Question #149095
Three identical coils, each of resistance 15 Ω and inductance 42 mH are
connected (a) in star and (b) in delta to a 415 V, 50 Hz, 3-phase supply.
Determine the total power dissipated in each case.
1
Expert's answer
2020-12-06T17:18:58-0500

(a) Star connection

Inductive reactance

XL=2πfLXL=2π×50×42×103=13.19  ΩX_L = 2πfL \\ X_L = 2π \times 50 \times 42 \times 10^{-3} = 13.19 \;Ω

Phase impedance

Zph=R2+XL2Zph=152+13.192=19.97Z_{ph} = \sqrt{R^2 + X_L^2} \\ Z_{ph} = \sqrt{15^2 + 13.19^2} = 19.97

Line voltage

Vline=3VphVph=Vline3=4153=240  voltsV_{line} = \sqrt{3}V_{ph} \\ V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{415}{\sqrt{3}} = 240 \;volts

Phase current

Iph=VphZphIph=24019.97=12.02  AI_{ph } = \frac{V_{ph}}{Z_{ph}} \\ I_{ph} = \frac{240}{19.97} = 12.02 \;A

Line current

Iline=Iph=12.02  AI_{line} = I_{ph} = 12.02 \;A

Dissipated power

P=3Iline2RP=3×(12.02)2×15=6.6  kWP = 3I_{line}^2R \\ P = 3 \times (12.02)^2 \times 15 = 6.6 \;kW

(b) Delta connection

Vline=Vph=415  VZph=19.97Iph=VphZphIph=41519.97=20.78  AV_{line} = V_{ph} = 415 \;V \\ Z_{ph} = 19.97 \\ I_{ph} = \frac{V_{ph}}{Z_{ph}} \\ I_{ph} = \frac{415}{19.97} = 20.78 \;A

Dissipated power

P=3Iph2RP=3×20.782×15=11.22  kWP = \sqrt{3}I_{ph}^2R \\ P = \sqrt{3} \times 20.78^2 \times 15 = 11.22 \;kW


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Comments

Saidi mhone
15.07.23, 15:29

The way the problem solved is very good and l appreciate it. Thanks

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