Chapter 10
Sinusoidal response of RC circuits
When both resistance and capacitance 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.
VR
VC
V R leads VS
V C lags VS
R
C
VS
I
I leads V S
Impedance of series RC circuits
In a series RC circuit, the total impedance is the phasor
sum of R and XC.
R is plotted along the positive x-axis.
XC is plotted along the negative y-axis.
 XC 

 R 
R
θ = tan −1 
R
θ
θ
Z is the diagonal
XC
XC
Z
It is convenient to reposition the
phasors into the impedance triangle.
Z
Impedance of series RC circuits
Sketch the impedance triangle and show the
values for R = 1.2 kΩ and XC = 960 Ω.
Z=
(1.2 kΩ )
2
+ ( 0.96 kΩ )
= 1.33 kΩ
θ = tan −1
= 39°
0.96 kΩ
1.2 kΩ
2
R = 1.2 kΩ
θ
39o
Z = 1.33 kΩ
XC =
960 Ω
Analysis of series RC circuits
Ohm’s law is applied to series RC circuits using Z,
V, and I.
V = IZ
V
I=
Z
V
Z=
I
Because I is the same everywhere in a series circuit,
you can obtain the voltages across different
components by multiplying the impedance of that
component by the current as shown in the following
example.
Analysis of series RC circuits
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasor diagram. The impedance
triangle from the previous example is shown for reference.
The voltage phasor diagram can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
R = 1.2 kΩ
θ
39o
Z = 1.33 kΩ
x 10 mA
=
XC =
960 Ω
VR = 12 V
θ
39o
VS = 13.3 V
VC =
9.6 V
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can only
be drawn for a single frequency because X is a
function of frequency.
R
As frequency changes,
Increasing f
θ
θ
θ
the impedance triangle
Z
X
f
for an RC circuit changes
Z
as illustrated here
because XC decreases
f
X
Z
with increasing f. This
determines the frequency
f
X
response of RC circuits.
3
2
1
3
C3
3
C2
2
C1
1
2
1
Applications
For a given frequency, a series RC circuit can be used to
produce a phase lag by a specific amount between an
input voltage and an output by taking the output across
the capacitor. This circuit is also a basic low-pass filter, a
circuit that passes low frequencies and rejects all others.
V
R
VR
Vout
θ
Vin
C
Vout
φ
(phase lag)
Vout
φ
Vin
(phase lag)
Vin
Applications
Reversing the components in the previous circuit produces
a circuit that is a basic lead network. This circuit is also a
basic high-pass filter, a circuit that passes high frequencies
and rejects all others. This filter passes high frequencies
down to a frequency called the cutoff frequency.
C
V
Vout
θ
Vin
(phase lead)
Vin
R
Vout
Vout
VC
Vin
θ
(phase lead)
Applications
An application showing how the phase-shift network is
useful is the phase-shift oscillator, which uses a
combination of RC networks to produce the required 180o
phase shift for the oscillator.
Amplifier
Rf
Phase-shift network
C
C
C
R
R
R
Sinusoidal response of parallel RC circuits
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
Conductance is the reciprocal of resistance.
Capacitive susceptance is the reciprocal
of capacitive reactance.
1
G=
R
1
BC =
XC
1
Admittance is the reciprocal of impedance. Y =
Z
Sinusoidal response of parallel RC circuits
In a parallel RC circuit, the admittance phasor is the sum
of the conductance and capacitive susceptance phasors.
The magnitude can be expressed as Y = G 2 + BC 2
 BC 

G 
From the diagram, the phase angle is θ = tan −1 
BC
Y
VS
G
BC
θ
G
Sinusoidal response of parallel RC circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
 BC 

G 
θ = tan −1 
Y is the diagonal
BC
Y
VS
G
BC
θ
G
Sinusoidal response of parallel RC circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G=
1
1
=
= 1.0 mS
R 1.0 kΩ
Y = G 2 + BC 2 =
BC = 2π (10 kHz )( 0.01 µ F ) = 0.628 mS
(1.0 mS)
2
2
+ ( 0.628 mS) = 1.18 mS
BC = 0.628 mS
VS
R
C
f = 10 kHz
1.0 kΩ
0.01 µF
Y=
1.18 mS
G = 1.0 mS
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using Y,
V, and I.
V=
I
Y
I = VY Y =
I
V
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.
Analysis of parallel RC circuits
If the voltage in the previous example is 10 V, sketch the
current phasor diagram. The admittance diagram from the
previous example is shown for reference.
The current phasor diagram can be found from
Ohm’s law. Multiply each admittance phasor by 10 V.
BC = 0.628 mS
Y=
1.18 mS
G = 1.0 mS
x 10 V
=
IC = 6.28 mA
IS =
11.8 mA
IR = 10 mA
Phase angle of parallel RC circuits
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: BC = 2π fC
As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.
IC
IS
θ
IR
Equivalent series and parallel RC circuits
For every parallel RC circuit there is an equivalent
series RC circuit at a given frequency.
The equivalent resistance and capacitive
reactance are shown on the impedance triangle:
Req = Z cos θ
θ
Z
XC(eq) = Z sin θ
Series-Parallel RC circuits
Series-parallel RC circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
series and parallel circuits.
Z1
Z2
For example, the
R1
C1
components in the
R2
C2
green box are in
series: Z1 = R12 + X C21
The components in
the yellow box are
R2 X C 2
in parallel: Z 2 =
R22 + X C2 2
The total impedance can be found by
converting the parallel components to an
equivalent series combination, then
adding the result to R1 and XC1 to get the
total reactance.
The power triangle
Recall that in a series RC circuit, you could multiply
the impedance phasors by the current to obtain the
voltage phasors. The earlier example is shown for
review:
R = 1.2 kΩ
θ
39o
Z = 1.33 kΩ
x 10 mA
=
XC =
960 Ω
VR = 12 V
θ
39o
VS = 13.3 V
VC =
9.6 V
The power triangle
Multiplying the voltage phasors by Irms gives the power triangle
(equivalent to multiplying the impedance phasors by I2). Apparent
power is the product of the magnitude of the current and magnitude of
the voltage and is plotted along the hypotenuse of the power triangle.
The rms current in the earlier example was 10 mA.
Show the power triangle.
VR = 12 V
x 10 mA
=
θ
39o
VS = 13.3 V
VC =
9.6 V
Ptrue = 120 mW
θ
39o
Pa = 133 mVA
Pr = 96
mVAR
Power factor
The power factor is the relationship between the
apparent power in volt-amperes and true power in
watts. Volt-amperes multiplied by the power factor
equals true power.
Power factor is defined mathematically as
PF = cos θ
The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.