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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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A single-phase transmission line of impedance j 0.8 Ω supplies a resistive load of 500 A at 300 V · The sending-end power factor is
Unity
0.8 lagging
0.8 leading
0.6 lagging
0.6 lagging
Given: Load voltage VL = 300 V, Load current I = 500 A, Line impedance ZL = j0.8 Ω, Load power factor = 1 (resistive).
Load voltage VL = 300 V, Load current I = 500 A, Line impedance ZL = j0.8 Ω, Load power factor = 1 (resistive).
VS=VL+I×ZL
Establish reference and determine Load Voltage
Since the load is resistive, the current is in phase with the load voltage · Let the load voltage be the reference: VL=300+j0 V.
VL=300∠0° V
Calculate Sending-end Voltage
The sending-end voltage VS is the sum of the load voltage and the voltage drop across the line impedance: VS=VL+I⋅ZL=300+(500∠0°)(j0.8).
VS=300+j400
Calculate the Phase Angle of Sending-end Voltage
Convert VS to polar form: VS=3002+4002∠tan−1(300400)=500∠53.13° V.
ϕS=53.13°
Determine Power Factor
The sending-end power factor is cos(ϕS)=cos(53.13°)=0.6. Since the current lags the sending-end voltage (or the impedance is inductive), it is lagging.
PFS=cos(53.13°)=0.6 lagging
D is correct because the sending-end voltage leads the current by an angle whose cosine is 0.6, resulting in a 0.6 lagging power factor.
In transmission line analysis, the sending-end power factor is always affected by the line reactance; an inductive line always contributes to a lagging power factor at the sending end regardless of the load type.