For the unbalanced ∆-connected load in Fig. 24.33 with two properly connected wattmeters:

a. Determine the magnitude and angle of the phase currents.
b. Calculate the magnitude and angle of the line currents.
c. Determine the power reading of each wattmeter.
d. Calculate the total power absorbed by the load.
e. Compare the result of part (d) with the total power calculated using the phase currents and the resistive elements.

Question

For the unbalanced ∆-connected load in Fig. 24.33 with two properly connected wattmeters:

a. Determine the magnitude and angle of the phase currents.
b. Calculate the magnitude and angle of the line currents.
c. Determine the power reading of each wattmeter.
d. Calculate the total power absorbed by the load.
e. Compare the result of part (d) with the total power calculated using the phase currents and the resistive elements.

Accepted Answer

a. [latex]I_{ab}=\frac{V_{ab}}{Z_{ab}}=\frac{E_{AB}}{Z_{ab}}=\frac{208 V ∠0°}{208 V ∠0°}=20.8 A ∠0° \[0.5cm] I_{bc}=\frac{V_{bc}}{Z_{bc}}=\frac{E_{BC}}{Z_{bc}}=\frac{208 V ∠-120°}{15 Ω + j 20 Ω}=\frac{208 V ∠-120°}{25 Ω ∠53.13°}=8.32 A ∠−173.13° \[0.5cm] I_{ca}=\frac{V_{ca}}{Z_{ca}}=\frac{E_{CA}}{Z_{ca}}=\frac{208 V ∠+120°}{12 Ω + j 12 Ω}=\frac{208 V ∠+120°}{16.97 Ω ∠-45°}=12.26 A ∠165° [/latex]

b. [latex]I_{Aa}=I_{ab}-I_{ca}\[0.5cm] =20.8 A ∠0° - 12.26 A ∠165°\[0.5cm] =20.8 A - (-11.84 A + j 3.17 A)\[0.5cm] =20.8 A + 11.84 A - j 3.17 A = 32.64 A - j 3.17 A=32.79 A ∠−5.55°\[0.5cm] I_{Bb}=I_{bc}-I_{ab}\[0.5cm] =8.32 A ∠-173.13° - 20.8 A ∠0°\[0.5cm] =(-8.26 A - j 1 A) - 20.8 A\[0.5cm] =-8.26 A - 20.8 A - j 1 A = -29.06 A - j 1 A=29.08 A ∠−178.03°\[0.5cm] I_{Cc}=I_{ca}-I_{bc}\[0.5cm] =12.26 A ∠165° - 8.32 A ∠-173.13°\[0.5cm] =(-11.84 A + j 3.17 A) - (-8.26 A - j 1 A)\[0.5cm] =-11.84 A + 8.26 A + j (3.17 A + 1 A) = -3.58 A + j 4.17 A=5.5 A ∠130.65°[/latex]

c. [latex]P_1=V_{ab}I_{Aa}cos\theta^{V_{ab}}_{I_{Aa}}\qquad V_{ab}=208 V ∠0°\[0.5cm] I_{Aa}=32.79 A ∠-5.55°\[0.5cm] =(208 V)(32.79 A) cos 5.55°=6788.35 W\[0.5cm] V_{bc}=E_{BC}=208 V ∠-120°\[0.5cm] but\qquad V_{cb}=E_{CB}=208 V ∠-120° + 180°=208 V ∠60°\[0.5cm] with\qquad I_{Cc}=5.5 A ∠130.65°\[0.5cm] P_2=V_{cb}I_{Cc}cos\theta^{V_{cb}}_{I_{Cc}}\[0.5cm] =(208 V)(5.5 A) cos 70.65°=379.1 W[/latex]

d. [latex]P_T = P_1 + P_2 = 6788.35 W + 379.1 W=7167.45 W[/latex]

e. [latex]P_T=(I_{ab})^2R_1+(I_{bc})^2R_2+(I_{ca})^2R_3\[0.5cm] =(20.8 A)^2 10 Ω + (8.32 A)^2 15 Ω + (12.26 A)^2 12 Ω\[0.5cm] =4326.4 W + 1038.34 W + 1803.69 W=7168.43 W[/latex]

(The slight difference is due to the level of accuracy carried through the calculations.)

Question 24.8: For the unbalanced ∆-connected load in Fig. 24.33 with two p...

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