AC current solved problems

Problem alternating current N°1

Calculate Vp, Vp-p, Vrms, and Vavg of a 16 V peak sine wave.

Solution 1 :

Vp = 16 V
Vp-p = 2 x VP = 2 x 16V = 32V
Vrms = 0.707 x Vp = 0.707 x 16 V = 11.3 V
Vavg = 0.637 x Vp = 0.637 x 16 V = 10.2 V

Problem  N°2

Calculate Vp, Vp-p, and Vavg of a 120 V (rms) ac main supply.

Solution 2 :

Vp = rms x 1.414 = 120 V x 1.414 = 169.68 V

Vp p = 2 x Vp = 2 x 169.68 V = 339.36 V
Vavg = 0.637 x Vp = 0.637 x 169.68 V = 108.09 V

The 120 V (rms) that is delivered to every home and business has a peak of 169.68 V. This ac value value will deliver the same power as 120 V dc.

Problem N°3

if sin wave has a period of 400us, what is its frequency ?

Solution 3 :

frequency (f) = (1/t) = 1/400us = 2.5 KHz or 2500 cycle/second

Problem N°4

If it takes a sin wave 25 ms to Complete two cycles, how many of the cycles will be received in 1 s?

Solution 4:

If the period of two cycles is 25 ms, one cycle period will equal 12.5 ms. The number of cy¬cles per second or frequency will equal :
frequency (f) = (1/t) = 1/12.5ms = 80 Hz or 80 cycle/second

Problem N°5

Calculate the period of the following:

1. 100 MHz
2. 40 cycles every 5 seconds
3. 4.2 kiloycles/second
4. 500 kHz

Solution 5 :

1. t=(1/100MHz) = 10 nanoseconds (ns)
2. 40 cycles /5 s = 8 cycles/second (8Hz)
   t = ( 1 / 8 Hz ) = 125 ms
3. t = ( 1 / 4.2 KHz ) = 238 us
4. t = ( 1 / 500 KHz ) = 2 us

Problem N°6

Calculate the duty cycle and Vavg of a square wave of 0 to 5 V.

Solution 6 :

The duty cycle of a square wave is always 0.5 or 50%. The averag