Answer to Question #153582 in Electrical Engineering for ahmed

Question #153582

A balanced Y load having a 15+j25 Ω load impedance in each leg is connected to a three-phase, four-wire, Y-connected generator having a line voltage of 230 V. The line impedance per phase is 0.5+ j1.5 Ω . Answer the following:

a. Draw the circuit showing the given values.

b. Calculate the line voltages of the load.

c. Calculate the phase currents of the load.

d. Calculate the power consumed by the 3-phase load.

Expert's answer

The image attached shows the schematic.

b) Phase voltage of source

"=\\frac{230}{\\sqrt{3}} =132.79 V."

Phase voltage on load side"=132.79 \\times \\frac{15+j25}{(15+j25) +(0.5+j1.5) }=126.1-j1.41=126.1\\angle-0.64\\degree"

Hence line voltage on load"=126.1\\times \\sqrt{3}= 218.41V"

C) phase current of the load "=\\frac{132.79}{(15+j25) +(0.5+j1.5) } = 4.32\\angle-59.68A"

d) power consumed by 3 phase load="3\\times i^2\\times Rload=3\\times4.32^2\\times15=839.81W"

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