Area under half cycle = Area of Δ OAB+ Area of ACDB
Example: 6
An alternating current of frequency 50 Hz has a maximum value of 120 A. Write the instantaneous equation.
f = 50 Hz
Im = 120 A
Instantaneous current
р = Im sin ωt
ω = 2πf = 2×π×50
= 314.16 r/s
i = 120 sin 314.16 t
Example: 7
Find the average value of the voltage waveform shown in figure.
Average value = Area under half cycle /Period (half cycle)
Area under half cycle = Area of Δ OAB+ Area of ACDB
Area of Δ OAB = ½ bh = ½ × 2 × 100
= 100
Area of ACDB = b x h
= 3 × 100 = 300
Period = 5
Average value = 100+300 /5 = 400/5
= 80 V
Example: 8
Calculate the form factor and peak factor of the current wave of period 10 sec.
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits : Tag: : Electrical Circuits - Problem Related to Waveforms and Definitions of Different Factor with Example
Basic Electrical and Electronics Engineering
BE3251 2nd semester Mechanical Dept | 2021 Regulation | 2nd Semester Mechanical Dept 2021 Regulation
Basic Electrical and Electronics Engineering
BE3251 2nd Semester CSE Dept 2021 | Regulation | 2nd Semester CSE Dept 2021 Regulation