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.

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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2..) Please choose the correct answers from the choices below. Show your complete solution
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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