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 V S
R
C
VS
I
I leads V S
Impedance
• 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 
  tan 1 
R
R


• Z is the diagonal
XC
XC
Z
• It is convenient to reposition the phasors
into the impedance triangle.
Z
Impediance
Impediance
Impediance
Sketch the impedance triangle and show the
values for R = 1.2 kW and XC = 960 W.
Z
1.2 kW  +  0.96 kW 
2
 1.33 kW
  tan 1
 39
0.96 kW
1.2 kW
2
R = 1.2 kW

39o
Z = 1.33 kW
XC =
960 W
Examples
Examples
Analysis of Series RC
• Ohm’s law is applied to series RC circuits using Z, V, and
I.
V  IZ
I
V
Z
Z
V
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.
Examples
Examples
Analysis of Series RC
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 kW

39o
Z = 1.33 kW
x 10 mA
=
XC =
960 W
VR = 12 V

39o
VS = 13.3 V
VC =
9.6 V
Phase Relationships
Variation of Phase Angle
• 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, the
Increasing f



impedance triangle for an RC
Z
X
f
circuit changes as illustrated
Z
here because |XC| decreases
with increasing f. This
f
X
Z
determines the frequency
response of RC circuits.
f
X
3
2
1
3
C3
3
C2
2
C1
1
2
1
Examples
Examples
Examples
Examples
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

Vin
C
Vout
VR
Vout
f
(phase lag)
Vout
f
Vin
(phase lag)
Vin
Examples
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

Vin
(phase lead)
Vin
R
Vout
Vout
VC
Vin

(phase lead)
Examples
Parallel Capacitor Impedance
• Explain why the capacitance of parallel capacitors is the sum
of capacitance of each capacitor.
1

jX C
1
1
1
1


jX C1 jX C2 jX C3
C1
C2
C3
Sinusoidal Response of Parallel RC
• For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance) and to review
conductance.
G
• Conductance is the reciprocal of resistance.
• Capacitive susceptance is the reciprocal of
capacitive reactance.
BC 
1
XC
• Admittance is the reciprocal of impedance.
Y
1
Z
1
R
Admittance
Admittance
Sinusoidal Response of Parallel RC
• 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
  tan 1 
• From the diagram, the phase angle is
BC
Y
VS
G
BC

G
Sinusoidal Response of Parallel RC
• Some important points to notice are:
• G is plotted along the positive x-axis.
• BC is plotted along the positive y-axis.
 BC 

G
  tan 1 
• Y is the diagonal
BC
Y
VS
G
BC

G
Sinusoidal Response of Parallel RC
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 kW
Y  G 2 + BC 2 
BC  2 10 kHz  0.01 mF  0.628 mS
1.0 mS
2
+  0.628 mS  1.18 mS
2
BC = 0.628 mS
VS
f = 10 kHz
R
1.0 kW
C
0.01 mF
Y=
1.18 mS
G = 1.0 mS
Analysis of Parallel RC
• Ohm’s law is applied to parallel RC circuits using Y, V,
and I.
I
V
Y
I
I  VY Y 
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
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
• Notice that the formula for capacitive susceptance is the
reciprocal of capacitive reactance. Thus BC and IC are
directly proportional to f: BC  2 fC
• As frequency increases, BC and
IC must also increase, so the
angle between IR and IS must
increase.
IC
IS

IR
Equivalent Series and Parallel RC
• 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 

Z
XC(eq) = Z sin 
Examples
Examples
Phase Relationships
Examples
Examples
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
components in the green
box are in series: Z1  R12  X C21
The components in the
yellow box are in parallel:
R2 X C 2
Z2 
R22  X C2 2
R1
C1
R2
C2
• 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
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 kW

39o
Z = 1.33 kW
x 10 mA
=
XC =
960 W
VR = 12 V

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
=

39o
VS = 13.3 V
VC =
9.6 V
Ptrue = 120 mW

39o
Pa = 133 mVA
Pr = 96
mVAR
Examples
Examples
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 
• The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.
Apparent
• Apparent power consists of two components; a true
power component, that does the work, and a reactive
power component, that is simply power shuttled back
and forth between source and load.
• Some components such as
transformers, motors, and
generators are rated in VA
rather than watts.
Ptrue (W)

Pa (VA)
Pr (VAR)
Quiz
Impedance The total opposition to sinusoidal current expressed
in ohms.
Phase angle
The angle between the source voltage and the total
current in a reactive circuit.
Capacitive The ability of a capacitor to permit current; the
suceptance (BC) reciprocal of capacitive reactance. The unit is the
siemens (S).
Admittance (Y) A measure of the ability of a reactive circuit to
permit current; the reciprocal of impedance. The
unit is the siemens (S).
Selected Key Terms
Power factor The relationship between volt-amperes and
true power or watts. Volt-amperes multiplied
by the power factor equals true power.
Frequency In electric circuits, the variation of the output
response voltage (or current) over a specified range of
frequencies.
Cutoff The frequency at which the output voltage of
frequency a filter is 70.7% of the maximum output
voltage.
Quiz
1. If you know what the impedance phasor diagram
looks like in a series RC circuit, you can find the voltage
phasor diagram by
a. multiplying each phasor by the current
b. multiplying each phasor by the source voltage
c. dividing each phasor by the source voltage
d. dividing each phasor by the current
Quiz
2. A series RC circuit is driven with a sine wave. If the
output voltage is taken across the resistor, the
output will
a. be in phase with the input.
b. lead the input voltage.
c. lag the input voltage.
d. none of the above
Quiz
3. A series RC circuit is driven with a sine wave. If you
measure 7.07 V across the capacitor and 7.07 V across
the resistor, the voltage across both components is
a. 0 V
b. 5 V
c. 10 V
d. 14.1 V
Quiz
4. If you increase the frequency in a series RC circuit,
a. the total impedance will increase
b. the reactance will not change
c. the phase angle will decrease
d. none of the above
Quiz
5. Admittance is the reciprocal of
a. reactance
b. resistance
c. conductance
d. impedance
Quiz
6. Given the admittance phasor diagram of a parallel
RC circuit, you could obtain the current phasor
diagram by
a. multiplying each phasor by the voltage
b. multiplying each phasor by the total current
c. dividing each phasor by the voltage
d. dividing each phasor by the total current
Quiz
7. If you increase the frequency in a parallel RC circuit,
a. the total admittance will decrease
b. the total current will not change
c. the phase angle between IR and IS will decrease
d. none of the above
Quiz
8. The magnitude of the admittance in a parallel RC
circuit will be larger if
a. the resistance is larger
b. the capacitance is larger
c. both a and b
d. none of the above
Quiz
9. The maximum power factor occurs when the phase
angle is
a. 0o
b. 30o
c. 45o
d. 90o
Quiz
10. When power is calculated from voltage and current for
an ac circuit, the voltage and current should be expressed
as
a. average values
b. rms values
c. peak values
d. peak-to-peak values
Quiz
Answers:
1. a
6. a
2. b
7. d
3. c
8. d
4. c
9. a
5. d
10. b