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Problem: A 250 resistor is connected inductance in parallel with an of 2.5 mH across a Vrms = 75<0° V, f = 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed.

Problem: A 250 resistor is connected inductance in parallel with an of 2.5 mH across a Vrms = 75<0° V, f = 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed.

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter17: Resistive-inductive Series Circuits
Section: Chapter Questions
Problem 2PA: You are a journeyman electrician working in an industrial plant. Your task is to connect an inductor...
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Circuit diagram V Phasor diagram IR d. 12.54<55.43° Q, e. P = 225 W a. IR = 3<0° A, IL = 4.0<-90° A a. IR = 3<0° A, IL = 4.77<-90° A d. 11.32<61.39° Q, e. P = 180 W c. phase angle or power factor angle = 61.39°, or power factor = cos(61.39°) = 0.479 lagging d. 13.32<57.83° Q, e. P = 224.82 W c. phase angle or power factor angle = 57.83°, or power factor = cos (57.83°) = 0.532 lagging b. l-total = 5<-53.13° A c. phase angle or power factor angle = 53.13°, or power factor = cos(53.13°) = 0.60 lagging b. l-total = 6.26<-61.39° A None of the above b. l-total = 5.63<-57.83° A a. IR 3<0° A, IL = 5.5<-90° A
Transcribed Image Text:Circuit diagram V Phasor diagram IR d. 12.54<55.43° Q, e. P = 225 W a. IR = 3<0° A, IL = 4.0<-90° A a. IR = 3<0° A, IL = 4.77<-90° A d. 11.32<61.39° Q, e. P = 180 W c. phase angle or power factor angle = 61.39°, or power factor = cos(61.39°) = 0.479 lagging d. 13.32<57.83° Q, e. P = 224.82 W c. phase angle or power factor angle = 57.83°, or power factor = cos (57.83°) = 0.532 lagging b. l-total = 5<-53.13° A c. phase angle or power factor angle = 53.13°, or power factor = cos(53.13°) = 0.60 lagging b. l-total = 6.26<-61.39° A None of the above b. l-total = 5.63<-57.83° A a. IR 3<0° A, IL = 5.5<-90° A
A 20 Q resistor is connected in parallel with an inductance of 2.387 mH (2.387E-03 H) across a 60 V, 1 kHz (1E+03 Hz) supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed. Solution: Using Phasors First, assume the supply voltage is RMS and as reference at angle 0. Vrms = 60<0° V a. Calculate the branch currents, IR and IL IR = Vrms/R = 60<0° V/20 Q = 3<0° A IL = Vrms/XL = 60<0° V/(2*1E+03 Hz*2.387E-03 H) = 4.0<-90° A b. l-total, vector sum of IR and IL I-total = 3<0° A + 4.0<-90° A = 5<-53.13° A c/d. You may determine first the impedance, Z Z = Vrms / I-total = 60<0° V/ 5<-53.13° A = 12<53.13° 0 the angle of impedance indicates the phase angle between the voltage and the current, answer is = 53.13°. It is also positive so the power factor is lagging (see the phasor diagram, V is at zero angle, I is at -53.13°, looking at counter clockwise motion of V and 1). Power factor = cos = cos(53.13°) = 0.60 lagging e. power consumed, true power P = |Vrms|*|Irms|*cos P = (60V)(5A)cos(53.13°) = 180 W or P = |IR|^2*R = (3^2) *20 Q = 180 W remember the average power, P for reactances XL and Xc are zero. Problem: A 25 Q resistor is connected in parallel with an inductance of 2.5 mH across a Vrms = 75<0° V, f = 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed.
Transcribed Image Text:A 20 Q resistor is connected in parallel with an inductance of 2.387 mH (2.387E-03 H) across a 60 V, 1 kHz (1E+03 Hz) supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed. Solution: Using Phasors First, assume the supply voltage is RMS and as reference at angle 0. Vrms = 60<0° V a. Calculate the branch currents, IR and IL IR = Vrms/R = 60<0° V/20 Q = 3<0° A IL = Vrms/XL = 60<0° V/(2*1E+03 Hz*2.387E-03 H) = 4.0<-90° A b. l-total, vector sum of IR and IL I-total = 3<0° A + 4.0<-90° A = 5<-53.13° A c/d. You may determine first the impedance, Z Z = Vrms / I-total = 60<0° V/ 5<-53.13° A = 12<53.13° 0 the angle of impedance indicates the phase angle between the voltage and the current, answer is = 53.13°. It is also positive so the power factor is lagging (see the phasor diagram, V is at zero angle, I is at -53.13°, looking at counter clockwise motion of V and 1). Power factor = cos = cos(53.13°) = 0.60 lagging e. power consumed, true power P = |Vrms|*|Irms|*cos P = (60V)(5A)cos(53.13°) = 180 W or P = |IR|^2*R = (3^2) *20 Q = 180 W remember the average power, P for reactances XL and Xc are zero. Problem: A 25 Q resistor is connected in parallel with an inductance of 2.5 mH across a Vrms = 75<0° V, f = 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed.
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