Analyzing a Series RLC Circuit A series RLC circuit has R = 410 2, L = 1.10 H, C = 3 µF. It is connected to an AC source with f = 60.0 Hz and AV = 150 V. max A series circuit consisting of a resistor, an inductor, and a capacitor connected to an AC source. EAUR AvL -Avc all R
Analyzing a Series RLC Circuit A series RLC circuit has R = 410 2, L = 1.10 H, C = 3 µF. It is connected to an AC source with f = 60.0 Hz and AV = 150 V. max A series circuit consisting of a resistor, an inductor, and a capacitor connected to an AC source. EAUR AvL -Avc all R
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Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![Analyzing a Series RLC Circuit
A series RLC circuit has R = 410 N, L = 1.10 H, C = 3 µF. It is connected to an AC source with f = 60.0 Hz and AV,
max
150 V.
A series circuit consisting of a resistor,
an inductor, and a capacitor connected
to an AC source.
ll
R
L
C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2b9fbfb-992e-4207-8e5e-c107bb952d65%2F1c66c2fd-e591-4de9-baf6-305cb9920505%2Fog2oxku_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Analyzing a Series RLC Circuit
A series RLC circuit has R = 410 N, L = 1.10 H, C = 3 µF. It is connected to an AC source with f = 60.0 Hz and AV,
max
150 V.
A series circuit consisting of a resistor,
an inductor, and a capacitor connected
to an AC source.
ll
R
L
C
![(e) What replacement value of
circuit stay the same.
should an engineer analyzing the circuit choose such that the current leads the applied voltage by 30.0°? All other values in the
SOLUTION
(For the next three answers, do not substitute numerical values; use variables only.)
-1
7.
Solve o
tan
for the inductive reactance (Use the following as necessary: Xc, R, w and o.)
R
X =
Substitute expressions for inductive reactance and capacitive reactance into your expression (Use the following as necessary: C, R, and w.):
wL =
+ R tan(o)
Solve for L (Use the following as necessary: C, R,
W,
and p.):
L =
Substitute the given values (Give your answer in H.):
L =
H
Finalize Because the capacitive reactance is larger than the inductive reactance, the circuit is more capacitive than inductive. In this case, the phase angle o is
negative, so the current leads
the applied voltage. By applying the phase relationship to its instantaneous voltage, the instantaneous voltages (in
volts) across the three elements are:
j
v )sin(377t)
AVR
AVL
V )cos(377t)
Avc =
v cos(377t)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2b9fbfb-992e-4207-8e5e-c107bb952d65%2F1c66c2fd-e591-4de9-baf6-305cb9920505%2Fs3a3beo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(e) What replacement value of
circuit stay the same.
should an engineer analyzing the circuit choose such that the current leads the applied voltage by 30.0°? All other values in the
SOLUTION
(For the next three answers, do not substitute numerical values; use variables only.)
-1
7.
Solve o
tan
for the inductive reactance (Use the following as necessary: Xc, R, w and o.)
R
X =
Substitute expressions for inductive reactance and capacitive reactance into your expression (Use the following as necessary: C, R, and w.):
wL =
+ R tan(o)
Solve for L (Use the following as necessary: C, R,
W,
and p.):
L =
Substitute the given values (Give your answer in H.):
L =
H
Finalize Because the capacitive reactance is larger than the inductive reactance, the circuit is more capacitive than inductive. In this case, the phase angle o is
negative, so the current leads
the applied voltage. By applying the phase relationship to its instantaneous voltage, the instantaneous voltages (in
volts) across the three elements are:
j
v )sin(377t)
AVR
AVL
V )cos(377t)
Avc =
v cos(377t)
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