EE-221-Review of DC Circuits

7/25/2019 EE-221-Review of DC Circuits

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Ohms Law

Ohms law states that the voltage acrossa resistor is directly proportional to the

current I flowing through the resistor.

Mathematical expression for Ohms Law :

Note: the current enters the positive

side and leaves the negative side of v.

Two extreme possible values of R: R = 0 (zero) v = 0 V -----short circuit

R =

(infinity) I = 0 A --- open circuit

2

iRv


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Ohms Law

Conductance is the ability of an element to conductelectric current; it is the reciprocal of resistance R and ismeasured in Siemens (S).

The power dissipated by a resistor:

R

vRivip

2

2

3

v

i

RG 1


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Nodes and Branches

A branchrepresents a single element such as a voltage sourceor a resistor (or a series connection of elements).

A nodeis the point of connection between two or more

branches.

4


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Kirchhoffs Current Law (KCL)

Kirchhoffs current law (KCL) states that the algebraicsum of currents entering a node (or a closedboundary) is zero.

5

01

N

n

niMathematically,


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Kirchhoffs Voltage Law (KVL)

Kirchhoffs voltage law (KVL) states that the algebraicsum of all voltages around a closed path (or loop) is

zero.

6

Mathematically, 0

1

M

m

nv


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Series Resistors and voltage Division

The equivalent resistance of any number of resistors

connected in a series is the sum of the individual

resistances.

The voltage divider can be expressed as

7

N

n

nNeq RRRRR1

21

vRRR

Rv

N

n

n

21
http://en.wikipedia.org/wiki/File:Resistors_in_series.svg


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Parallel Resistors and Current Division

The equivalent resistance of a circuit with N resistors inparallel is:

The total current i is shared by the resistors in inverseproportion to their resistances. The current divider can be

expressed as:

8

Neq RRRR

1111

21

n

eq

n

n

R

iR

R

vi
http://en.wikipedia.org/wiki/File:Resistors_in_parallel.svg


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Superposition Theorem

10

Example 1: Find v in the circuit shown below.

3A is discardedby open-circuit

6V is discardedby short-circuit

Answer: v= 10V


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Source Transformation

11

Source transformation is the process of replacing avoltage source vSin series with a resistor Rby anequivalent circuit that consists of a current sourceiSin parallel with a resistor R,or vice versa.

R

vi

Riv

s

s

ss


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Source Transformation

12

Example 1: Find ioin the circuit shown below.

Answer: io= 1.78A


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TheveninsTheorem

13

A linear two-terminalcircuit can be replaced byan equivalent circuitconsisting of a voltagesource VThin series witha resistor RTh,

where

VThis the open-circuitvoltage across terminals a-b.

RThis the equivalentresistance of the linearcircuit with the independentsources turned off.


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TheveninsTheorem

14

Example 1: Find the Thevenins equivalent circuit to theleft of the terminals in the circuit shown below.

Answer: VTh= 6V, RTh= 3W, i = 1.5A

6

4

(a)

RTh

6

2A

6

4

(b)

6 2A

+

VTh


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Maximum Power Transfer

L

Th

ThL

RVPRR4

2

max

L

LTh

Th

L R

RR

VRiP

2

2

15

The power consumed by the loadresistance RL is given by

Power transfer profile with differentvalues of RL

Maximum power is obtainedby dP/dRL= 0.


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Capacitors

A capacitor is a passive element designed to store energy

in its electric field. It consists of two conducting plates

separated by an insulator (or dielectric).

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Capacitors

17

Current-voltage relationship ofcapacitor according to aboveconvention:

td

vdCi )(

10

0

tvtdiC

vt

t

and

Energy stored in a capacitor:

2

2

1vCw

A capacitor acts as opencircuitunder constant voltage.

The voltage across a capacitorcannot change abruptly


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Capacitors

Example 1: The current flow into an initially discharged 1mF capacitoris shown below. Calculate the voltage across its terminals at t= 2 msand t= 5 ms.

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Answer: v(2ms) = 100 mV, v(5ms) = 500 mV


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Series and Parallel Capacitors

The equivalent capacitance of Nparallel-connected

capacitors is the sum of the individual capacitances.

19

Neq CCCC ...21

The equivalent capacitance of Nseries-connectedcapacitors is the reciprocal of the sum of the reciprocals ofthe individual capacitances.

Neq CCCC

1...

111

21


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Series and Parallel CapacitorsExample 2:Find the equivalent capacitance seen at the

terminals of the circuit in the circuit shown below:

20

Answer: Ceq= 40F


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Inductors

An inductor is a passive element designed to store

energy in its magnetic field. It consists of a coil of

conducting wire.

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InductorsExample 1: Determine vc, iL, and the energy stored in the

capacitor and inductor in the circuit of circuit shown below

under steady-state conditions.

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Answer: iL= 3A, vC= 3V,WL= 1.125J, WC= 9J


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Series and Parallel Inductors The equivalent inductance of series-connected

inductors is the sum of the individual inductances.

24

Neq LLLL ...21

The equivalent capacitance of parallelinductors is the

reciprocal of the sum of the reciprocals of the individual

inductances.

Neq LLLL

1...

111

21


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RC Circuit (source free)

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Currentflow inresistor R

Current flow incapacitor C

0dt

dvCR

v0CR

ii

By KCL

Applying Kirchhoffs laws to purely resistive circuit

results inalgebraic equations. Applying the laws to RC and RL circuitsproduces first-

first-order differential equations.

Vois the initial voltage= RC is the time constant


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RC Circuit (source free)

Example 1: Refer to the circuit below, determine vC, and io

for t 0. Assume that vC(0) = 30 V.

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Answer: vC = 30e0.25tV;

io = 2.5e0.25tA

Example 2: The switch in circuit below was closed for a long

time, then it opened at t = 0, find v(t) for t 0.

Answer: V(t) = 8e2tV


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RL Circuit (source free)

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0RL

vvBy KVL

0 iRdt

di

L

0)( iL

R

dt

di /0)(

teIti

Iois the initial current= L/R is the time constant


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RL Circuit (source free)

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Example 2: For the circuit, find i(t) for t > 0.

Answer: i(t) = 2e2tA


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Unit-Step Function

The unit step functionu(t) is 0 for negative values of

t,and 1 for positive values of t.

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0,1

0,0)(

t

t

tu

o

o

o

tt

tt

ttu

,1

,0)(

o

o

o

tt

tt

ttu

,1

,0)(


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Unit-Step Function

1. voltage source.

2. for current source:

31

Represent an abrupt change:


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Step-Response of an RC CircuitThe step response of a circuit is its behavior when theexcitation is the step function, which may be a voltage or a

current source.

32

Initial voltage (given):v(0-) = v(0+) = V0

Applying KCL,

or

Where u(t) is the unit-step function

0)(

R

tuVv

dt

dvc

s

RC

tuVv

RCdt

dvs

)()

1(


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The Step-Response of a RC Circuit

Integrating both sides and considering the initialconditions, the solution of the equation is:

33

0)(0)(

/

0

0

teVVV

tVtv

t

ss

Final valueat t ->

Initial valueat t = 0

Source-freeResponse


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Step-Response of a RC Circuit

Example 1: Find v(t) for t > 0 in the circuit in below.

Assume the switch has been open for a long time and isclosed at t = 0. Calculate v(t) at t = 0.5.

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Answer: and v(0.5) = 0.52 V515)( 2 tetv


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Step-response of a RL Circuit

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Initial current (given)

i(0-) = i(0+) = Io

Final inductor currenti() = Vs/R

Apply KVL:

Time constant = L/R

0)()(

teR

VI

R

Vti

t

so

s

L

tuVi

L

R

dt

dis

)()(

/

)]()0([)()(

teiiiti


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Second-Order Circuits

Chapter 8

8.1 Examples of 2nd order RCL circuit

8.2 The source-free series RLC circuit8.3 The source-free parallel RLC circuit

8.4 Step response of a series RLC circuit

8.5 Step response of a parallel RLC

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Examples of 2nd Order RLC circuits

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A second-order circuit is characterized by asecond-order differential equation.It consists ofresistorsand the equivalent of two energy

storage elements.

RLC Series RLC Parallel RL T-config RC Pi-config


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Source-Free Series RLC Circuits

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The solution of the source-freeseries RLC circuit is called as thenatural responseof the circuit.

The circuit is excitedby the energyinitially stored in the capacitor andinductor.

02

2

LC

i

dt

di

L

R

dt

id

How to derive and how to solve?


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02 2

02

2

idt

di

dt

id

40

There are three possible solutions for the following 2ndorder differential equation (where the constants A1,

A2, B1, B2 are found from the initial conditions.

1. If > o, over-damped casetsts

eAeAti 21 21)( 2

0

2

2,1 swhere

2. If = o

, critical damped case

tetAAti

)()( 12 2,1swhere

3. If < o, under-damped case

)sincos()( 21 tBtBeti ddt

where22

0

d


LCand

L

R 1

20 where


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Example 1

If R = 10 , L = 5 H, and

C = 2 mF, find and 0.

What type of natural

response will the circuithave?

41

Answer: = 1, 0 = 10, under-damped


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Example 2The circuit shown below

has reached steady state

at t = 0-. If the make-

before-break switch

moves to position b at t =

0, calculate i(t) for t > 0.

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Answer: i(t) = e2.5t[5cos1.66t 7.538sin1.66t] A


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Source-Free Parallel RLC Circuits

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011

2

2

vLCdt

dv

RCdt

vd

0

0 )(1

)0( dttvL

IiLet

v(0) = V0

Apply KCL to the top node:

t

dt

dvCvdt

LR

v0

1

Taking the derivative with

respect to t and dividing by C


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Source-Free Parallel RLC Circuits

LCRCv

dt

dv

dt

vd 1and

2

1where02

0

2

02

2

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There are three possible solutions for the following2nd order differential equation:

1. If > o, over-damped casetsts

eAeAtv 21 )( 21 2

0

2

2,1 swhere

2. If = o

, critical damped case

tetAAtv

)()( 12 2,1swhere

3. If < o, under-damped case

)sincos()( 21 tBtBetv ddt

where22

0

d


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Step-Response Series RLC Circuits

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The step responseis obtained by thesudden applicationof a dc source.

LC

v

LC

v

dt

dv

L

R

dt

vd s

2

2


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The solution of the equation should have two components:the transient response vt(t)& the steady-state response vss(t):

)()()( tvtvtvsst

The transient response vtis the same as that for source-free case

The steady-state response is the final value of v(t).

vss(t) = v()

The values of A1and A2are obtained from the initial conditions:

v(0) and dv(0)/dt.

tsts

t eAeAtv 21 21)( (over-damped)

t

t etAAtv

)()( 21 (critically damped)

)sincos()( 21 tAtAetv

dd

t

t

(under-damped)


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Example 4Having been in position for a long time, the switch in thecircuit below is moved to position b at t = 0. Find v(t)and vR(t) for t > 0.

48

Answer: v(t) = 10 +e-2t(2cos3.46t 1.15sin3.46t) V

vR(t)= e2t

[2.31sin3.46t] V


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Step-Response Parallel RLC Circuits

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LC

I

LC

i

dt

di

RCdt

ids

1

2

2


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8.5 Step-Response Parallel RLC Circuits

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The solution of the equation should have two components:the transient response vt(t)& the steady-state response vss(t):

)()()( tititisst

The transient response itis the same as that for source-free case

The steady-state response is the final value of i(t).

iss(t) = i() = Is

The values of A1and A2are obtained from the initial conditions:

i(0) and di(0)/dt.

tsts

t eAeAti 21 21)( (over-damped)

t

t etAAti

)()( 21 (critical damped)

)sincos()(21

tAtAetidd

t

t

(under-damped)


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Step-Response Parallel RLC Circuits

Find i(t) and v(t) for t > 0 in the circuit shown in circuit

shown below:

Answer: v(t) = Ldi/dt = 5x20sint = 100sint V

EE-221-Review of DC Circuits

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