Chapter OneBasic Concepts

(1) Electron :electron is a mobile charge carrier.

•The electron is measured in coulumb [ C ]

• e = 1.6*10-19 C

• Multiple of electrons constitute charge (q).

• This course basically deals with the analysis of electric circuits.• The most basic quantity used in the analysis of electrical circuits

is the electric charge (electron).Basic Quantities

•The movement of charge (q) over time causes current.

(2) Current :the time rate of change of charge produces an electrical current

• the electric current is measured in Amper [A]

1 A = 1 C / 1 sec

•.current convention.

dtdq(t)i(t) = Or ∫

−∞==

t

τd τ)i( τq(t)

e e e

i

---

i(t)

time

There are 2 types of currents 1. Direct current (DC)

2. Alternating current (AC)

i (t)

time

(3) Voltage :The voltage is defined as the work or energy (in Joules) required per unit charge to move a test charge though an element

qWV = And

C 1J 11V =

• Since we are dealing with a changing charge and energy, we have

dqdwv =

(4) Power :

Power is the time rate of change of energy.

dtdw(t)P(t) =

dtdq

dqdw(t)

dtdw(t))(P ⋅==t

i(t)V(t))(P =t

•The unit of power is Watt [W].

• 1 W = 1 V * 1A

(5) Energy: energy can be expressed as

∫=

=∫=

=2

1

2

1

t

ttdti(t)v(t)

t

ttdtp(t)w(t)

Passive sign conventionCurrent flow from the positive to the negative terminal.

i(t) R

(+)

(-)

• Power can be absorbed or supplied by an element.

• Power is absorbed (or dissipated) by an element if the sign ofpower is (+)

• Power is supplied (delivered or generated) by an element if the sign of power is (-)

Circuit Active Elements:

There are 4 types of active elements (sources):

1. Independent voltage source: It is a 2-terminal sources that maintains a specific voltage across its terminals regardless of the current through it

+-

2. Independent current source:It is a 2-terminal sources that maintains a specific current through it regardless of the voltage across it terminals.

3. Dependent voltage source:It is a 2-terminal sources that generates a voltage that is determined by a voltage or current at a specified location in the circuit.

4. Dependent current source:It is a 2-terminal sources that generates a current that is determined by voltage or current at a specified location in the circuit.Example :Compute the power that is absorbed or supplied by each of the elements in the following circuit

R2

-

+-Vs = 36 V

Ix = 4 A R1

+ 12 V

++

--24 V 28 V

1 Ix

I R2 I R3=2 A

R3

)(supplies144W4)(36)(IVP xsvs −=−==

)(absorbs48W(12)(4)IVP xR1R1 ===

(absorbs)48W2)-(24)(4)I-(IVIVP R3xR2R2R2R2

====

(supplies)W8-(4)(-2)))(II(1IVP R3xR3DsDs ====

(absorbs) W56(28)(2)IVP R3R3R3 ===

Prefixes For Engineering Notation

POWER OF 10 PREFIX SYMBOL1012 tera T109 giga G106 mega M103 kilo k10-3 milli m10-6 micro µ10-9 nano n10-12 pico p

Chapter 2

Resistive CircuitsOhm’s law :The voltage across a resistor is directly proportional to the current flowing through it.

V (t) = R i(t) R ≥ 0 R1

v (t)

i (t)The symbol of ohm is Ω( )

A1V1Ω1 =

R(t)v(t)iRi(t)v(t)P(t)

R(t)v

Rv(t)v(t)(t)iR

i(t)i(t)Ri(t)v(t)P(t)

22

22

===∴

===

==

Note: Last equation says that the power at a resistor is always positive

Resistors always absorb power.

The instantaneous power P (t):

The inverse of resistance is conductance

R1G =

The unit of conductance is Siemens (S)

1V1AS1 =

The current can be also expressed as

V(t)Gi(t)=

And the instantaneous power is

G(t)ii(t)

Gi(t)i(t)v(t)P(t)

2

===

(t)G vG(t)ii(t)v(t)

(t)vGv(t)Gv(t)i(t)v(t)P(t)

22

2

===⇒

===

Open and short CircuitsOpen circuit ( R = ) ∞ G = 0

R=∞circuit circuit Open circuit

0v(t)R

v(t)i(t) =∞

==

R= 0circuit circuit Short circuit

0)(*0)(v(t) === titRi

Short circuit ( R = 0) G = ∞

Example :Consider the circuit:

Find the current and power absorbed by the resistor

+-

I

Ωvs=12 v R = 2 k

Am6Ωk2v12

Rv

I s ===

wm72m)(6(12)IvP R ===

Example:The power absorbed by a 10 k resistor in the circuit is 3.6 mW. Find voltage and current in the resistor.

RIIVP 2s ==

Ω

( )

mA0.610*3.6I

)10*(1010*3.6RPI

RPI

7

33

2

==

==

=

−

−

V6V)k(10A)m(0.6RIV

=Ω==

+-

I

vsΩk10R=

Example :

Find the value of the voltage source and the power absorbed by the resistance

G = 50 Sµ R=1/G =2*104

Wm5mA)(0.5V)(10IVPV10Ω)10*(20A)m(0.5RIV

R

4s

======

+-

I=0.5 m A

VsSµ50G =

Example :

Find R and the voltage acrossThe resistor?

Ω===

==

=

==

=

−

−

−

−

k

V

5A10*4

V. 20IVR

RA)10*(4RIV.20V

A10*4W10*80

IsPV

Is VP

3

3

3

3

P=80mW RIs=4mA

Kirchoff’s Laws:

(1) kirchoff’s current law (KCL) :the sum of all currents entering any node is zero.

∑=

=N

1kk 0(t)i

Where N= number of currents.

Here we have (4) nodes:

At node (1) :

At node (2) :

At node (3) :

At node (4) :

(t)i(t)i(t)i 521 =+

(t)i(t)i50(t)i 322 =+

(t)i(t)i(t)i50 142 =+

(t)i(t)i(t)i 543 =+

Example:Write the KCL equation

50 i2

i4 R4

i1 R1

i3

R3

i2R2

+

-

i5

Vs

(1)(2)

(3)

(4)

(2) Kirchoff’s voltage Law (KVL):

The sum of the voltage around any loop is zero.

0(t)vN

1kk∑

=

= N = # of voltage

Example:

Find VR3 ? using KVL

-30+18-5+12-15+ VR3 = 0

VR3 = 20 V

-- VR3 +

R3

VR2= 12 V R2

VR1=18 V

R1

+ -

Vs3= 15 V

+-

Vs2= 5 V

+

-Vs1=30 V

Example :Find the KVL equation for the two paths abda and bcdb

0vvv sR2R1 =−+

0vvv20 R2R3R1 =−+

Path abda:

Path bcdb:

+-

a b c

d

R1

R2 R3

VR1

VR2 VR3

20 VR1

--

-

+

+

+

Vs

Single Loop circuitsWe will discuss (2) issues :

1. Voltage divider rule:Voltage is divided between resistor in direct proportion to their resistance

v(t)RR

R(t)v

v(t)RR

R(t)v

21

22

21

11

+=

+

vRR

Rv

)RR

v(RiRv

21

11

21111

+=

+==

+-

+

-R1

R2

V1

V2-

+V(t)

How?

Multi Sources / resistors :

•Source can be added v=v1+v2+……•Resistors can be added R= R1+R2+…..

Where: v = v1 + v2

R = R1 + R2 + R3

+-

+-

R1 R2

R3v1

v2

+- RV

Single Node-Pair circuits :

We will discuss (2) issues:

1. Current-divider Rule .

(t)iRR

R(t)i

(t)iRR

R(t)i

21

12

21

21

+=

+=

1221

21

21

2211

iiiiii

iRR

i

RiRiv

−=⇒+=

=∴

==Why ??

i (t) R1 R2

i1(t) i2(t)

iRR

Ri

iRR

)R

RR(i

iRR

)RR

1(i

)i(iRR

i

21

21

1

2

1

211

1

2

1

21

11

21

+=

=+

=+

−=

2. Multiple sources/resistors :

•Current source can be added.

•Resistors can added as reciprocals

R1 R2i1(t) i2(t) R3

321

21

R1

R1

R1

R1

(t)i(t)ii(t)

++=

+=

Ri (t)

Series and parallel resistors :

Series :

Parallel

∑=

=⇒+++=N

1kksN21 RRRRRR K

∑=

=

+++=

N

1k kP

N21P

R1

R1

R1

R1

R1

R1

K

Example :Find equivalent resistance

Ωk10

Ωk1Ωk6

Ωk6Ωk6

Ωk2

Ωk4

Ωk2

Ωk2

Ωk9

( ) ( ) Ωk10Ωk6||]Ωk2Ωk1[R1 ++=

Ωk6

Ωk2

Ωk4

Ωk2

Ωk9

Ωk6 Ωk12R 1 =

Ωk2

Ωk4

Ωk9

Ωk6 k6R2=

[ ] 6kk2k6//k12R2 =+=

( ) k3k6//k6R3 ==

Ω

Ω

Ωk2

Ωk4

Ωk9

k21R3 =

( ) k5k2k4||k12Req =+=

Ωk5R eq =

3 kΩ

Example :Find all currents and voltages

The equivalent circuit is :

Ωk9

Ωk4Ωk6+- Ωk3

I2

I1 I5

I4

I3

Va Vb Vc

Ωk9

+ + +

-- -12 V

Ωk3

eqRVa+

-

12 V

Ωk9I1+

-

Am21

k63

Ωk6V

I

V3IRVAm1k3k9

V12I

a2

ieqa1

===∴

==⇒=+

=

Am21Am

21Am1IIII 3213 =−=⇒−=

( )[ ][ ]kΩ3

k6//k3k4//k9k3R eq

=

++=

V1.5V01.53V

0VVV

bb

Ωk3ab

=⇒=+−

=+−∴

Am81

k121.5

k3k9V

I b5 ==

+=

V83)(3kΩAm

81)(3kΩIV 5c ===

V1.5)Ωk(3A)m21()Ωk(3IV 3Ωk3 ===

Example :Find the source voltage Vo if I4=1/2 m A ?

Am1.5Am21Am1III

Am1k3

3k3

VI

V3Ω)k(6A)m21((6k)IV

432

b3

4b

=+=+=

===∴

===

k6

k1k2

+-

I2

I1 I5

Va+

-

Ωk3

Vo

k3 k6I3 I4

Vb

k4

+

-

k6

k1k2

+-

I2

I1 I5

Va+

-

Ωk3

Vo

k3 k6I3 I4

Vb

k4

+

-

V36V36630V

33m)k(310)(Ik4VVI)Ωk(6V

0

0

1ba10

==+=

++=+++=

Am3Am1.5Am1.5III

Am1.54k

331k3kVV

I

V3m)(1.5k)(2IΩk2V

521

ba5

2a

=+=+=∴

=+

=++

=

===∴

Example :Find V0 ? Using KVL

V10Vm)(2k)(5)(IΩ)k(5V

Am2I126kI0)(Ik5I2000)(Ik312

0

10

11

111

==

=⇒==+−+−

+-

2000 I1

R25k ohm

R13k ohm

+

-

Vs112 V

I1 +

--

Vo

Example:Find V0 using KCL:

V8(12)32V

k4k2k4V

V12V

Am10k3

4k3

1k6

1V

0)k3

V(4

k3V

k6V

m10

s0

s

s

sss

==+

=

=

−=

−+

=−++

VS

+

-

4 I0

I0

3 k

2 k

4 k V0

+

-

10 m A

Example:Find V0 using KVL

V6V12V3V

121kV(3k)V

k1VI

012I(3k)V0VV2Ik)(312

0

00

00

0

0

00

==+−

=

+−

=

=−+−=+−+−

+-

12 V

I 3 k 2 V0

1 k V0

+

-

Example:Find V0 in the network

( )

21

21

23211

210

i21i

i)Ωk(3i)Ωk(6AlsoiRRiR

0Am2ii2000V

=∴

=+=

=+−−

i1 i2

R1 R2

R3

Ωk1Ωk6Am2

V0

+

-Ωk2

2000V0

V8VAm2k2

V21

0Am2k2

V23

k2V

00

00

=⇒=

=+

−∴

k2V

iiRV

0Am2i23

2000V

0Am2ii21

2000V

02230

20

220

=⇒=

=+−

=+−−∴

Q