Test 10.1 Three-Phase Generator

In a balanced 3-phase generator, the incremental phase difference between adjacent voltage sources is

45°
60°
90°
120°

Test 10.2 Y-Δ Configurations

What is V₁₂ between nodes 1 and 2?

V₁₂ = 208∠30°
V₁₂ = 120∠30°
V₁₂ = 208∠-60°
V₁₂ = 120∠-30°

Test 10.3 Three-Phase Power

In the balanced Y-Y configuration, the total power delivered to the Y- load varies with time as

cos(ωt).
cos(2ωt).
cos(2ωt + φ), where φ is the phase of Z_Y.
does not vary with time.

Test 10.4 Power-Factor Compensation

In 3-phase networks, power-factor compensation is used to

increase the average power consumed by the loads.
increase the reactive power in the circuit.
decrease the reactive power in the circuit.
decrease the average power delivered to the loads.

Test 10.5 Power-Factor Compensation

Given that ω = 377 rad/s, V_aN = 60∠30° V, V_bN = 60∠60° V, and V_cN = 60∠120° V, determine υ_ab(t).

υ_ab = 73.48 cos(377t - 45°) V
υ_ab = 31.06 cos(377t - 45°) V
υ_ab = 73.48 cos(377t + 45°) V
υ_ab = 31.06 cos(377t + 45°) V

Test 10.6 3-Phase Configuration

Given that Z_a = Z_b = Z_c = (10 - j5) Ω, determine I_L1.

I_L1 = 17.89 ∠51.56° A
I_L1 = 20 ∠26.56° A
I_L1 = 20 ∠51.56° A
I_L1 = 17.89 ∠-26.56° A

Test 10.7 Power

Compute total complex power supplied by the 3-phase source.

S_T = 3.42 ∠-8.13° kVA
S_T = 10.27 ∠8.13° kVA
S_T = 10.27 ∠-8.13° kVA
S_T = 3.42 ∠8.13° kVA

Test 10.8 Power Factor

A balanced 3-phase source provides 1200 V (rms) to three loads at ω = 377 rad/s. The total complex power delivered by the source is
S_T = (15000 + j12000) VA.
What is the power factor?

pf = 0.42
pf = 0.55
pf = 0.62
pf = 0.78

Test 10.9 Power Factor

A balanced 3-phase source provides 1200 V (rms) to three loads at ω = 377 rad/s. The total complex power delivered by the source is
S_T = (15000 + j12000) VA.
To raise the power factor to 0.9, a capacitor C is added across each load. What is the magnitude of C?

C = 11.64 μF
C = 5.82 μF
C = 2.91 μF
C = 1.45 μF

Test 10.10 Power Meter

Given that I_L1 = (7.87 + j3.16) A and Z_y = (15 - j5) Ω, what would power meter P₁ measure?

P₁ = 1.51 kW
P₁ = 0.75 kW
P₁ = 3.02 kW
P₁ = 6.04 kW

Test 10.11 Power Meter

Given that I_L3 = (-8.2 + j2.2) A and Z_y = (15 - j5) Ω, what would power meter P₂ measure?

P₂ = 2.16 kW
P₂ = 1.08 kW
P₂ = 0.54 kW
P₂ = 4.32 kW

Test 10.12 Y-Δ Transformation

The Y-load impedance and the Δ-load impedance are related by:

Z_Δ = (1/3)Z_y
Z_Δ = Z_y
Z_Δ = √3 Z_y
Z_Δ = 3Z_y

Test 10.13 3-Phase Motor

A balanced 3-phase source is connected to a 3-phase motor designed as a balanced Y-load. If the powers measured using the two-wattmeter method are P₁ = 1000 W and P₂ = 600 W, what is the complex power delivered by the source?

S_T = (400 - j346.4) VA
S_T = (1600 - j346.4) VA
S_T = (1600 - j692.82) VA
S_T = (1600 + j692.82) VA

Test 10.14 Impedance Phase Angle

A balanced 3-phase source is connected to a balanced Y-load. The two- wattmeter method yielded P₁ = 1200 W and P₂ = 1800 W. What is the phase angle of the load impedance?

φ = 76.4°
φ = 57.3°
φ = 19.1°
φ = 38.2°

Test 10.15 Impedance Phase Angle

A balanced 3-phase source is connected to a balanced Y-load. The two- wattmeter method yielded P₁ = 1200 W and P₂ = 1800 W. What is the complex power delivered by the source?

S_T = (3000 + j1039.2) VA
S_T = (3000 - j1039.2) VA
S_T = (600 + j519.6) VA
S_T = (600 - j519.6) VA