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Problem: A series RLC circuit is compose of R = 50 , XL = j40 , Xc = -j15 Q. The supply voltage is Vrms = 240<0° V, find the impedance, current, and power factor of the circuit. O pf = cos(45°) = 0.707 lagging O 1=4.92<26.57° Q OZ = 55.90<26.57° 0 1=4.92<-90° 2 O NFTOC OZ = 55.90<-26.57° Opf = cos(26.57°) = 0.894 leading pf = cos(0°) = 1, unity (the current and voltage are in-phase) O 1=4.92<-26.57° 0 O pf = cos(26.57°) = 0.894 lagging O Z = 55.90<90° Q

Problem: A series RLC circuit is compose of R = 50 , XL = j40 , Xc = -j15 Q. The supply voltage is Vrms = 240<0° V, find the impedance, current, and power factor of the circuit. O pf = cos(45°) = 0.707 lagging O 1=4.92<26.57° Q OZ = 55.90<26.57° 0 1=4.92<-90° 2 O NFTOC OZ = 55.90<-26.57° Opf = cos(26.57°) = 0.894 leading pf = cos(0°) = 1, unity (the current and voltage are in-phase) O 1=4.92<-26.57° 0 O pf = cos(26.57°) = 0.894 lagging O Z = 55.90<90° Q

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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The standard form of inductance reactance in phasor XL = jwL = j(2πfL) Q but j is 1<90°, also XL = 2πfL<90° The standard form of capacitive reactance in phasor Xc = 1/(jwC) = 1 / (j*2πfC) = -j/ (2TTfC) but -j is 1<-90°, so Xc = 1<-90°/(2TfC) Ω Example: 3 inductive reactance are connected in series, XL1 = j40 , XL2 = j25 Q2, XL3 = j35 Q, the total reactance is j40 + j25 +j35 Q = j100 Ω Example: 3 capacitive reactance are connected in series, Xc1 = -j50 , Xc2 = -j60 Q2, Xc3 = -j40 , the total reactance is-j50 Ω -j60 Ω,-j40 Ω = - j150 Ω Possibly, impedance Z can have resistance and reactances. Z = R + jX; R is the real part and represents the value of resistance, ±jX represents the value of reactance (it can be inductive when positive or capacitive when negative) Example, A series circuit is composed of R = 5 Q2, XL = j20 02, Xc = -j30 Q Z = 5+j20-j30 = 5-j10 02 (this means the circuit is capacitive), In polar form Z = 11.18<-63.43° Q Notes: 1. When the angle of the impedance Z (in polar form) is positive, the circuit is inductive and the power factor is lagging (the current lags behind the voltage by a certain angle). Example: Vrms = 230<0° V, Z = 30<35°, find I. Irms = Vrms/Z = 230<0° V/30<35° = 7.67<-35° A The angle between the voltage and the current is 35° (absolute), but the current lags behind the voltage (V is along positive x-axis, while I is in 4th quadrant) 2. When the angle of the impedance Z (in polar form) is negative, the circuit is capacitive and the power is leading (the current leads the voltage by a certain angle). Example: Vrms = 230<0° V, Z = 25<-55°, find I. Irms = Vrms / Z = 230<0° V / 25<-55° = 9.2<55° A The angle between the voltage and the current is 55° (absolute), but the cur the voltage (V is along positive x-axis, while I is in 1st quadrant) Problem: A series RLC circuit is compose of R = 50 Q2, XL = j40 Q2, Xc = -j15 2. The supply voltage is Vrms = 240<0° V, find the impedance, current, and power factor of the circuit. Opf = cos(45°) = 0.707 lagging 1 = 4.92<26.57° Q OZ = 55.90<26.57° Q 14.92<-90° Q ONFTOC OZ = 55.90<-26.57° Q Opf = cos(26.57°) = 0.894 leading pf = cos(0°) = 1, unity (the current and voltage are in-phase) 0 1 = 4.92<-26.57° Ω Opf = cos(26.57°) = 0.894 lagging OZ = 55.90<90° Q
Transcribed Image Text:The standard form of inductance reactance in phasor XL = jwL = j(2πfL) Q but j is 1<90°, also XL = 2πfL<90° The standard form of capacitive reactance in phasor Xc = 1/(jwC) = 1 / (j*2πfC) = -j/ (2TTfC) but -j is 1<-90°, so Xc = 1<-90°/(2TfC) Ω Example: 3 inductive reactance are connected in series, XL1 = j40 , XL2 = j25 Q2, XL3 = j35 Q, the total reactance is j40 + j25 +j35 Q = j100 Ω Example: 3 capacitive reactance are connected in series, Xc1 = -j50 , Xc2 = -j60 Q2, Xc3 = -j40 , the total reactance is-j50 Ω -j60 Ω,-j40 Ω = - j150 Ω Possibly, impedance Z can have resistance and reactances. Z = R + jX; R is the real part and represents the value of resistance, ±jX represents the value of reactance (it can be inductive when positive or capacitive when negative) Example, A series circuit is composed of R = 5 Q2, XL = j20 02, Xc = -j30 Q Z = 5+j20-j30 = 5-j10 02 (this means the circuit is capacitive), In polar form Z = 11.18<-63.43° Q Notes: 1. When the angle of the impedance Z (in polar form) is positive, the circuit is inductive and the power factor is lagging (the current lags behind the voltage by a certain angle). Example: Vrms = 230<0° V, Z = 30<35°, find I. Irms = Vrms/Z = 230<0° V/30<35° = 7.67<-35° A The angle between the voltage and the current is 35° (absolute), but the current lags behind the voltage (V is along positive x-axis, while I is in 4th quadrant) 2. When the angle of the impedance Z (in polar form) is negative, the circuit is capacitive and the power is leading (the current leads the voltage by a certain angle). Example: Vrms = 230<0° V, Z = 25<-55°, find I. Irms = Vrms / Z = 230<0° V / 25<-55° = 9.2<55° A The angle between the voltage and the current is 55° (absolute), but the cur the voltage (V is along positive x-axis, while I is in 1st quadrant) Problem: A series RLC circuit is compose of R = 50 Q2, XL = j40 Q2, Xc = -j15 2. The supply voltage is Vrms = 240<0° V, find the impedance, current, and power factor of the circuit. Opf = cos(45°) = 0.707 lagging 1 = 4.92<26.57° Q OZ = 55.90<26.57° Q 14.92<-90° Q ONFTOC OZ = 55.90<-26.57° Q Opf = cos(26.57°) = 0.894 leading pf = cos(0°) = 1, unity (the current and voltage are in-phase) 0 1 = 4.92<-26.57° Ω Opf = cos(26.57°) = 0.894 lagging OZ = 55.90<90° Q
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