^2+ i_n^2}{n})joules

Now if Ieff is the effective current, then heat produced by this current in time T = I_{eff}^2 RT joules. By definition these two expressions are equal

I_{eff}^2RT = RT(\dfrac{i_1^2+ i_2^2+ i_3^2+ i_n^2}{n})

 

I_{eff}^2 = \dfrac{i_1^2+ i_2^2+ i_3^2+i_n^2}{n}

 

I_{eff}^2 = \sqrt{\dfrac{i_1^2+ i_2^2+ i_3^2+ i_n^2}{n}}

Hence the effective or virtual value of alternating current or voltage is equal to the square root of the mean of the squares of successive ordinates and that is why it is known as root-mean-square (rms) value.

Using the integral calculus, the root mean square (rms) or effective value of an alternating quantity over a time period is given by

F_{rms} = \sqrt{\dfrac{1}{T}\int_0^TF^2(t)dt}       ……(4)

AVERAGE AND EFFECTIVE (RMS) VALUES OF SINUSOIDAL CURRENT AND VOLTAGE

Average Value For Sinusoidal Current or Voltage.

The average value of a sine wave over a complete cycle is zero. Therefore, the half cycle average value is intended. Instantaneous value of sinusoidal current is given by

i = I_{max}\ sin\omega t

Average Value For Sinusoidal Current or Voltage

Consider first half cycle i.e. when \omega t varies from 0\ to\ \pi

We get,

I_{av} = \dfrac{Area\ of\ first\ half\ cycle}{\pi}

 

=\dfrac{1}{\pi}\int_0^{\pi}id(\omega t) = \dfrac{1}{\pi}\int_0^{\pi}I_{max}\ sin\omega t\ d(\omega t)

 

Or, I_{av} = \dfrac{I_{max}}{\pi}[-cos\omega t]_0^{\pi}

 

 =\dfrac{2}{\pi}I_{max} = 0.637\ I_{max}       …..(5)

Similarly, Eav = 0.637 Emax

Effective (RMS) Value For Sinusoidal Current or Voltage

A sinusoidal alternating current is represented by

i = I_{max} sin\omega t

 

I_{max}^2 = \dfrac{Area\ of\ first\ half\ cycle\ of\ i^2}{\pi}

 

=\dfrac{1}{\pi}\int_0^{\pi}i^2 d(\omega t) = \dfrac{1}{\pi}\int_0^{\pi}I_{max}^2 sin^2\omega t d(\omega t)

 

=\dfrac{I_{max}^2}{2\pi}\int_0^{\pi}(1-cos2\omega t)d(\omega t)

 

\dfrac{I_{max}^2}{2\pi}[\omega t-\dfrac{1}{2}sin 2\omega t]_0^{pi}

 

=\dfrac{I_{max}^2}{2\pi}\times\pi = \dfrac{I_{max}^2}{2}

 

Or, I_{rms} = \sqrt{\dfrac{I_{max}^2}{2}} = \dfrac{I_{max}}{\sqrt{2}}      …..(6)

 

Similarly, E_{rms} = \dfrac{E_{max}}{\sqrt{2}}

rms value of sinusoidal current and voltage

Example 1. A sinusoidal varying alternating current of frequency 60 Hz has a maximum value of 15 amperes.

  1. Write down the equation for instantaneous value,
  2. Find the value of current after 1/200 second,
  3. Find the time taken to reach 10 amperes for the first time, and
  4. Find its average value.

Solution:

Maximum value of current, Imax = 15A

Frequency of alternating current, f = 60 Hz

1    The equation for instantaneous value of sinusoidal alternating current with θ as zero when time is zero, is given as

I = Imax sin 2π f t = 15 sin 2π\times 60 t

= 15 sin 120 π t Ans.

2    Substituting t=\dfrac{1}{200} second in the expression

I = 15 sin 120 π t, we have value of current after \dfrac{1}{200} second.

= 15 sin 120 π\times\dfrac{1}{200}

= 15 sin 0.6 π

= 15\times0.951[/latex]

=14.266 A Ans.

3    Let the instantaneous value of current be 10 A for the first time t seconds after the instant the current is zero and becoming positive

10 = 15 sin 120 π t

Or t = \dfrac{1}{120\pi}sin^{-1}\dfrac{10}{15} = \dfrac{1}{120\pi}sin^{-1}0.6667

= 0.001936 seconds Ans.

   Average Value, I_{av} = \dfrac{2}{\pi}I_{max} = \dfrac{2}{\pi}\times 15 = 9.55 A Ans.

Example 2. An ac sinusoidal current has rms value of 40 A at 50 Hz frequency. Write expression of instantaneous current and obtain its value 0.002 sec after passing through maximum positive value.

Solution:

Expression for instantaneous value of an ac sinusoidal current with \theta as zero when time is zero is given as

i= I_{max} sin\omega t = I_{max} sin 2\pi f t = 40\sqrt{2} sin 2 \pi \times 50t

56.57 sin 314t Ans.

When time is measured from the positive maximum value, the above equation is modified into

v = 40\sqrt{2}sin(314t+\dfrac{\pi}{2}) = 40\sqrt{2}cos 314t 40\sqrt{2}cos 314 \times 0.002 40\sqrt{2}cos 0.626

40\sqrt{2}cos\dfrac{\pi}{5} = 54.77 A Ans.

 

Example 3. An alternating current when passed through a resistance immersed in water for 5 minutes, just raised the temperature of water to boiling point. When a direct current of 4 amperes was passed through the same resistance under identical conditions it took 8 minutes to boil the water. Find the rms value of the alternating current.

Solution:

Let the rms value of alternating current passed through the resistance be Irms amperes.

Heat produced when an alternating current of Irms amperes is passed through a resistance R immersed in water for 5 minutes

 = I_{rms}^2\times R\times 5\times 60 = 300 I_{rms}^2 R joules     …..(i)

Heat produced when a direct current of 4 A is passed through the same resistance R immersed in water for 8 minutes

= 4^2 \times R \times 8 \times 60 = 7,680 R joules     …..(ii)

Since heat produced in both of the case is same, equating expression (i) and (ii) we get

 300 I_{rms}^2 R = 7,680 R

Or  I_{rms} = \sqrt{\dfrac{7,680}{300}} = \sqrt{25.6} 5.06 A Ans.

Example 4. An alternating voltage is given by v 141.4 sin 314 t. Find

  1. Frequency
  2. Rms value
  3. Average value
  4. The instantaneous value of voltage when ‘t’ is 3 ms
  5. The time taken for the voltage to reach 100V for the first time after passing through zero value.

Solution:

Instantaneous value of alternating voltage is given by expression

 v = 141.4 sin 314 t

   The frequency is given by the coefficient of time divided by 2 π, therefore,

Frequency,  f = \dfrac{314}{2\pi} = 50 Hz.Ans.

Peak value of an alternating voltage is given by the coefficient of sine of time angle, therefore,

Vmax = 141.4 volts

2    RMS value of given alternating voltage,

 V_{max} = \dfrac{V_{max}}{\sqrt{2}} = \dfrac{141.5}{\sqrt{2}} = 100 V Ans.

3     Average value of given alternating voltage,

V_{av} = 0.637 V_{max} = 0.637\times141.4 = 90V Ans.

4     Instantaneous value of voltage at 3 ms(0.003 s) after the instant the voltage is zero and increasing in positive direction may be obtained by substituting t = 0.003 second in above equation. So, we have

v = 141.4 sin 314\times 0.003

=141.4 sin 0.942 = 114.4 V Ans.

   Time taken from t = 0 for the voltage to reach 100 V

 t = \dfrac{1}{315}sin^{-1}\dfrac{100}{141.4}

 =\dfrac{\dfrac{\pi}{2}}{314} second = \dfrac{1}{400}s = 2.5 ms Ans.