Question

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:

 

 Z_ab =29.41∠ 54.69°Ω ;      Z_bc=25.61∠-38.66° Ω ;    Z_ca =17.8 ∠ 51.84° Ω

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

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

Calculate the active power( P_ab, P_{bc ,} Pca) in phase AB of the load in phasor form.

Determine the line current,( I_a, I_{b &} I_c) 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,(W_a) connected in line a.

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

Answer 1

$e_C(t) =221.831 sin (wt+ \frac{5 \pi}{18})V\\ e_B(t)=e_C(t) \angle-240^0\\ e_B(t)=221.831 sin (wt+ \frac{5 \pi}{18}-\frac{2 \pi}{3})\\ e_B(t)=221.831 sin (wt- \frac{7 \pi}{18})\\ e_A(t)=e_C(t) \angle-240^0\\ e_A(t)=221.831 sin (wt+ \frac{5 \pi}{18}+\frac{2 \pi}{3})\\ e_A(t)=221.831 sin (wt+\frac{7 \pi}{18})\\ I_{ab}=\frac{E_A-E_B}{Z_{AB}}\\ E_A=\frac{221.831}{\sqrt{2}}\angle(\frac{17}{18}\pi)= 156.86 \angle 170^0\\ E_B=\frac{221.831}{\sqrt{2}}\angle(-\frac{7}{18}\pi)= 156.86 \angle -70^0\\ Z_{ab}=29.41\angle54.69^0 \Omega\\ I_{ab}=\frac{156.86 \angle 170^0-156.86 \angle -70^0}{29.41\angle54.69^0}\\ I_{ab}=9.238\angle85.31^0$