Test: Balanced Three Phase Circuits - Electrical Engineering (EE) MCQ

As the line voltage V_RY = V∠0⁰ is taken as a reference phasor. Then the source voltage V_YB is V∠-120⁰.

In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.

As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the Y impedance is I_Y = V_YB∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

Taking the line voltage V_RY = V∠0⁰ as a reference V_RY = 400∠0⁰V, V_YB = 400 ∠ -120 ⁰V and V_BR = 400∠-240⁰V. Impedance per phase = (4 + j8) Ω = 8.94∠63.4⁰Ω. Phase current I_R = (400∠0o)/(8.94∠63.4o )= 44.74 ∠ -63.4⁰A.

As the line voltage V_RY = V∠0⁰ is taken as a reference phasor. Then the source voltage V_BR is V∠-240⁰.

As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is I_R = V_BR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

In wye or star connection, similar ends of the three phases are joined together within the alternator. The common terminal so formed is referred to as the neutral point or neutral terminal.

The voltage between line and line is called line voltage. And the voltage between line and neutral point is called phase voltage. The currents flowing through the phases are called the phase currents, while those flowing in the lines are called the line currents.

Concept:
Power dissipation by resistor is given as:

Where,

V_rms = RMS value of voltage

I_rms = RMS value of current

R = Resistance

V_m = Peak or maximum value of voltage

Calculation:

Given-

R = 1 Ω, V₁ = 200∠0°, V₂ = 100∠120°, V₃ = 100∠-120°  

All three voltage sources are connected in series, therefore equivalent voltage is

V = V₁ + V₂ + V₃ = 200∠0° + 100∠120° + 100∠-120°  

V = 200(cos0° + j sin0°) + 100(cos120° + j sin120°) + 100[cos(-120°) + j sin(-120°)]

V = 100∠0° Volts

Since, R = 1 Ω:

∴ V_rms = I_rms
The power dissipation in the resistor will be:

P = 5000 W