Slide 3.0CE164 - 2024 F. Sepulveda - CSEE - Essex University
Foundations of Electronics – II
Lecture 3 (Week 18)
RC circuits - Part 2
 Parallel RC
 Power
 Analytical phasor method
Inductors
Francisco Sepulveda
E-mail: f.sepulveda
CE164
moodle.essex.ac.uk/course/view.php?id=3644
Slide 3.1CE164 - 2024 F. Sepulveda - CSEE - Essex University
Cont. from last week:
Sinusoidal Steady-state
& RC circuit power
Slide 3.2CE164 - 2024 F. Sepulveda - CSEE - Essex University
For parallel circuits, it is useful to introduce two new
quantities (susceptance and admittance), and to
revisit conductance.
Conductance is the reciprocal of
resistance.
1
G
R

Admittance is the reciprocal of
impedance.
Capacitive susceptance is the
reciprocal of capacitive
reactance.
1
C
C
B
X

1
Y
Z

Sinusoidal response of parallel RC circuits
Slide 3.3CE164 - 2024 F. Sepulveda - CSEE - Essex University
In a parallel RC circuit, the admittance phasor is the sum of
the conductance and capacitive susceptance phasors. The
magnitude can be expressed as 2 2 + CY G B
VS G
BC
Y
G
BC
q
From the diagram, the phase angle is 1tan C
B
G
q     
 
Sinusoidal response of parallel RC circuits
Slide 3.4CE164 - 2024 F. Sepulveda - CSEE - Essex University
VS G
BC
Y
G
BC
q
G is plotted along the positive horizontal axis
BC is plotted along the positive vertical axis
1tan C
B
G
q     
 
Y is the diagonal
Other important points to notice are:
Sinusoidal response of parallel RC circuits
Slide 3.5CE164 - 2024 F. Sepulveda - CSEE - Essex University
VS
C = 0.01 mF
Y =
1.18 mS
G = 1.0 mS
BC = 0.628 mS
Draw the admittance phasor diagram for the circuit below.
f = 10 kHz
R
1.0 kW
1 1
1.0 mS
1.0 k
G
R
  
W
   2 10 kHz 0.01 F 0.628 mSCB  m 
The magnitude of the conductance and susceptance are:
   2 22 2 + 1.0 mS + 0.628 mS 1.18 mSCY G B  
Sinusoidal response of parallel RC circuits
Slide 3.6CE164 - 2024 F. Sepulveda - CSEE - Essex University
Ohm’s law is applied to parallel RC circuits using Y, V,
and I.

I I
V I VY Y
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 circuits
remember: use rms for I and V in AC systems
Slide 3.7CE164 - 2024 F. Sepulveda - CSEE - Essex University
If the voltage in the previous example (slide 3.5) 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 10V.
x 10 V
=Y =
1.18 mS
G = 1.0 mS
BC = 0.628 mS
IR = 10 mA
IC = 6.28 mA
IS =
11.8 mA
Analysis of parallel RC circuits
Slide 3.8CE164 - 2024 F. Sepulveda - CSEE - Essex University
Notice that the formula for capacitive susceptance is the
reciprocal of capacitive reactance. Thus BC and IC are directly
proportional to f :
2CB fC
IR
IC
IS
q
As frequency increases, BC and IC
must also increase, so the angle
between IR and IS must increase.
Phase angle of parallel RC circuits
i.e.:
q increases as f increases
q decreases as f decreases
Slide 3.9CE164 - 2024 F. Sepulveda - CSEE - Essex University
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:
Z
Req = Z cos q
XC(eq) = Z sin q
q
Equivalent series and parallel RC circuits
R XC
2 2 + CY G B
Z = 1 / Y
Vs
XC (eq)
R eq
Slide 3.10CE164 - 2024 F. Sepulveda - CSEE - Essex University
Series-parallel RC circuits are combinations of both series
and parallel elements. These circuits can be solved by
methods from series and parallel circuits.
For example, the
components in the
green box are in
series:
The components in
the yellow box are in
parallel: 2 2
2
2 2
2 2
C
C
R X
Z
R X


R1 C1
R2 C2
Z1
Z2
The total impedance can be found by
converting the parallel components to an
equivalent series combination Z2, then
adding the result to Z1 to get the total
reactance.
2 2
1 1 1C
Z R X 
Series+Parallel RC circuits
Slide 3.11CE164 - 2024 F. Sepulveda - CSEE - Essex University
Recall that in a series RC circuit, you could multiply the impedance
phasors by the current to obtain the voltage phasors. An earlier
example (Lec. 2) is shown for review:
VR = 12 V
VC = 9.6 V
x 10 mA
=
R = 1.2 kW
XC =
960 WZ = 1.54 kW
39o
VS = 15.4 V
39o
q q
The power (phasor) triangle
Slide 3.12CE164 - 2024 F. Sepulveda - CSEE - Essex University
Multiplying the voltage phasors by Irms gives the power
triangle (equivalent to multiplying the impedance phasors
by I 2). Apparent power (Pa) 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.
Ptrue = 120 mW
Pr = 96
mVAR
x 10 mA
= 39
o
Pa = 154 mVA
VR = 12 V
VC =
9.6 VVS = 15.4 V
39o
q q
• reactive power Pr in VAR (volt-ampere-reactive)
• apparent power Pa in VA (volt-ampere)
The power (phasor) triangle
Slide 3.13CE164 - 2024 F. Sepulveda - CSEE - Essex University
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.
The power factor is defined mathematically as
PF = cos q
The power factor can vary from 0 for a purely
reactive circuit to 1 for a purely resistive circuit.
AC Power factor
Ptrue = Pa cos q
Slide 3.14CE164 - 2024 F. Sepulveda - CSEE - Essex University
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.
Ptrue (W)
Pr (VAR)
Some components
such as transformers,
motors, and generators
are rated in VA rather
than watts.
Pa (VA)
q
Apparent power
Slide 3.15CE164 - 2024 F. Sepulveda - CSEE - Essex University
Chp. 13, Boylestad (2003)
Phasor Angular Frequency
(for v and i, a.k.a. angular velocity*)
s = Am sin(w t ± q )
angular velocity w : how many wave cycles go through per second:
w  a / t  2  f
* For mechanical systems, angular velocity = Da/Dt, where Da may not be 2.
Converting phase difference to time:
Dt = Dq
2f
Slide 3.16CE164 - 2024 F. Sepulveda - CSEE - Essex University
RC Phasors:
Analytical Approach
i.e., using equations only
C
Z = R – j XC Z = |Z| ∠ q
Z = |Z| (cos(q) + j sin(q))
|Z| = sqrt( R2 + XC2)
VS G
BC Y = R + j BC Y = |Y| ∠ q
Y = |Y| (cos(q) + j sin(q))
|Y| = sqrt( G2 + BC
2)
Rectangular Polar
form form
Slide 3.17CE164 - 2024 F. Sepulveda - CSEE - Essex University
Examples:
Z = RT – j XCT
= R1 + R2 – j (XC1 + XC2)
= 147 kW – j (159 kW + 72 kW )
= 147 kW – j 231 kW
|Z| = Sqrt (RT
2 + XCT
2) = 274 kW
q = atan (XCT / RT) = 1.00 rad = 57.5o
Z = 274kW ∠ – 57.5o
Z = R – j XCT
= R – j / (1/XC1 + 1/XC2)
= 10kW – j ( 14.2kW // 14.2kW )
= 10kW – j 7.1 kW
|Z| = Sqrt (RT
2 + XCT
2) = 12.3 kW
q = atan (XCT / RT) = 26.8o
Z = 12.3 kW ∠ – 26.8o
Analytical solutions:
(from PS2)
Slide 3.18CE164 - 2024 F. Sepulveda - CSEE - Essex University
Y = G + j BC
= 1.0 mS + j (2 x  x f x C)
= 1.0 mS + j 0.63 mS
|Y| = Sqrt (G2 + BCT
2) = 1.2 mS
q = atan (BC / G) = 32 o
Y = 1.2 mS ∠ +32o (positive angle for Y)
|Z| = 1 / |Y| = 834 W
Z = 834 W ∠ – 32o (negative angle for Z)
P = Vs
2 / |Z| = 100 / 834 W = 120 mW ∠ –32o
VS
C = 0.01 mFf = 10 kHz
R
1.0 kW
Example: (from slide 3.5)
Find the impedance, phase, and power when Vs = 10Vrms
Analytical solution:
Slide 3.19CE164 - 2024 F. Sepulveda - CSEE - Essex University
Inductors:
The basics
Reading:
Chapter 11 in Floyd & Buchla (2014, 8th ed.)
Slide 3.20CE164 - 2024 F. Sepulveda - CSEE - Essex University
When a length of wire is formed into a coil, it becomes a
basic inductor. When there is current in the inductor, a
three-dimensional magnetic field is created.
A change in current
causes the magnetic
field to change. This in
turn induces a voltage
across the inductor that
opposes the original
change in current.
NS
The Basic Inductor
Slide 3.21CE164 - 2024 F. Sepulveda - CSEE - Essex University
The effect of inductance is
greatly magnified by adding
turns and winding them on
a magnetic material. Large
inductors and transformers
are wound on a core to
increase the inductance. Magnetic
core
One Henry is the inductance of a coil when a current,
changing at a rate of one ampere per second,
induces one volt across the coil. Most coils are much
smaller than 1 H.
The Basic Inductor
Slide 3.22CE164 - 2024 F. Sepulveda - CSEE - Essex University
Faraday’s law will be revisited in two weeks, but it
is briefly introduced here because of its importance
to inductors.
The amount of voltage induced in a coil is
directly proportional to the rate of change of the
magnetic field* with respect to the coil.
Faraday’s law
* note: the magnetic field in a non-magnetizable
conductor is created by the motion of electrons,
i.e., the current.
Slide 3.23CE164 - 2024 F. Sepulveda - CSEE - Essex University
Lenz’s law will also be revisited later in the module
and is an extension of Faraday’s law, defining the
direction of the induced voltage:
When the current through a coil changes and an
induced voltage is created as a result of the
changing magnetic field, the direction of the
induced voltage is such that it always opposes
the change in the current.
Lenz’s law
Slide 3.24CE164 - 2024 F. Sepulveda - CSEE - Essex University
A basic circuit to demonstrate Lenz’s law is shown.
Initially SW is open and there is a small
current in the circuit through L and R1.

R1
SW
R2VS
L
+
 +
Lenz’s law
I
Slide 3.25CE164 - 2024 F. Sepulveda - CSEE - Essex University
SW closes and immediately a voltage appears
across L that tends to oppose any change in
current.


R1
SW
R2VS
L
+
+
 +
Initially, the meter
reads the same
current as before
the switch was
closed.
Lenz’s law
I
Slide 3.26CE164 - 2024 F. Sepulveda - CSEE - Essex University
After some time, the current stabilizes at a
higher level (due to I2) as the voltage decays
across the coil.
+

R1
SW
R2VS
L
+
Later, the meter
reads a higher
current because of
the load change.
Lenz’s law
I
Slide 3.27CE164 - 2024 F. Sepulveda - CSEE - Essex University
In addition to inductance, actual inductors have winding
resistance (RW) due to the resistance of the wire and
winding capacitance (CW) between turns. An equivalent
circuit for a practical inductor including these effects is
shown:
L
RW
CW
Notice that the winding
resistance is in series with
the coil and the winding
capacitance is in parallel
with both.
Practical inductors
The effects of RW will be discussed later today and next week w.r.t. RCL filters
Slide 3.28CE164 - 2024 F. Sepulveda - CSEE - Essex University
Common symbols for inductors (coils) are
Air core Iron core Ferrite core Variable
There are a variety of inductors, depending on the amount of
inductance required and the application. Some, with fine
wires, are encapsulated and may appear like a resistor.
Types of inductors
Slide 3.29CE164 - 2024 F. Sepulveda - CSEE - Essex University
Inductors come in a variety of sizes. A few common
ones are shown here.
Encapsulated Torroid coil Variable
Practical inductors
Slide 3.30CE164 - 2024 F. Sepulveda - CSEE - Essex University
Four factors affect the amount of inductance for a
coil. The equation for the inductance of a coil is
2
N A
L
l
m

where
L = inductance in Henries
N = number of turns of wire
m = permeability in H/m (same as Wb/(A m))
l = coil length (i.e., the wound cylinder length) (m)
Factors affecting inductance
Slide 3.31CE164 - 2024 F. Sepulveda - CSEE - Essex University
What is the inductance of a 2cm long, 150 turn
coil wrapped on an low carbon steel core that is 5mm
diameter?
The permeability of low carbon steel is ~3x104 H/m
(Wb/(A m)).
 22 5 2π π 0.0025 m 7.85 10 mA r    
2N A
L
l
m

= 6.6 mH
1 96
Slide 3.32CE164 - 2024 F. Sepulveda - CSEE - Essex University
When inductors are connected in series, the total
inductance is the sum of the individual inductors. The
general equation for inductors in series is
2.18 mH
T 1 2 3 ... nL L L L L   
If a 1.5 mH inductor is
connected in series with
an 680 mH inductor, the
total inductance is
L1 L2
1.5 mH 680 Hm
Series inductors
Slide 3.33CE164 - 2024 F. Sepulveda - CSEE - Essex University
When inductors are connected in parallel, the total
inductance is smaller than the smallest one. The
general equation for inductors in parallel is
The total inductance of two inductors is
…or you can use the product-over-sum rule.
T
1 2 3 T
1
1 1 1 1
...
L
L L L L

   
T
1 2
1
1 1
L
L L


Parallel inductors
Slide 3.34CE164 - 2024 F. Sepulveda - CSEE - Essex University
If a 1.5 mH inductor is connected in parallel with
an 680 mH inductor, the total inductance is 468 mH
L1 L2
1.5 mH 680 Hm
Parallel inductors