Answer to Question #223618 in Electrical Engineering for TAPION

Question #223618

In an unbalanced, three-phase, star-delta system has a frequency of 50 Hz, and CBA phase rotation.The phase impedances

of the load are as follows:

 

 Zab =29.41∠ 54.69°Ω ;      Zbc=25.61∠-38.66° Ω ;    Zca =17.8 ∠ 51.84° Ω


The emf of the source is represented by:ec(t)=221.831 𝐬𝐢𝐧 (𝛚𝐭 +(𝟓𝝅/𝟏𝟖) ) V



Calculate the phase current,( Iab, Ibc & Ica) of the load in polar form.

Calculate the active powerPab, Pbc , Pca) in phase AB of the load in phasor form.

Determine the line current,( Ia, Ib & Ic) in polar form.

If two wattmeters are connected in lines a and b respectively of the above circuit to

measure the active power

Determine the reading of the wattmeter,(Wa) connected in line a.

Determine the reading of the wattmeter,(Wb) connected in line b.



1
Expert's answer
2021-08-06T04:30:49-0400

"e_C(t) =221.831 sin (wt+ \\frac{5 \\pi}{18})V\\\\\ne_B(t)=e_C(t) \\angle-240^0\\\\\ne_B(t)=221.831 sin (wt+ \\frac{5 \\pi}{18}-\\frac{2 \\pi}{3})\\\\\ne_B(t)=221.831 sin (wt- \\frac{7 \\pi}{18})\\\\\ne_A(t)=e_C(t) \\angle-240^0\\\\\ne_A(t)=221.831 sin (wt+ \\frac{5 \\pi}{18}+\\frac{2 \\pi}{3})\\\\\ne_A(t)=221.831 sin (wt+\\frac{7 \\pi}{18})\\\\\nI_{ab}=\\frac{E_A-E_B}{Z_{AB}}\\\\\nE_A=\\frac{221.831}{\\sqrt{2}}\\angle(\\frac{17}{18}\\pi)= 156.86 \\angle 170^0\\\\\n E_B=\\frac{221.831}{\\sqrt{2}}\\angle(-\\frac{7}{18}\\pi)= 156.86 \\angle -70^0\\\\\nZ_{ab}=29.41\\angle54.69^0 \\Omega\\\\\nI_{ab}=\\frac{156.86 \\angle 170^0-156.86 \\angle -70^0}{29.41\\angle54.69^0}\\\\\nI_{ab}=9.238\\angle85.31^0"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS