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:
The instantaneous voltage ([tex]V_{s}[/tex]) across the half-wave rectifier is given by
[tex]V_{s}[/tex](t) = [tex]V_{m}[/tex]cos(wt + θ) V ------------ (i)
Where [tex]V_{m}[/tex] is the peak voltage and Θ is the phase angle.
Given:
[tex]V_{s}[/tex](t) = 170sin(377t) V -------------------------(ii)
Load Resistance R = 15Ω
Comparing equations (i) and (ii)
[tex]V_{m}[/tex] = 170V
(a) Average load current [tex]I_{0}[/tex] = [tex]\frac{V_{0}}{R}[/tex] = [tex]\frac{V_{m} }{\pi R}[/tex]
Taking pi as 22/7 and substituting the values of R and [tex]V_{m}[/tex] into the above equation, we have;
[tex]I_{0}[/tex] = [tex]\frac{170}{\pi * 15}[/tex] = 3.6A
(b) The rms load current [tex]I_{rms}[/tex] = [tex]\frac{V_{m}}{2R}[/tex]
Substituting the values of R and [tex]V_{m}[/tex] into the equation above gives;
[tex]I_{rms}[/tex] = [tex]\frac{170}{2 * 15}[/tex] = 5.67A
(c). Power absorbed by the load is given by the ac and the dc.
Dc Power absorbed = [tex]\frac{V_{m} ^{2} }{\pi^{2} * R }[/tex] = [tex]\frac{170^{2} }{\pi ^{2} * 15}[/tex] = 195W
Ac Power absorbed = [tex]\frac{V_{rms} ^{2} }{R}[/tex]
where [tex]V_{rms}[/tex] = [tex]\frac{V_{m} }{2}[/tex] = [tex]\frac{170}{2}[/tex] = 85V
Therefore, Ac Power absorbed = [tex]\frac{85^{2} }{15}[/tex] = 481.67W
(d) Apparent Power is the product of the rms values of the current and the voltage.
Apparent Power = [tex]I_{rms}[/tex] * [tex]V_{rms}[/tex]
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 = [tex]\frac{Power absorbed}{Apparent Power}[/tex]
Power factor = [tex]\frac{481.67}{481.95}[/tex] = 0.999
PS: Power absorbed could also be called the real power
Hope this helps
