FORM FACTOR AND PEAK FACTOR

There exists a definite relation among the average value, rms value, and peak value of an alternating quantity. The relationship is expressed by the two factors, namely form factor and peak factor.

1. Form factor: The ratio of rms value to average value of an alternating quantity is called form factor.Mathematically, form factor For the current varying sinusoidallyForm factor 
2. Peak factor: The ratio of maximum value to rms value of an alternating quantity is called peak factor.Mathematically, peak factor For current varying sinusoidallyPeak factor 

Example 6.1

The equation of an alternating current is i = 42.42 sin 628 t

Determine (i) its maximum value; (ii) Frequency; (iii) rms value; (iv) Average value; and (v) Form factor.

(U.P.T.U. 2005–2006)

Solution:

Given equation is

 

i = 42.42 sin 628 t

Comparing above equation with the standard equation, i = I_m sin ωt

 

i_m = 42.42 A and ω = 628 rad/sec

1. Maximum value, I_m = 42.42 A (Ans.)
2. Here, ω = 628 rad/sec or 2πf = 628Frequency, 
3. RMS value, 
4. Average value, 
5. Form factor 

Example 6.2

A supply voltage of 230 V, 50 Hz is fed to a residential building. Write down its equation for instantaneous value.

Solution:

 

ν = V_m sin ω t

where

ω = 2 × π × f = 2 × π × 50 = 314.16 radians

∴

 

ν = 325.27 sin 314.16 t (Ans.)

Example 6.3

An alternating voltage is given by ν = 141.1 sin 314t. Find the following:

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

(U.P.T.U. 2006–2007)

Solution:

1. Given that, ν = 141.14 sin 314 tComparing above equation withν = V_in sin ωtω = 314 rad/sec.∴ Frequency,
2. RMS value of the voltage, 
3. Average value, 
4. ν = 141.1 sin 314t. At t = 3 ms = 3 × 10⁻³ sν = 141.4 sin 314 × 3 × 10⁻³ = 114.4 V (Ans.)
5. ν = 141.4 sin 314tor

Example 6.4

A sinusoidally alternating current of frequency 60 Hz has a maximum value of 15 A.

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 A for the first time and
4. Find its average value.(U.P.T.U., July 2003)

Solution:

Here, I_m = 15 A f = 10 Hz

1. The equation for instantaneous value of sinusoidal alternating current, i = I_m sin 2π f t = 15 sin 2 π × 60 t = 15 sin 120 π t (Ans.)
2. When,  
3. When the current is zero and becoming positive, i = 15 sin 120 πtWhere i = 10 A∴ 10 = 15 sin 120 πtor
4. Average value, 

Example 6.5

Find the rms value, average value, and form factor of the voltage waveform shown in Figure 6.12.

(U.P.T.U. 2002–2003)

Fig. 6.12  Full rectified wave

Solution:

Average value of voltage over one cycle

RMS value of voltage over one cycle

Example 6.6

Find the average and rms values of the current, I(t) = 10 + 10 sin 314t.

Solution:

The wave diagram of given current i(t) = 10 + 10 sin 314t is shown in Figure 6.13.

Fig. 6.13  Wave shape as per data

Average value of the given current,

I_av = I_dc + I_av of ac component.

RMS value of the given current

I_rms = I_dc + I_rms of AC component

Example 6.7

A transmission line carries a DC voltage of 50 V and a half-wave sinusoidal voltage as shown in Figure 6.14. Calculate the following:

(i) rms value   (ii) the average value, and   (iii) form factor

Fig. 6.14  Wave shape as per data

Solution:

Peak value of half rectified wave (neglected DC).

 

V_m = 100 V

RMS value, 

RMS value of resultant wave,

 

V_rms = 50 +16 = 66 V (Ans.)

Average value of half rectified wave, rejecting DC,

Average value of resultant wave,

 

V_av = 50 +16 = 66 V (Ans.)

Example 6.8

An alternating current of frequency 60 Hz has a maximum value of 120 A. Write down the equation for its instantaneous value. Reckoning time from the instant, the current is zero and is becoming positive, find (i) the instantaneous value after 1/360 seconds and (ii) the time taken to reach 96 A for the first time.

Solution:

Instantaneous value of current is given by the following equation.

 

i = I_m sin ωt = I_m sin 2 π f t = 120 sin 2 π × 60 t = 120 sin 377 t (Ans.)

When the reckoning time is taken from the instant, the current is zero and becoming positive, equation for current is given as follows:

 

i = 120 sin 377 t

1. At t = 1/360 s; i = 120 sin 2π × 60 ×  = 120 sin π/3 = 120 × 0.866 = 103.92 (Ans.)
2. Let t seconds be the time taken to reach the current to 96 A for the first time. Then, 96 = 120sin 2 × π × 60 × t or sin120 × π × t or sin120 ×180° × t = 0.8or 120 ×180° × t = sin⁻¹ 0.8or120 ×180° × t = 53.13°; 

Note: On both the sides, the angle must be in the same units, that is, either in degrees or in radians.

Example 6.9

Calculate the average value, rms value, form factor, and peak factor of a periodic current wave having values for equal time interval changing suddenly from one value to next: 0, 30, 45, 70, 90, 70, 45, 30, 0, −30, −45, −70, etc. in ampere. What would be the average and the rms value of a sine wave having the same peak value?

Solution:

The periodic wave form of the alternating current is shown in Figure 6.15.

Fig. 6.15  Periodic wave as per data

Average value of current,

= 47.5 A (Ans.)

RMS value of current,

Form factor 

Peak factor 

Average value of sinusoidal AC having peak value of 90 A;

 

I_av = 0.637 × I_m = 0.637 × 90 = 57.33 A (Ans.)

RMS value,

Example 6.10

An alternating voltage e = 200 sin 314 t is applied to a device which offers an ohmic resistance of 20 Ω to the flow of current in one direction, while preventing the flow of current in opposite direction. Calculate rms value, average value, and form factor for the current over one cycle.

(Nagpur University, 1992)

Fig. 6.16  (a) Give sine wave (b) Half rectified wave

Solution:

The wave diagram of an ac supply voltage is shown in Figure 6.16(a). The wave diagram of the rectified current that flows in the circuit is shown in Figure 6.16(b). The instantaneous value of applied

 

e = 200 sin 314 t

Maximum value of applied voltage, E_m= 200 V

Resistance of rectifying device, R = 20 Ω

Maximum value of half-wave rectified alternating current,

RMS value of half-wave rectified alternating current,

Average value of the half-wave rectified alternating current,

Form factor of the half-wave rectified alternating current

Example 6.11

Find the average and effective values of voltage for sinusoidal waveform shown in Figure 6.17.

(Allahabad University, 1991)

Fig. 6.17  Give wave shape

Solution:

The equation of the given sinusoidal waveform is ν = 100 sin θ.

or

 

V_rms = 47.67 V (Ans.)

Example 6.12

Find the average value of the periodic function shown in Figure 6.18.

Fig. 6.18  Given wave shape

Solution:

Average value,

Area of alternation 

Base = π

Average value,

Example 6.13

For a half-wave rectified alternating current, find (i) average value, (ii) rms value, (iii) form factor, and (iv) peak factor.

Solution:

1. A half-wave rectified AC is shown in Figure 6.19.The negative half cycle is suppressed in this case, that is, current flows only during positive half cycle.The wave is unsymmetrical, and therefore, one complete cycle is to be considered.Area of complete cycle = Area of half cycleBase = 2π∴ Fig. 6.19  Half wave rectified ac
2. The squared wave of a half wave rectified wave is shown in Figure 6.19.RMS value,Area of squared wave over a complete cycle= Area of squared wave in half cycleBase = 2π ; I_rms 
3. Form factor 
4. Peak factor 

Example 6.14

Determine the form factor of voltage waveform shown in Figure 6.20.

(U.P.T.U. 2004–2005)

Solution:

As per the wave diagram shown in Figure 6.21, the point A(1, 10) shows, t = 1 and V = 10.

 

θ(t) = m × t + 0 = 10 × t

Form factor = 

Fig. 6.20  Given voltage wave

Fig. 6.21  Voltage wave as per data