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### 2018_EBook_QRN_Basic_CIrcuit_Analysis_Afandi_v2-edaran

Basic Circuit Analysis

Notes

Department of Computer Engineering
Faculty of Electrical & Electronic Engineering

Prepared By

Associate Professor Dr Afandi Bin Ahmad
Department of Computer Engineering
Faculty of Electrical & Electronic Engineering
Universiti Tun Hussein Onn Malaysia
afandia@ut hm.edu.my

Graphic & Revision

Mohd Shafie Bin Nemmang

All Right Reserved 2018
No part of this publication may be reproduced, distributed, or transmitted in
any form or by any means, including photocopying, recording, or other
electronic or mechanical methods, w ithout the prior w ritten permission of the
author and Univ ersiti Tun Hussein Onn Malaysia.

CONTENT 07
08
Circuit Variables & Elements 09
Resistive Circuits 10
Circuit Analysis 11
Network Theorem 12
Capacitor 13
Inductor 15
First Order Circuit
Second Order Circuit

Circuit Variables & Elements

Important Terminology SI Units C harg e

 Coulomb (C): The basic unit QUANTITY BASIC UNIT SYMBOL  Is an electrical property of the atomic
used to measure electric charge particles of which matter consist,
Length Meter M measured in Coulomb (C).
 Joule (J): A joule is the work
done by a constant 1-N force Mass Kilogram Kg q  to idt
applied through a 1-m distance. t
Time Second S
 Ampere (A): One ampere or amp Electric Current
is the current that flows when 1 Electric current Ampere A
Coulomb of charge passes each  Electric current is the time rate of
second (1 A = 1 C/s) Thermodynamic temperature Kelvin K change of charge, measured in
amperes (A).
 Volt (V): If a charge of 1 Coulomb Luminous intensity Candela cd
may be moved between two i  dq
points in space with expenditure Topic 1 dt
of 1 Joule of work, 1 Volt is said to Circuit Variables &
be a potential difference existing  Two common type of currents:
between these points (1 V = 1 J/ Elements
C)

 Watt (W): The rate at which work
is done or energy expended. The
watt is defined as 1 Joule per
second (1 J/s).

Direct current Alternating current

Passive and Active Elements Voltage

 The voltage vab between two points a and
b in an electric circuit is the energy (or

work) needed to move a unit charge from

a to b; mathematically,
 Passive element is an electrical element that
absorbs or stores energy.

 Examples of passive element: resistor,
inductor and capacitor.

[email protected] Currentdirection ofa passive elementstarts
Passive Sign Convention

from positive terminal to negative terminal.  Whenever reference direction  dw
for current direction in an dq
i element is in the direction of vab
reference voltage drop across
+ Passive element _ the element, use a positive

+v Ener gy

– sign in any expression that  Energy is the fundamental ability to do
relates voltage to current.
 Active element is an electrical element that

supplies energy to other elements in a circuit.  Otherwise, use a negative work and produce action.
 Examples of active element: voltage source, sign.  Power is a measure of how fast energy is

current source, transistor. being used.
 Current direction for an active element is  In other words, power is the rate at which

going out of positive terminal into the energy is used.

negative terminal. p  energy  W
time t
i

+ Active element _

+v Power

Electrical Sources  Power is the time rate of expending or
absorbing energy, measured in watts (W).
 An electrical source is a device that is converting
non-electrical energy to electrical energy and vice  +power absorbed = - power supplied
versa.  Power can be delivered or absorbed as

 The sources are either deliver or absorb electrical defined by the polarity of the voltage and the
power in order to maintain the voltage or current. direction of the current.

 There are two types of electrical sources; voltage -+ Power delivered or supplied
and current sources. V by voltage source

 There are four types of electrical sources used in I Power absorbed by resistor
circuit analysis.
1.Independent Sources +-
2.Ideal Voltage Sources V

 Ideal Current Sources Dependent (Controlled)
Sources
1.Voltage Sources
2.Current Sources

7 [email protected]

Resistive Circuits

Resistor Ohm Law Nodes, Branches, Loops

 Resistance: physical property or  Ohm's law states that the voltage, v A branch:
ability to resist current, behavior of across a resistor is directly proportional A single element such as a voltage source or a
resisting the flow of electric to the current, i flowing through the resistor
charge. resistor. Represents any two-terminal element

 Resistance factor:- vαi A branches:
The point of connection between two or more

length, l  R is the material property can change branches.
if the internal or external condition of Usually indicated by a dot in a circuit.

the element are altered. A Loops:
 To apply v=iR, the current and voltage Any closed path in circuit

must conform with the passive sign

cross - sectional area, A material with resistivity,  convention.
 This implies that the current flows from

Rρ l a higher potential to a lower potential Current Divider
A (v = iR).
 If current flows from a lower to high I
potential, v = -iR

 Possible value of R:- + I1 + I2 I1  R2 I
 If R = 0, short circuit V1 R1 V2 R2 (R1  R 2 )
 v = iR = 0, voltage is zero, current VS
- - I2  R1 I
could be anything Topic 2 (R1  R 2 )
 In practical, short circuit perfect Resistive Circuits

conductor resistance approaching

zero
 If R = infinity, open circuit
 Current is zero, voltage could be
Voltage Divider
anything
 Open circuit perfect conductor Equivalent Resistance R1 R2 R3 RN Rseries
+V-
resistance approaching infinity R1 R2 R3 R6
R7 (b)
[email protected]
+ V1 - + V2 - + V3 - + VN - Vs
Vs
Requivalent R4 R5
(a)

 Kirchoff's Current Law (KCL) VX  RX VS
states that the algebraic sum RT
of current s entering a node (or Req3 Req2 Req1
a closed loop boundary) is
zero. R eq1  (R 6  R 7 )Ω Source Transformation
R eq2  (R eq1//R 5 )Ω
 The sum of the currents R eq3  (R eq2//R 4 )Ω x VS  IS  R b
entering a node is equal to the  R equivalent  (R eq3  R1  R 2  R 3 )Ω Ra
sum of the currents leaving the VS Is Rb IS  VS
node. y Ra

I6 I1

I5 I2 Ra  Rb
I4 I3
 Ientering  Ileaving

Wye-Delta Transformation

I1 - I2 - I3 + I4 - I5 + 16=0 @ I1 + I4 + 16 = I2 + I3 +I5 a RaRc
Rb
R1  Ra  Rc 

current entering R1 Ra R2 RaRb
Rc Rb Rc
 Kirchoff's Voltage Law (KVL) current leaving R2  Ra  Rb 
states that the algebraic sum of all
voltages around a closed path (or b R3  Rb  Rc  RbRc
loop) is zero. Ra
R3

c

R1

+ V1 - R2 +

V2 Delta-Wye Transformation
E1 -
- V3 +
E2
a

R3 Ra  R1R 2
 R2  R3
R1 Ra R2 R1
Rc Rb
Rb  R1 R 2R 3 R3
R2 
E1 – V1 – V2 – E2 – V3 = 0 @ E1 – E2 = V1 + V2 + V3

b R3 Rc  R1R 3
 R2  R3
R1

c

afandia@ut hm.edu.my 8

Circuit Analysis

Example

Nodal Analysis Step 1: KCL at node a b IB2
R3
 Nodal analysis - based on the systematic application of KCL. a IB1  I1  I2
 Analyze any linear circuit by: R2 IB1 I2 IB1  I1  I2  0
I1 0  I1  I2  IB1
1. Obtaining a set of simultaneous equations IB1 R1
2. Solved to obtain the required values (voltage or current) c
3. Solve the simultaneous equation either using Cramer's
Rule or any other Va  Vc  Va  Vb  IB1  0
 Nodal analysis provides a general procedure for analyzing circuit R1 R2
using node voltages as a circuit variables.
 Important key idea resistance is a passive element, by the Va  0  Va  Vb  IB1 ...............[1]
passive sign convention, current must always flow from a higher R1 R2
potential to a lower potential

i  vhigher  vlower
R

Supernode Step 2: KCL at node b Step 3: Simplify equation [1] and [2]

 Formed by enclosing a (dependent or Vb  Vc  Vb  Va  IB2  0 Step 4: Solve the equation
independent) voltage source connected R3 R2
between two nonreference nodes and any
elements connected in parallel with it. Vb  0  Vb  Va  IB2 ...............[2] Va  Va  Vb  IB1 ...............[3]
4 R3 R2 R1 R2 R2

Supernode Vb  Vb  Va  IB2 ...............[4]
R3 R2 R2

5V  1  1    1  Va   IB1 
12 2 3 R1 R2 R2   
2ix   
[email protected] 1    
Topic 3    R3   Vb IB2 
4 Circuits Analysis  1 1  
 R2 R2 

How to deal with Supernode? Mesh Analysis
 Supernode equation combination of
 Mesh analysis provides another general procedure
KCL equation for the respective for analyzing circuits.
nodes.
 Support equation equation for the  Recall that a loop is a closed path with no node
voltage drop in between the combined passed more than once.
nodes compared to the voltage
source.  A mesh is a loop that does not contain any other
loop within it.
Example
 Mesh analysis apply KVL to find unknown currents

Write the supermesh and support Example R5
equation for the following circuit.
R1 R3
R1 R3

Mesh Analysis with Current Sources VB1 R2 R4 VB2

VB1 I1 R2 I2 VB2 Case 1: Loop 1,
When a current source exists only in one mesh I1R1  (I1 - I2 )R 2 - VB1  0
Supermesh equation, I1R1  (I1 - I2 )R2  VB1 ................ [1]
I1R1  I2R3 - VB2 - VB1  0 ............... [1] 10Ω 4Ω
Support equation,
I1 - I2  Ib (I1  I2  IB) ............... [2] 24V I1 12Ω I2 3A

9 [email protected] Loop 2, ................. [2]
(I2 - I1)R 2  I2R 3  (I2I3 )R 4  0
I2R 2 - I1R 2  I2R 3  I2R 4  I3R 4  0

I2=-3A Loop 3, ................. [3]
(I3 - I2 )R 4  I3R 5  -VB2
Case 2: I3R 4 - I2R 4  I3R 5  -VB2
When a current source exists in between of two meshes

create supermesh.
Set the following equation:

1. Supermesh equation
2. Support equation

Network Theorem

Superposition Theorem Thevenin's Theorem Norton's Theorem

Definition: Definition: Definiton:
Superposition theorem states that the voltage Thevenin's Theorem states that a linear two- Norton's Theorem states that a linear two-
across (or current through) an element in a terminal circuit can be replaced by an terminal circuit can be replaced by equivalent
circuit is the algebraic sum of the voltages equivalent circuit consisting of: circuit consisting of:
across (or current through) that elements A voltage source VTh in SERIES with A current source IN in PARALLEL with
due to each independent source acting A resistor RTh A resistor RN
alone.
Where; Where;
How to apply? VTh is the open-circuit voltage at the terminals. IN is the SC current through the terminals.
1. Consider one independent source at a RTh is the input or equivalent resistance at the RN is the input or equivalent resistance at the
time, while other independent sources are terminals when the independent sources are terminals when the independent sources are
turned-off. [short-circuit for voltage source and turned off. turned-off.
open-circuit the current source]
2. Dependent source are left intact because
they are controlled by the other circuit
variables.

Topic 4
Network Theorems

Thevenin's and Norton's Theorems with Maximum Power Transfer Theorem
Dependent Sources

Thevenin equivalent - useful in finding the maximum

power a linear circuit can deliver to a load.

Consider the following circuit for maximum power
transfer.

R Th a
Notes:

1. To get VTh and IN, we can analyze the circuit
using any method same with independent

sources.

a[email protected]. But, for RTh and RN, we can not go directly
(because the behavior of dependent sources). To VTh i p  i2R   VTh 2 R L
deal with this situation, let's go through the RL R Th  R L
following ideas.

How to deal with dependent source in Thevenin's b
and Norton's Theorems?

Method 1 Maximum power is transferred to the load when the load
Find the VTh and IN. Then: resistance equals the Thevenin resistance as seen from
RN  R Th  VTh
IN To prove the maximum power transfer differentiate p with
respect to RL and set the results =0

Method 2 The power is maximum when,
1.Introduce one independent voltage/current
source with any value at respective terminal. For dP  0
any dependent sources in the circuit must be dR L
turned-off first.
2.Consider the following circuit, for introducing  dP  VT h 2  (R Th  R L )2  2R L (R Th  R L ) 
any dependent sources. dR L  (R Th  R L )2 

 

 VT h 2  (R Th  R L  2R L )   0
 (R Th  R L )3 
a Introduce VT ,  

Linear circuit VT R Th  RN  VT ()
I
This implies that
b 0  (R Th  R L  2R L )  (R Th  R L ),[R Th  R L ]
Showing that, the maximum power takes place when R L  RTh.

a Introduce IT , With R L  R Th ,

Linear circuit Vab IT R Th  RN  Vab () Pmax  VT h 2  VT h 2 (watt)
IT 4R L 4R L

b

10afandia@ut hm. edu. my

Capacitor

Definition

 Passive elements designed to store energy in its electric field . Working Principles

 In general a capacitor is constructed by two plates separated  The amount of charge stored = q (directly
by an insulating (dielectric) material
proportional to the applied voltage)
 So, C = constant of proportionality called

capacitance of the capacitor
 Capacitance is the ratio of the charge on one
Properties of a capacitor
plate of a capacitor to the voltage difference

 When the voltage across a between the two plates. Measured in Farad (F)
capacitor is not changing with
time (eg. DC voltage), the 1 Farad = 1 Coulomb/Volt
 Capacitance:
 Ratio of the charge, q per plate to the applied

current through the capacitor voltage, V
is zero.  It does not depend on q or V, but depends on
 A capacitor is open circuit to
the physical dimensions of the capacitor

DC
 However, if a battery (DC

voltage) is connected across a

capacitor, the capacitor Topic 5

charges.
 The voltage on the capacitor
must be continuous. Energy Storage Elements

 The voltage on a capacitor (Capacitor)

cannot change abruptly.

Current-voltage relationship for a capacitor

Basically, i  C dv V(t)  1 t 
q  CV dt C
dq  dCV idt V(t o )
dv  i
[email protected] dt dt C to

wh er e V(t o )  q(t o )  the voltage across
C

dq  C dV , since i  dq dv  1  idt the capacitor at time t o Parallel and Series Capacitor
C * shown that capacitor voltage
dt dt dt

i  C dv  dv  1  idt depends on the past current
dt C
V(t)  1 t  q(t o)

idt i1 i2 i3
1 C to C
v  C  idt I C1 C2 C3 iN 

CN V

Instantaneous power and energy stored

P  Vi  VC dv  CV dv E  dw  CV dv
dt dt dt dt

i  i1 i2 i3 ...  iN

dw  CV dv i  C dv
dt dt dt

 dw   CV dv dt i  C1ddvt  C2 ddvt  C3 dv  ...  CNddvt
dt dt

w  t dv
-CV dt
where,
t
w  C -Vdv Ce  C1 C2 C3 ...  CN
q
w  C V2 t ,we note that v(- )  0 because the capacitor
2 t- i C1 C2 C3 CN V  V1  V2  V3  ...  VN
V  V1   V2   V3   VN 
was uncharged at t  -, thus 1 t
Ck
w  1 CV 2(t) ButVk  i(t)  Vk (to ).
2
to

Therefore,

w(t)  1 CV 2(t), q  CV  V1 t   1 t 
2 C1 V1(to ) C2 V2 (to )
i(t) i(t)
w  qV , V  q
2C to to

W  q2   1 t  1 t 
2C C3 V3 (to )  ...  CN VN (to )
i(t) i(t)

to to

where,

1  1  1  1  ...  1
Ceq C1 C2 C3 CN

11 [email protected]

Inductor

Definition

 Passive element designed to store energy in its magnetic field. Working Principles

 To enhance the inductive effect: a practical inductor is usually  If current is allowed to pass through an inductor,
formed into a cylindrical coil with many turns of conducting wire.

voltage across the inductor is directly proportional to

the time rate of change of the current.
 Inductance is the property whereby an inductor

Properties of an inductor exhibits opposition to the change of current flowing
through it, measured in henrys (H).
 Inductance depends on its physical dimension and

 When the voltage across a construction.

capacitor is not changing with v  L di
time (eg. DC voltage), the dt
current through the capacitor
is zero. L  constant of proportionality

 A capacitor is open circuit to L  inductance (Unit  Henry)

DC
 However, if a battery (DC

voltage) is connected across a

capacitor, the capacitor

charges. Topic 5
 The voltage on the capacitor Energy Storage Elements

must be continuous.
 The voltage on a capacitor
cannot change abruptly. (Inductor)

Voltage-current relationship for an inductor

Basically,

v  L di
dt

vdt  Ldi

[email protected]1vdt
L
where Parallel and Series Inductor
i(t0 )  total current for    t  t0
and i()  0. i L1 L2 L3 LN
The idea i()  0 is reasonable, because   V1   V2   V3   VN 
there must be a time in the past when there was

1 no current in the inductor. V
L
 di   vdt

1 Energy _ stored 
L
i   vdt p  vi   L di i Based on KVL,
 dt 
i  1 t v(t)dt  i(t0 )       di
L t0   w  pdt  t pdt  t Li di dt V V1 V2 V3 ... VN ; V L dt
dtt0 t0
V  L1 di  L2 di  L3 di  ...  LN di
t L t 1 1 dt dt dt dt
  2 i2 t0  2 Li2 (t)  2 Li2 (t0  )
L idi N di di
V  k1 Lk dt  Leq dt
t0

 1 Li2
2 Leq  L1  L2  L3  ...  LN

Important Properties i i1 i2 i3 iN

 Voltage across an inductor is zero when the current is V L1 L2 L3 LN
constant.

 Inductor acts like a short-circuit to DC
 The current through an inductor cannot change Based on KCL,

instantaneously. ii1  i2  i3  ...   1 Vdt  i(t0 )
 The ideal capacitor does not dissipate energy. The iN;i L

inductor takes power from the circuit when it storing  i1 t  1 t
energy and delivers power to the circuit when  L1 t0 Vdt  i1(t0 ) L2 t0 Vdt  i2(t0)
returning previously stored energy.

 1 t 1 t  iN(t0)
L3 t0 Vdt  i3(t0)  ...  LN t0 Vdt

i 1 t Vdt  i(t0 )
Leq t0

1  1  1  1  ...  1
Leq L1 L2 L3 LN

12afandia@ut hm. edu. my

First Order Circuit

Definition Source-Free RC Circuit

 RC and RL circuits are circuits that combined resistors with Since the capacitor is fully charged, the v(0)  V0
capacitors and resistors with inductors respectively. initial voltage and energy are given as
w(0)  1 CV0
 The analysis techniques used in RC and RL circuits are the Applying KCL at top node of the 2
same as in resistive circuits but they involve differential circuit gives
equations. iC iR  0
By definition, i = Cdv/dt and iR = v/R,
 Both RC and RL circuits produce first order differential thus C dv  v  0
equations, thus these circuits are known as first-order dt R
circuits. Rearranging the equations and
integrating both sides ln v   t  ln A
 There are two ways of supplying energy to these circuits: RC
Initial condition of storage elements in the circuit itself known
as source-free circuit or natural response.

 Applying independent sources to the circuit known as step
response.

 Combining two type of first order circuits and two ways of
exciting the circuit give four different types of first order circuit
variations.

Source-Free RL Circuit Putting variable v on right hand side t
gives
v(t)  Ae RC

Topic 6 From initial condition, A = v(0) = V0 t

v(t)  V0 e RC

First-Order Circuits t

Natural response for RC circuit v(t)  V0e RC

Note:
 From the equation obtained, it is obvious that the voltage

In RL circuit, inductor current is taken as the i(0)  I0 response of RC circuit is reduced exponentially from its
response. Its current and energy are given as initial value.
 1  Since there is no electrical source involved, therefore the
w(0) 2 LI 0 response is called the natural response.
 In other words, natural response of an RC circuit solely
depends on the energy stored in the capacitor.
Applying KVL around the loop of the circuit
gives vL  vR  0 τ  RC v(t)  V0e  t
L di  iR  0 τ
[email protected] definition, vL = Ldi/dt and vR = iR,
dt  It is clear that the smaller the time constant, the faster
the response, i.e. the faster the voltage decays to reach

Rearranging the equations and integrating ln i   Rt  ln A the steady state response.
both sides L  The response during process of decaying from its initial

value is noted as transient response.
 The natural response of RC circuit in the following

 Rt figure.

Putting variable i on right hand side gives i(t)  Ae L

From initial condition, A = i(0) = I0  Rt

i(t)  I0e L

Natural response for RL circuit  Rt

i(t)  I0e L Current and power dissipated in resistor

Note: iR (t)  v(t)  V0 e  t
 The current response of RL circuit is reduced R R τ

exponentially from its initial value. p  V02 e 2t
 Since the energy is only coming from inductor, τ

thus the response is known as natural
response.

τ L i(t)   t R
R τ
I0e

 The smaller the time constant, the faster the Summary of Source-Free RC Circuit
inductor current decays from its initial value.
 Determine the initial voltage, V0, of the capacitor.
 The inductor current decays much faster during  Find the time constant of the circuit , τ = RC.
the transient response before it reaches the  Calculate the capacitor voltage to complete the RC natural
response.
 The step response of inductor current is  Use natural response of capacitor voltage to find expression
illustrated in the following figure.
of capacitor current iC, resistor voltage vR and resistor
current iR.

Current and power dissipated in resistor

 t Summary of Source-Free RL Circuit
τ
vR (t)  iR  I Re 
0  Determine the initial inductor current, I0.
Calculate the time constant of the circuit, τ
2t = L/R
τ  Finalise the RL natural response by finding the inductor current
p  I 2 Re 
0
response.
 From natural response of inductor current, determine the inductor

voltage vL, resistor voltage vR and resistor current iR.

13 [email protected]

First Order Circuit

Step Response RC Circuit

Assume that the initial capacitor voltage just before v0  v0   V0
and after the voltage source is applied
C dv  v  VSut  0
Applying KCL in the circuit produces
dt R

Rearranging the equation yields dv   dt
v  VS RC

Integrating both sides and putting initial ln v  VS   1
conditions V0  VS RC

Simplify the equation by taking v to the right hand vt  VS  V0  VS e  t τ t  0
side and substitutes RC term with τ

Uncharged capacitor Charged capacitor A complete step response of an RC circuit is vt  V0 , t 0
given as  t0
VS  V0  VS e  t τ ,

complete step response If the capacitor is assumed not charged vt   0, 1 - e tτ , t0
initially, therefore V0=0 VS t0
Complete response = natural response + forced response

v = vn + vf

where vn  V0e t τ vt  VS 1- e  t τ  ut
vf  VS1 e t τ 
It also can be written as

Note: The current through the uncharged capacitor is it  VS e  t τ ut
 Natural response, vn, is actually the given as
R
response due to capacitor itself it will Step Response RL Circuit

continue to decay and eventually

dies.
 Forced response, vf, is the response
produced due to independent voltage

source.
 Natural response normally exists

along with the transient response

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Topic 6
First-Order Circuits

along with steady-state response. i 0  i0   I0

Complete response = transient response + steady-state response Assume that the initial inductor current just di VSu t 
before and after the current source is applied dt
v = vt + vss Applying KVL around the loop produces

 vt  V0 - VS e t τ Rearranging the equation yields L  iR   0

where vss  VS

di   Rdt
i - VS L

step response form in general R

vt  v v0 ve t τ i  VS R
ln R
where v   is the final capacitor voltage Integrating both sides and putting initial VS L
v 0  is the initial capacitor voltage conditions R
I0 
τ is the time constant

Simplify the equation by taking i to the right i(t)  VS   I0  VS e  t t0
hand side and substitutes R/L term with τ R  R 

Summary of Step Response RC Circuit If the capacitor is assumed not charged 
initially, therefore I0 = 0
 Initial capacitor voltage, V0, if the 0, t  0
capacitor has been charged 
it   VS tτ
 Final capacitor voltage, VS  R 1 - e , t0
 Time constant, τ

Summary of Step Response RL Circuit It can also be written as it  VS 1 - e tτ  ut
R
 Initial inductor current, I0, if the inductor The voltage across the uncharged inductor is
has been charged given as vt  VS e t τ ut

 Final inductor current, IS general form for step response of RL circuit
 Time constant, τ
 The equation of step response can be it   i   i0  i  e t τ

used to find the instantaneous current at
any time.

where i  is the final inductor current Uncharged inductor Charged inductor
i0  is the initial inductor current

τ is the time constant

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Second Order Circuit

Source Free Series RLC Circuit

Initial and Final Values With initial capacitor voltage V0 and initial inductor current I0.
Thus, at initial condition (t = 0):

v0, i0, dv0, di0, v, i v0  V0 i0  I0

dt dt Ri  L di  1 t
Applying KVL around the loop in the circuit
idt  0
dt C 

 where v denotes the capacitor

voltage and i denotes the the Differentiate each term to eliminate the d2i  R di  i  0
inductor current. integral and rearrange them dt 2 L dt LC

Note:
 Another important point to
remember is that both variables v To solve the second order equation, there must be initial Ri0  L di0  V0  0
and i must be defined according values of i and di/dt or initial values of some v and i.
to passive sign convention. The initial value of i is I0 and to find initial value of di/dt is dt

 Capacitor voltage is always di0   1 RI0  V0 
continuous, the same goes to L
inductor current First order solution is i  Aest dt

v 0   v0 and i 0   i 0  As2est  AR sest  A est  0

Characteristic equation L LC

Aest s2  R s  1 
 L LC 
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Expressing the roots

s2  R s  1  0 s1   R   R 2  1 s1  α  α2  ω02 s2  α  α2  ω02
L LC 2L  2L  LC where α  R ,
1
 R 2  1 2L ω0  LC
 2L  LC
s2   R 
2L

Second-Order Circuits s1 and s2 are called natural frequencies since they are associated
with natural response,
0 is the resonant frequency and
 is the damping factor.

Solution of natural response for series RLC i1  A1es1t i2  A2es2t

If  > 0, the solution is overdamped,  i t  A1es1t  A2es2t
If  = 0, the solution is critically damped, and
If  < 0, the solution is underdamped.

Overdamped case Critically damped case Underdamped case

This occurs when a > w0, or when C > 4L/R2. This occurs when  = 0, or when C = 4L/R2 It happens when  < 0, or when C < 4L/R2
Both roots s1 and s2 are negative and real. The natural response solution: The natural response solution:

The response is it  A1t  A2 eαt it  eαt B1cosdt  B2sindt

 i t  A1es1t  A2es2t where d is damped natural frequency

15 [email protected]

Second Order Circuit

Source Free Parallel RLC Circuit

With initial voltage is given as V0 and initial inductor current I0.

Applying KCL at the top node yields

v  1 t vtdt  C dv  0
R L  dt

Differentiate each element w.r.t. t gives

d2v  1 dv  1 v  0
dt 2 RC dt LC

Characteristic equation

s2  1 s  1  0
RC LC

Expressing the roots

  1   1 2  1 s1  α  α2  ω02 s2  α  α2  ω02
2RC  2RC  LC where α  R ,

2L
[email protected] 1  1 2  1 ω0  1
2RC  2RC  LC LC
s2   

s1 and s2 are called natural frequencies since they are associated
with natural response,
0 is the resonant frequency and
 is the damping factor.

Topic 7
Second-Order Circuits

Overdamped case Critically damped case Underdamped case

This occurs when a > w0, or when C > 4L/R2. This occurs when  = 0, or when C = 4L/R2 It happens when  < 0, or when C < 4L/R2
Both roots s1 and s2 are negative and real. The natural response solution: The natural response solution:
The response is
vt  A1t  A2 eαt vt  eαt A1cosdt  A2sindt
 v t  A1es1t  A2es2t
where d is damped natural frequency

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Second Order Circuit

Step Response Series RLC Circuit Applying KVL around the loop in the circuit

L di  iR  v  VS
dt

It is known that i  C dv , therefore d2v  R dv  v  VS
dt dt 2 L dt LC LC

Solution for step response vt  vt  vss

The solution of transient response is the same that are
obtained in natural response, which are

 v t  A 1e s1t  A 2 e s2t Overdamped case

v t   A 1t  A 2 e  αt Critically damped case

vt   eαt A1cosd t  A 2sind t  Underdamped case

Steady state response is the same as the final value of v(t)

vss t  v  VS

Complete solution for step response for series RLC

 v t  VS  A1es1t  A es2t Overdamped case

2

Topic 7    v t  VS  A1t  A2 eαt Critically damped case

[email protected] Circuits
   v t  VS  eαt A1cosd t  A2sind t Underdamped case

Note:
The values for A1 and A2 are obtained from
the initial conditions v(0) and dv(0)/dt.

Step Response Parallel RLC Circuit Applying KCL at the top node for t > 0 yields

Complete solution for step response C dv i  v  IS
for series RLC circuit dt R

 i t  IS  A1es1t  A2es2t Overdamped case It is known that v  L di , therefore d2i  R di  i  IS
   i t  IS  A1t  A2 eαt Critically damped case dt dt 2 L dt LC LC
   i t  IS  eαt A1cosdt  A2sindt Underdamped case
The form of the equation obtained is the same as
17 [email protected] natural response parallel RLC but with variable i.

 Solution for step response i t  it  iss

The solution of transient response

 i t  A1es1t  A2es2t Overdamped case

   i t  A1t  A2 eαt Critically damped case
   i t  eαt A1cosdt  A2sindt Underdamped case

Steady state response is the same as the final value of i(t)

iss t  i  IS

Basic Circuit Analysis – Quick Reference Notes
Disember 2018