Review of

EE 220 Circuits I

1

Overview

1. Current, voltage, power, energy

2. Circuit elements

3. Ohm’s Law

4. Node, branches, andloops

5. Kirchhoff’s Laws.

6. Series resistors

7. Parallel resistors current

8. Wye-Delta transformations

9. Nodal Analysis

10. Mesh Analysis

11. Superposition

12. Source Transformation

13. Thevenin Equivalent

14. Norton Equivalent

15. Maximum Power Transfer

16. Capacitors

17. Inductors

18. RC Circuits– source free

19. RL Circuits – source free

20. Unit Step Function

21. Step Response of RC Circuits

22. Step Response of RL Circuits

23. Series RLC circuit - source free

24. Parallel RLC circuit – source free

25. Step Response of series RLC ckt

26. Step Response of Parallel RLC ckt

2

1. Current

Electric current is defined as the rate of flow of

electric charge, i.e., i = dq/dt.

The unit of ampere can be derived as 1 A = 1C/s.

A direct current (DC) is a current that remains

constant with time.

An alternating current (AC) is a current that

varies sinusoidally with time

3

1. Voltage

Voltage (or potential difference) is the energy required to move a unit charge through an element.

Mathematically, (volt)

where w is energy in joules (J) and q is charge (C).

voltage, vab, is always defined across a circuit element or

between two points in a circuit.

◦ vab > 0 means the potential of a is higher than potential of b.

◦ vab < 0 means the potential of a is lower than potential of b.

4

dqdwvab /

1. Power

Power is the time rate of expending or absorbing energy, measured in watts (W).

Mathematical expression:

5

vidt

dq

dq

dw

dt

dwp

i

+

–

v

i

+

–

v

Passive sign convention P = +vi p = –vi

absorbing power supplying power

1. Energy

6

t

t

t

tvidtpdtw

0 0

• Energy is the capacity to do work,

measured in joules (J).

• Mathematical expression

2. Circuit Elements

7

Active Elements Passive Elements

Independent sources

Dependant sources

• A dependent source is an active element in which the source quantity is controlled by another voltage or current.

• They have four different types: VCVS,

CCVS, VCCS, CCCS. Keep in minds the signs of dependent sources.

3. Ohm’s Law

Ohm’s law states that the voltage across

a resistor is directly proportional to the

current I flowing through the resistor.

Mathematical expression for Ohm’s 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

8

iRv

3. Ohm’s Law

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

The power dissipated by a resistor:

R

vRivip

22

9

v

i

RG

1

4. Nodes, Branches, Loops and Meshes

A branch represents a single element such as a voltage source

or a resistor (or a series connection of elements).

A node is the point of connection between two or more

branches.

A loop is any closed path in a circuit.

A mesh is a loop that contains no elements inside it.

Example: How many notes, branches , loops and meshed are

there in the circuit below?

10

5. Kirchhoff’s Current Law (KCL)

Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero.

11

01

N

n

niMathematically,

5. Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is

zero.

12

Mathematically, 01

M

m

nv

6. 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

13

N

n

nNeq RRRRR1

21

vRRR

Rv

N

n

n

21

7. Parallel Resistors and Current Division

The equivalent resistance of a circuit with N resistors in

parallel is:

The total current i is shared by the resistors in inverse

proportion to their resistances. The current divider can be

expressed as:

14

Neq RRRR

1111

21

n

eq

n

nR

iR

R

vi

8. Wye-Delta Transformations

15

)(1

cba

cb

RRR

RRR

)(2

cba

ac

RRR

RRR

)(3

cba

ba

RRR

RRR

1

133221

R

RRRRRRRa

2

133221

R

RRRRRRRb

3

133221

R

RRRRRRRc

Delta → Wye Wye → Delta

9. Nodal Analysis

16

Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables.

• Steps

1.Select a node as the reference node.

2.Assign voltages v1,v2,…,vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.

3.Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express the branch currents in terms of node voltages.

4.Solve the resulting simultaneous equations to obtain the unknown node voltages.

9. Nodal Analysis

17

v1 v2

3

answer: v1 = -2V, v2 = -14V, P2 =2W, P7= 28W, P6 = 24W

Apply KCl at node 1 and 2

Example 1: Use nodal analysis to determine the power consumed by each resistor in the circuit below.

9. Nodal Analysis

18

Example 2 – Use nodal analysis to find the power supplied by the 2 V source (hint: use supernode)

Answer: 2= v1/2 +v2/4 +7

v2 – v1 = 2

P = ???

10. Mesh Analysis

19

Mesh analysis provides another general procedure for analyzing circuits using mesh currents as the circuit variables. Mesh analysis applies KVL to find unknown currents.

Steps

1. Assign mesh currents i1, i2, …, all the n meshes.

2. Apply KCL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.

3. Solve the resulting n simultaneous equations to get the mesh currents.

10. Mesh Analysis

20

Example 1 – Use mesh analysis to find the voltage across the dependent source.

Answer: 4Io = 6 V

10. Mesh Analysis

21

Example 2 – Use mesh analysis to find i1 and i2 (hint: use supermesh)

Answer: -20 +6i1+10i2+4i2 = 0 i2 – i1 = 6

11. Superposition Theorem

22

The voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to each independent source acting alone. Steps: 1. Turn off all independent sources except one source.

Independent voltage sources are replaced by 0 V (i.e.,

short circuit) Independent current sources are replaced by 0 A (i.e.,

open circuit) Dependent sources are left intact .

2. Find the output (voltage or current) due to that active source using nodal or mesh analysis.

3. Repeat step 1 for each of the other independent sources.

4. Find the total contribution by adding algebraically all the contributions due to the independent sources.

11. Superposition Theorem

23

Example 1: Use the superposition theorem to find v in the circuit shown below.

3A is discarded by open-circuit

6V is discarded by short-circuit

Answer: v = 10V

11. Superposition Theorem

24

Example 2: Use superposition to find vx in the circuit below.

Answer: Vx = 12.5V

2A is discarded by open-circuit

20 v1

4 10 V

+

(a)

0.1v1 4

2 A

(b)

20

0.1v2

v2

10V is discarded by open-circuit

Dependant source keep unchanged

12. Source Transformation

25

Source transformation is the process of replacing a voltage source vS in series with a resistor R by an equivalent circuit that consists of a current source iS in parallel with a resistor R, or vice versa.

R

vi

Riv

ss

ss

12. Source Transformation

26

Example 1: Find io in the circuit shown below using source transformation.

Answer: io = 1.78A

13. Thevenin’s Theorem

27

A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where • VTh is the open-circuit

voltage across terminals a-b.

• RTh is the equivalent resistance of the linear circuit with the independent sources turned off.

13. Thevenin’s Theorem

28

Example 1: Find the Thevenin’s equivalent circuit to the left of the terminals in the circuit shown below.

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

6

4

(a)

RTh

6

2A

6

4

(b)

6 2A

+ VT

h

13. Thevenin’s Theorem

29

Example 2: Find the Thevenin equivalent circuit of the circuit shown below to the left of the terminals.

Answer: VTh = 5.33V, RTh = 3

6 V

5 Ix

4

+

1.5Ix

i1

i2

i1 i2

3

o

+ VT

h

b

a

1.5I

x 1 V

+

3 0.5Ix

5

a

b

4

Ix i

14. Norton’s Theorem

30

A linear two-terminal circuit can be replaced by an equivalent circuit of a current source IN in parallel with a resistor RN, Where • IN is the short circuit current

through the terminals. • RN is the equivalent resistance o

the circuit with the independent sources turned off.

The Thevenin and Norton equivalent circuits are related by a source transformation.

14. Norton’s Theorem

31

Example 1: Find the Norton equivalent of the circuit shown below.

Answer: RN = 1, IN = 10A.

2

(a)

6

2vx

+ +

vx

+ vx

1V

+ ix

i

2

(b)

6 10 A

2vx

+ +

vx

Isc

15. Maximum Power Transfer

L

ThThL

R

VPRR

4

2

max

L

LTh

ThL R

RR

VRiP

2

2

32

The power consumed by the load resistance RL is given by

Power transfer profile with different values of RL

Maximum power is obtained by dP/dRL = 0.

15. Maximum Power Transfer

33

Example 1: Determine the value of RL that will draw the maximum power from the rest of the circuit to the right. Then calculate the maximum power.

Answer: RL = 4.22 , Pmax = 2.9 W

16. 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).

34

16. Capacitors

Capacitance C is the ratio of the charge q on one plate of a capacitor to the voltage difference v between the two plates, measured in farads (F).

• Capacitance can be calculated in

terms of - the permittivity of the

dielectric material between the plates, A - the surface area of each plate, and d - the distance between the plates.

• Units of capacitance: pF (10–12), nF (10–9), μF (10-6), mF (10-3), F

35

v

qC

d

AC

16. Capacitors

36

Current-voltage relationship of capacitor according to above convention:

td

vdCi )(

10

0

tvtdiC

vt

t and

Energy stored in a capacitor:

2

2

1vCw

– A capacitor acts as open circuit under constant voltage.

– The voltage across a capacitor cannot change abruptly

16. Capacitors

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

37

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

16. Series and Parallel Capacitors

The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.

38

Neq CCCC ...21

The equivalent capacitance of N series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

Neq CCCC

1...

111

21

16. Series and Parallel Capacitors Example 2: Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:

39

Answer: Ceq = 40F

Example 3: Find the voltage across each of the capacitors in the circuit shown below:

Answer:

v1 = 30V

v2 = 30V

v3 = 10V

v4 = 20V

17. Inductors

An inductor is a passive element designed to store

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

conducting wire.

40

17. Inductors

41

• Inductance L is the ratio of flux linkage λ within the coil to the current flow in the coil, measured in farads (F).

• Inductance L of a solenoid can be calculated in terms of μ - the

magnetic constant of the material inside the coil, A - the cross sectional surface of the coil, N – the number of turns, and l – the length of the coil.

• Units of inductance: pH (10–12), nH (10–9), μH (10-6), mH (10-3), H

iL

l

ANL

2

17. Inductors

Voltage-current relation in an inductor:

42

Energy stored in an inductor.

td

idLv and )()(

10

0

titdtvL

it

t

2

2

1iLw

– An inductor acts as short circuit under constant current.

– The current flow through an inductor cannot change abruptly

17. Inductors Example 1: Determine vc, iL, and the energy stored in the

capacitor and inductor in the circuit of circuit shown below

under steady-state conditions.

43

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

17. Series and Parallel Inductors The equivalent inductance of series-connected

inductors is the sum of the individual inductances.

44

Neq LLLL ...21

The equivalent capacitance of parallel inductors is the

reciprocal of the sum of the reciprocals of the individual

inductances.

Neq LLLL

1...

111

21

Recap: voltage-current relation and power in

passive circuit elements

45

+

+

+

18. RC Circuit (source free)

46

Current flow in resistor R

Current flow in capacitor C

0 dt

dvC

R

v0 CR ii

By KCL

• Applying Kirchhoff’s laws to purely resistive circuit results in algebraic equations.

• Applying the laws to RC and RL circuits produces differential equations.

Vo is the initial voltage τ = RC is the time constant

18. RC Circuit (source free)

The natural response of a circuit refers to the behavior

(in terms of voltages and currents) of the circuit itself, with

no external sources of excitation.

47

• The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.

CRTime constant

Decays more slowly

Decays faster

18. RC Circuit (source free)

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

for t ≥ 0. Assume that vC(0) = 30 V.

48

Answer: vC = 30e–0.25t V ;

vx = 10e–0.25t ;

io = –2.5e–0.25t A

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) = 8e–2t V

19. RL Circuit (source free)

49

0 RL vvBy KVL

0 iRdt

diL

0)( iL

R

dt

di /0 )( teIti

Io is the initial current τ = L/R is the time constant

19. RL Circuit (source free)

Example 1: Find i and vx in the circuit. Assume i(0) = 5A.

50

Answer: i(t) = 5e–53t A

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

Answer: i(t) = 2e–2t A

20. Unit-Step Function The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

51

0,1

0,0)(

t

ttu

o

o

ott

ttttu

,1

,0)(

Represent an abrupt change in voltage and current:

21. Step-Response of an RC Circuit The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

52

• 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

dv s )()

1(

21. Step-Response of a RC Circuit

53

0)1(

0)(

//

0

0

teVeV

tVtv

t

s

t

Complete Response = Natural response + Forced Response (stored energy) (independent source)

/ )]( )0( [ )( )( tevvvtv General Solution:

21. 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 is closed at t = 0. Calculate v(t) at t = 0.5.

54

Answer: and v(0.5) = 0.52 V 515)( 2 tetv

22. Step-response of a RL Circuit

55

• Initial current (given) i(0-) = i(0+) = Io

• Final inductor current i(∞) = Vs/R

• Apply KVL:

• Time constant = L/R

0)()(

teR

VI

R

Vti

t

so

s

L

tuVi

L

R

dt

di s )()(

/ )]( )0( [ )( )( teiiiti

22. Step-Response of a RL Circuit

Example 1: The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0.

56

Answer: teti 102)(

23. Series RLC Circuit (source free)

57

• The solution of the source-free series RLC circuit is called as the natural response of the circuit.

• The circuit is excited by the energy initially stored in the capacitor and inductor.

• Expression of current (using KVL):

02 2

02

2

idt

di

dt

id

LCand

L

R 1

20

02 2

02

2

idt

di

dt

id

58

There are three possible solutions for the following 2nd order differential equation:

1. If > o, over-damped case

tstseAeAti 21

21)( 2

0

2

2,1 swhere

2. If = o, critical damped case

tetAAti )()( 122,1swhere

3. If < o, under-damped case

)sincos()( 21 tAtAeti dd

t where

22

0 d

23. Series RLC Circuit (source free)

The constants A1 and A2 are obtained from the initial conditions

23. Series RLC Circuit (source free)

Example 1: The 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.

59

Answer: i(t) = e–2.5t[5 cos1.6583t – 7.538 sin1.6583t] A

24. Parallel RLC Circuit (source free)

60

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

LCRCv

dt

dv

dt

vd 1and

2

1 where0 2 0

2

02

2

24. Parallel RLC Circuit (source free)

Example 3: Refer to the circuit shown below where the switch has been closed for a long time. Find v(t) for t > 0.

61

Answer: v(t) = 66.67(e–10t – e–2.5t) V

25. Step-Response of Series RLC Circuit

62

LC

vv

dt

dv

dt

vd s 2

02

2

2

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

)()()( tvtvtv sst

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

vss(t) = v(∞) = Vs.

• The transient response is the same as the source-free

response

Depends on the values of α and ωo.

LCand

L

R 1

20

25. Step-Response of Series RLC Circuit

Example 1: Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) for t > 0.

63

Answer: v(t) = 10 +e-2t (–2cos3.464t – 1.1547sin3.464t) V

26. Step-Response of Parallel RLC Circuit

64

LC

I

LC

i

dt

di

RCdt

id s1

2

2

LCand

RC

1

2

10

The solution of the above equation has two components: transient response it(t) & steady-state response iss(t):

)()()( tititi sst

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

iss(t) = i(∞) = Is

• The transient response is the same as the source-free

response

Depends on the values of α and ωo.

26. Step-Response of Parallel RLC Circuit

Example 1: Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below. Assume that there was no energy stored in C and L.

65

Answer:iss(t) = Is =20 A, α =0, and ω0 = 1, oscillatory case i(0) = 0 A, di(0)/dt = 0 A/sec → A1 = -20, A2 = 0.

i(t) = -20 cost A

v(t) = Ldi/dt = 100 sint V