Answer :
Answer:
16.67 + j11.08 Ω
Explanation:
To determine the load impedance per phase if the line impedance, we need to find the Line to Line voltage of a balanced system and the real power absorbed by the load. So lets take it one after the other.
The formula for Line to Line Voltage for this balanced system can be represented as:
|[tex]V_{an[/tex]| =[tex]\frac{|V_{ab}|}{\sqrt{3} }[/tex]
where |[tex]V_{ab}[/tex]| = 208 V rms
∴ |[tex]V_{an[/tex]| = [tex]\frac{208}{\sqrt{3} }[/tex]
|[tex]V_{an[/tex]| = 120.09 V rms
Now, to the real power absorbed by the load
P = 3 ([tex]V_{an}[/tex])([tex]I_{aA[/tex]) cos θ
where:
P = 1800 W
[tex]V_{an}[/tex] = 120.09 V rms
[tex]I_{aA[/tex] = 6 A
cos θ =???
substituting our known data to determine our unknown data; we have:
1800 = 3(120.09 × 6 × cos θ)
cos θ = [tex]\frac{1800}{3(120.09*6)}[/tex]
cos θ = [tex]\frac{1800}{2161.62}[/tex]
cos θ = 0.8327
θ = cos⁻¹ (0.8327)
θ = 33.62°
angle θ depicts the value of the real power factor
To determine the load impedance per phase; we have:
[tex]Z_{load}=\frac{V_{an}}{I_{aA}}< \theta[/tex]
[tex]Z_{load}=\frac{120.09}{6}<33.62^0[/tex]
[tex]Z_{load}=20.02<33.62 ^0[/tex]
[tex]Z_{load}=[/tex] 16.67 + j11.08 Ω
Hence, the load impedance per phase if the line impedance is negligible = 16.67 + j11.08 Ω