Question

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.

Answer 1

Answer:

a  average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by  P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

Answer 2

Answer:

The instantaneous voltage ($V_{s}$) across the half-wave rectifier is given by

$V_{s}$(t) = $V_{m}$cos(wt + θ) V    ------------ (i)

Where $V_{m}$ is the peak voltage and Θ is the phase angle.

Given:

$V_{s}$(t) = 170sin(377t) V    -------------------------(ii)

Load Resistance R = 15Ω

Comparing equations (i) and (ii)

$V_{m}$ = 170V

(a) Average load current $I_{0}$ = $\frac{V_{0}}{R}$ = $\frac{V_{m} }{\pi R}$

Taking pi as 22/7 and substituting the values of R and $V_{m}$ into the above equation, we have;

$I_{0}$ = $\frac{170}{\pi * 15}$ = 3.6A

(b) The rms load current $I_{rms}$ = $\frac{V_{m}}{2R}$

Substituting the values of R and $V_{m}$ into the equation above gives;

$I_{rms}$ = $\frac{170}{2 * 15}$ = 5.67A

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

Dc Power absorbed = $\frac{V_{m} ^{2} }{\pi^{2} * R }$ = $\frac{170^{2} }{\pi ^{2} * 15}$ = 195W

Ac Power absorbed = $\frac{V_{rms} ^{2} }{R}$

where $V_{rms}$ = $\frac{V_{m} }{2}$ = $\frac{170}{2}$ = 85V

Therefore, Ac Power absorbed = $\frac{85^{2} }{15}$ = 481.67W

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

Apparent Power = $I_{rms}$ * $V_{rms}$

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 = $\frac{Power absorbed}{Apparent Power}$

Power factor = $\frac{481.67}{481.95}$ = 0.999

PS: Power absorbed could also be called the real power

Hope this helps