AC Series RL Circuit
Sections 16-1, 16-2, 16-3 and 16-7
Sinusoidal response of series RL circuits
Series RL Voltages
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Series RL Voltages
- Phasors
Phasor – a vector used to
represent a value that
constantly changes phase
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Example
Calculate the source voltage
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Example
Calculate the source voltage
𝑉𝑉𝑠𝑠 =
𝑉𝑉 2 + 𝑉𝑉 2 =
𝐿𝐿
𝑅𝑅
32 + 42 =
25 = 5𝑉𝑉
Example
Calculate the source voltage
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Sinusoidal response of series RL circuits
When both resistance and inductance are in a series circuit, the
phase angle between the applied voltage and total current is
between 0° and 90°, depending on the values of resistance and
reactance.
Impedance of series RL circuits (1 of 2)
In a series RL circuit, the total impedance is the phasor sum of R and jXL.
R is plotted along the positive x-axis.
XL is plotted along the positive y-axis (+j).
 XL 

 R 
θ = tan −1 
Z is the diagonal
It is convenient to reposition the
phasors into the impedance triangle.
Impedance of series RL circuits (2 of 2)
Example
Sketch the impedance triangle and show the values for R = 1.2 kΩ and XL =
960 Ω.
Z=
(1.2 kΩ ) + ( 0.96 kΩ )
2
= 1.54 kΩ
0.96 kΩ
1.2 kΩ
=
+38.7°
θ = tan −1
2
Analysis of series RL circuits
Ohm’s law is applied to series RL circuits using phasor quantities
of Z, V, and I.
=
V IZ
=
I
V
Z
Z=
V
I
Analysis of series RL circuits
Example
Assume the supply voltage Vs in the previous example is 10 V.
Sketch the phasor diagram.
Analysis of series RL circuits
Example
Assume the supply voltage Vs in the previous example is 10 V.
Sketch the phasor diagram.
𝐼𝐼 =
𝑉𝑉𝑆𝑆
= 𝟔𝟔. 𝟒𝟒𝟒𝟒 𝒎𝒎𝒎𝒎 ∠ − 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕
𝑍𝑍
𝑉𝑉𝐿𝐿 = 𝐼𝐼 ∗ 𝑗𝑗𝑋𝑋𝐿𝐿 = 𝐼𝐼 ∗ 𝑋𝑋𝐿𝐿 ∠90° = 𝟔𝟔. 𝟐𝟐𝟐𝟐𝟐𝟐 ∠𝟓𝟓𝟓𝟓. 𝟑𝟑𝟑
𝑉𝑉𝑅𝑅 = 𝐼𝐼 ∗ 𝑅𝑅 = 𝟕𝟕. 𝟕𝟕𝟕𝟕𝟕𝟕 ∠ − 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕
Example
Calculate the total impedance for the circuit
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Example
Calculate the total impedance for the circuit
𝑋𝑋𝐿𝐿 = 2𝜋𝜋𝑓𝑓𝐿𝐿 = 2𝜋𝜋 5𝐾𝐾𝐻𝐻𝑧𝑧 33𝑚𝑚𝐻𝐻 ≅ 1.04 𝑘𝑘Ω
𝑍𝑍𝑇𝑇 =
𝜃𝜃
𝑋𝑋𝐿𝐿2 + 𝑅𝑅2 = 1.83𝑘𝑘Ω
𝑋𝑋𝐿𝐿
−1
= tan
𝑅𝑅
= 34.7°
𝑍𝑍𝑇𝑇 = 1.83 𝑘𝑘Ω∠34.7°
Example
Calculate the circuit current
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Example
Calculate the circuit current
𝑍𝑍𝑇𝑇 =
𝜃𝜃
𝑋𝑋𝐿𝐿2 + 𝑅𝑅2 = 19.6𝑘𝑘Ω
𝑋𝑋𝐿𝐿
−1
= tan
𝑅𝑅
= 40°
𝑍𝑍𝑇𝑇 = 19.6 𝑘𝑘Ω ∠40°
𝐼𝐼 =
𝑉𝑉𝑆𝑆
𝑍𝑍𝑇𝑇
=
20𝑉𝑉
19.6𝑘𝑘∠40°
= 1.02 𝑚𝑚𝐴𝐴∠ − 40°
Exercise
Determine the voltage, current and impedance values for the circuit shown.
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Exercise
Determine the voltage, current and impedance values for the circuit shown.
𝑍𝑍𝑇𝑇 = 4.17𝑘𝑘 ∠58.1°
𝐼𝐼 = 1.44 𝑚𝑚𝐴𝐴 ∠ − 58.1°
VR=3.17V∠ − 58.1°
VL=5.19V∠31.9°
Or, if you take current as the reference:
𝑍𝑍𝑇𝑇 = 4.17𝑘𝑘 ∠58.1°
𝐼𝐼 = 1.44 𝑚𝑚𝐴𝐴 ∠0°
VR=3.17V∠0°
VL=5.19V∠90°
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can be drawn for only a single
frequency because XL is a function of frequency.
As frequency changes, the impedance triangle for an RL circuit changes as
illustrated here because XL increases with increasing f. This determines the
frequency response of RL circuits.
Example 16-3 in the Text
Determine the magnitude of total impedance and the phase
angle for each of the following frequencies:
(a) 10 kHz
(b) 20 kHz
(c) 30 kHz
Applications (1 of 2)
For a given frequency, a series RL circuit can be used to produce
a phase lead by a specific amount between an input voltage and
an output by taking the output across the inductor. This circuit is
also a basic high-pass filter, a circuit that passes high frequencies
and rejects all others.
Applications (2 of 2)
Reversing the components in the previous circuit produces a
circuit that is a basic lag network. This circuit is also a basic lowpass filter, a circuit that passes low frequencies and rejects all
others.
Summary
Power in AC Circuits
i
i
+
R v
p
VmIm
p
L
-
Resistive Load
-
Inductive Load
p(t)
0.5VmIm
i(t)
T
v(t)
p (t)
L
VI
Ave.
power
T
2
+
L v
i(t)
t
T
2
T
v(t)
-V I
Energy
released
Energy
stored
Energy
released
Energy
stored
t
AC Power
• For a purely resistive load, the active power, P = VRIR (W)
• Reactive power (VAR), QL = VLIL = IL2XL = VL2/XL
• Note: P’s and Q’s are added algebraically in series and/or parallel
circuits.
Power Triangle
Ι
+
E
-
R
+
V
-
RL Circuit
•
•
•
•
•
S
L
QL
θ
P
Power triangle
P = I2R, and Q L = I2XL
Apparent power, S = VI* = I2Z = V2/Z (VA)
In RL circuit, S = P + jQ L
P = S cos θ or VI cos θ
Q = S sin θ or VI sin θ
• cos θ is referred to as the power factor, pf =cos θ =
𝑷𝑷
𝑺𝑺
2
| S |= P + QL
2
Example 16-16 in the Text
Determine the true power P, reactive power Q, apparent power S and the power factor pf. Draw the
power triangle.
Textbook End of Chapter Problems
Section 16-2
• Problems 3, 4, 5, 6
Section 16-3
• Problems 10, 11, 12
Section 16-7
• Problems 38, 39