The half-wave rectifier circuit of vs(t) = 170 sin(377t) v and a load resistance r = 15ω. determine: a. the average load current. b. the rms load current. c. the power absorbed by the load. d. the apparent power supplied by the source. e. the power factor of the circuit.

The instantaneous voltage () across the half-wave rectifier is given by

(t) = cos(wt + θ) V    ------------ (i)

Where is the peak voltage and Θ is the phase angle.

Given:

(t) = 170sin(377t) V    -------------------------(ii)

Load Resistance R = 15Ω

Comparing equations (i) and (ii)

= 170V

(a) Average load current = =

Taking pi as 22/7 and substituting the values of R and into the above equation, we have;

= = 3.6A

(b) The rms load current =

Substituting the values of R and into the equation above gives;

= = 5.67A

(c). Power absorbed by the load is given by the ac and the dc.

Dc Power absorbed = = = 195W

Ac Power absorbed =

where = = = 85V

Therefore, Ac Power absorbed = = 481.67W

(d) Apparent Power is the product of the rms values of the current and the voltage.

Apparent Power = *

Apparent Power = 5.67 * 85 = 481.95W

(e) Power factor is the ratio of the Power absorbed by load to the Apparent Power

Therefore Power factor =

Power factor = = 0.999

PS: Power absorbed could also be called the real power

Hope this helps

are there options? cause you ain’t. real specific

the answer would most likely be c, because if it's a comical piece of information, it may likely not be true, if it's a site that allows anybody to edit the content, it could mean a five year old is typing the info, and if it's a tv show produced by a celeb, it may have a biased opinion.