Chapter 12
RL Circuits
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C.Y. Lee
Objectives
Describe the relationship between current and
voltage in an RL circuit
Determine impedance and phase angle in a series
RL circuit
Analyze a series RL circuit
Determine impedance and phase angle in a parallel
RL circuit
Analyze a parallel RL circuit
Analyze series-parallel RL circuits
Determine power in RL circuits
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Sinusoidal Response of
RL Circuits
The inductor voltage leads the source voltage
Inductance causes a phase shift between voltage
and current that depends on the relative values of
the resistance and the inductive reactance
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Impedance and Phase Angle of
Series RL Circuits
The phase angle is the phase difference between
the total current and the source voltage
The impedance of a series RL circuit is
determined by the resistance (R) and the inductive
reactance (XL)
(ZL= jωL= jXL)
(Z= R+ZL= R+jXL)
(Z = R + jXL)
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Impedance and Phase Angle of
Series RL Circuits
In the series RL circuit, the total impedance is the
phasor sum of R and jXL
Impedance magnitude: Z = √ R2 + X2L
Phase angle: θ = tan-1(XL/R)
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Impedance and Phase Angle of
Series RL Circuits
Example: Determine the impedance and the phase angle
Z = √(5.6)2 + (10)2 = 11.5 kΩ
θ = tan-1(10/5.6) = tan-1(1.786) = 60.8°
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Analysis of Series RL Circuits
Application of Ohm’s Law to series RL circuits
involves the use of the quantities Z, V, and I as:
V = IZ
V
I=
Z
V
Z=
I
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Analysis of Series RL Circuits
Example: If the current is 0.2 mA, determine the source
voltage and the phase angle
XL = 2π(10×103)(100×10-3) = 6.28 kΩ
Z = √(10×103)2 + (6.28×103)2 = 11.8 kΩ
VS = IZ = (200µA)(11.8kΩ) = 2.36 V
θ = tan-1(6.28k/10k) = 34.2°
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Relationships of I and V in a
Series RL Circuit
In a series circuit, the current is the same through
both the resistor and the inductor
The resistor voltage is in phase with the current, and
the inductor voltage leads the current by 90°
I
VR
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VL
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KVL in a Series RL Circuit
I
From KVL, the sum of the
voltage drops must equal
the source voltage (VS)
Since VR and VL are 90°
out of phase with each
other, they must be added
as phasor quantities
VR
VL
VL= I(jXL)
VS= IZ = I(R+jXL)
VS = √V2R + V2L
θ = tan-1(VL/VR)
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VR= IR
I
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KVL in a Series RL Circuit
Example: Determine the source voltage and the phase angle
VS = √(50)2 + (35)2 = 61 V
θ = tan-1(35/50) = tan-1(0.7) = 35°
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Variation of Impedance and
Phase Angle with Frequency
For a series RL circuit;
as frequency increases:
– R remains constant
– XL increases
– Z increases
– θ increases
Z
f
R
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Impedance and Phase Angle of
Parallel RL Circuits
Total impedance in parallel RL circuit:
Z = (RXL) / (√ R2 +X2L)
Phase angle between the applied V and the total I:
1
1
1
-1
=
+
θ = tan (R/XL)
Z
R jX
L
1
jX L + R
=
Z
R ( jX L )
Z =
RX L
X L − jR
⎛ RX L ⎞
⎟⎟
V = IZ = I ⎜⎜
⎝ X L − jR ⎠
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Conductance, Susceptance and
Admittance
Conductance is the reciprocal of resistance:
G = 1/R
Inductive susceptance is the reciprocal of
inductive reactance:
BC = 1/XL
Admittance is the reciprocal of impedance:
Y = 1/Z
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Ohm’s Law
Application of Ohm’s Law to parallel RL circuits
using impedance can be rewritten for admittance
(Y=1/Z):
I
V=
Y
I = VY
I
Y=
V
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Relationships of the I and V in a
Parallel RL Circuit
The applied voltage, VS, appears across both the
resistive and the inductive branches
Total current, Itot, divides at the junction into the
two branch current, IR and IL
IR= V/R
Itot= V/Z
= V((XL−jR)/RXL)
IL= V/(jXL)
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KCL in a Parallel RL Circuit
From KCL, Total current
(IS) is the phasor sum of the
two branch currents
Since IR and IL are 90°
out of phase with each
other, they must be added
as phasor quantities
Itot = √
I2
R
+
I2
Itot= V/Z
= V((XL−jR)/RXL)
L
θ = tan-1(IL/IR)
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IR= V/R
IL= V/(jXL)
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KCL in a Parallel RL Circuit
Example: Determine the value of each current, and describe
the phase relationship of each with the source voltage
Vs
IR = 12/220 = 54.5 mA
IC = 12/150 = 80 mA
Itot = √(54.5)2 + (80)2 = 96.8 mA
θ = tan-1(80/54.5) = 55.7°
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Series-Parallel RL Circuits
An approach to analyzing circuits with combinations
of both series and parallel R and L elements is to:
– Calculate the magnitudes of capacitive reactances (XL)
– Find the impedance (Z) of the series portion and the
impedance of the parallel portion and combine them to get
the total impedance
– …
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RL Circuit as a Low-Pass Filter
An inductor acts as a short to dc
As the frequency is increased, so
does the inductive reactance
– As inductive reactance increases,
the output voltage across the
resistor decreases
– A series RL circuit, where output
is taken across the resistor, finds
application as a low-pass filter
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RL Circuit as a Low-Pass Filter
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RL Circuit as a High-Pass Filter
For the case when output voltage
is measured across the inductor
– At dc, the inductor acts a short,
so the output voltage is zero
– As frequency increases, so does
inductive reactance, resulting in
more voltage being dropped
across the inductor
– The result is a high-pass filter
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RL Circuit as a High-Pass Filter
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Summary
When a sinusoidal voltage is applied to an RL
circuit, the current and all the voltage drops are
also sine waves
Total current in an RL circuit always lags the
source voltage
The resistor voltage is always in phase with the
current
In an ideal inductor, the voltage always leads the
current by 90°
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Summary
In an RL circuit, the impedance is determined by
both the resistance and the inductive reactance
combined
The impedance of an RL circuit varies directly with
frequency
The phase angle (θ) if a series RL circuit varies
directly with frequency
In an RL lag network, the output voltage lags the
input voltage in phase
In an RL lead network, the output voltage leads the
input voltage in phase
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