elec 1103 lec 2

ELEC1103 44

ELEC1103 – Semester 1 Chapter 2. Introduction

Dr. Craig Jin

12 March 2012

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BASIC CIRCUIT ELEMENTS

Ch2. I. Ideal Basic Circuit Elements

› Five Ideal Basic Circuit Elements + Ideal Wire: - Ideal Wire

- Ideal Resistor

- Ideal Current Source

- Ideal Voltage Source

- Ideal Capacitor

- Ideal Inductor

› We will first study resistive networks, leaving capacitors and inductors for the second half of the semester because their operation is governed by differential equations.

We will study these first.

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› Ideal Wire: - In an ideal wire, electrons can move freely without any change in energy. In a

real wire there will be some resistance, but this resistance is usually negligible when designing or analysing the circuit.

- Voltage difference is a measure of energy difference between two points in a circuit. Because there can be no change in energy along an ideal wire, the entire wire must be at the same voltage.

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› Ideal Independent Voltage Source: - An ideal independent voltage source maintains a specified voltage across its

terminals regardless of the current flowing through it and other elements connected to it.

- The current flowing through an ideal independent voltage source is determined by the rest of the circuit.

dc voltage source ac voltage source

12 V +_ 5 cos(2πt) +_

Batteries can be modeled as a dc voltage source

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› Ideal Independent Current Source: - An ideal independent current source maintains a specified current flow through

its terminals regardless of the voltage across it and other elements connected to it.

- The voltage across an ideal independent current source is determined by the rest of the circuit.

dc current source ac current source

2 A 3 sin(100πt)

Keithley current source

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› Ideal Dependent Sources or Controlled Sources: - Drawn with a diamond shape around it.

- Come in four varieties, discussed in following slides.

Independent Sources

Dependent Sources

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vx

+

_2 vx

+_

› Voltage Controlled Voltage Source (VCVS) › A VCVS maintains a voltage across its terminals that is proportional to the

voltage across a pair of terminals elsewhere in the circuit.

› In general the voltage across a VCVS is where is a gain with units (V/V) so it is really unitless.

xvµ µ

VCVS

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› Current Controlled Voltage Source (CCVS) › A CCVS maintains a voltage across its terminals that is proportional to the

current through a circuit element elsewhere in the circuit.

› In general the voltage across a VCVS is where is a gain with units (V/A).

xiρ ρ

CCVS

3 ix+_

ix

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› Voltage Controlled Current Source (VCCS) › A VCCS maintains a current through its terminals that is proportional to the

voltage across a pair of terminals elsewhere in the circuit.

› In general the current through a VCCS is where is a gain with units (A/V).

xvα α

VCCS

vx

+

_3 vx

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› Current Controlled Current Source (CCCS) › A CCCS maintains a current through its terminals that is proportional to the

current through an element elsewhere in the circuit.

› In general the current through a CCCS is where is a gain with units (A/A) so it is unitless.

xiβ β

CCCS

iy

2 iy

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› Valid or Not Valid ?

Yes No

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Yes No

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Yes No

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RESISTOR AND OHM’S LAW

Ch2. II. Resistor and Ohm’s Law

› Ohm’s Law: The voltage across an ideal resistor is proportional to the current through it.

› The circuit symbol for a resistor is: i(t)

+

_v(t) R

( ) ( )v t i t R=

i-v characteristic of an ideal resistor

i

vv = i R

( ) ( ) / ( )i t v t R G v t= =

1GR

=

The constant of proportionality is the resistance R. The units of resistance are V/A, which are called ohms (Ω).

G is called the conductance G. The units of conductance are A/V, which are called siemens (S).

Ohms’ Law

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› Passive sign convention and Ohm’s Law: (Current flows into the positive terminal of the device, then use + sign to right Ohm’s law.)

v = iR v = - iR

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› Real Resistors:

Area

AreaLR ρ

=

Resistivity Resistance increases with length, but decreases with cross-sectional area.

The units of resistivity are Ωm.

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› Resistivity of Materials at 300 K:

Conductors Aluminium 2.73 ×10–8

Carbon (amorphous) 3.5 ×10–5

Copper 1.72 ×10–8

Gold 2.27 ×10–8

Nichrome 1.12 ×10–6

Silver 1.63 ×10–8

Tungsten 5.44 ×10–8

Semiconductors Silicon (device grade) 10–5 to 1 depends on impurity concentration

Insulators Fused quartz > 1021

Glass (typical) 1 ×1012

Teflon 1 ×1019

Smaller the resistivity, the better the conductor. Resistivity depends on the temperature. A short circuit (i.e. an ideal conductor) is a resistor of 0 Ω resistance (no such material exists,

but a thick copper wire can be a good approximation). An open circuit is a resistor of 0 S conductance (or equivalently ∞ Ω resistance). Again no

such material exists, but some insulators can be a good approximation.

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› Resistance Calculation:

Calculate the resistance of a copper wire having a diameter of 2.05 mm and a length of 10 m.

2 3 26 2(2.05 10 ) 3.3 10 m

4 4dA π π −

−×= = = ×

8

6

(1.72 10 )(10) 0.052(3.3 10 )

LRAρ −

−

×⇒ = = = Ω

×

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› Real Resistors:

http://www.electronics-tutorials.ws/resistor/res_1.html




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› Real Resistors:



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› Chip Resistors:

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› Chip Resistors:

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› Power in a Resistor:

› We can calculate the power in a resistor in terms of voltage using Ohm’s Law (v=iR) and the expression for power (p=vi).

2v vP vi vR R

= = =

( ) 2P vi iR i i R= = =

› We can calculate the power in a resistor in terms of current using Ohm’s Law (v=iR) and the expression for power (p=vi).

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› Using Ohm’s Law:

Given current and resistanceFind the voltage

AI 2=Ω= 5R

Given current and resistanceFind the voltage

AI 2=Ω= 5R

−=

+][10 VV

v iR=

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Given Current and VoltageFind Resistance

−

+][20 V

][4 AI =




vRi

=

Given Current and VoltageFind Resistance

−

+][20 V

][4 AI =

Ω= 5R

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Given Voltage and ResistanceCompute Current

−

+][12 V Ω=3R




viR

=

Given Voltage and ResistanceCompute Current

−

+][12 V Ω=3R

][4 AI =

Note: we determine the direction of current (blue arrow) using the passive sign convention.

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› Using Ohm’s Law and Keeping Units Straight:

Table 1 Keeping Units Straight

Voltage Current Resistance

Volts Amps Ohms

Volts mA kΩ

mV A mΩ

mV mA Ω

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› More Ohm’s Law:

› Find the current and power absorbed by the resistor ?

[ ][ ] [ ]12 V

6 mA2 k

I = =Ω

[ ]( ) [ ]( ) [ ]12 V 6 mA 72 mWP VI= = =

+

-

Use passive sign convention.

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› More Ohm’s Law:

› Find the voltage of the voltage source and power absorbed by the resistor ?

3

S 6

0.5mA 0.5 10 A 10V50 S 50 10 S

IV IRG µ

−

−

⋅= = = = =

⋅

GIRIP

22 ==

( )232

6

0.5 10 [ ]0.5 10 [ ] 5[ ]

50 10 [ ]

AP W mW

S

−−

−

×= = × =

×

+

-

Use passive sign convention.

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NODES, LOOPS, AND BRANCHES

Ch2. III. Nodes, Loops, and Branches

› A branch represents a single element such as a voltage source or a resistor.

› A node is the point of connection between two or more branches.

› A loop is any closed path in a circuit.

› A network with b branches, n nodes, and L independent loops will satisfy the fundamental theorem of network topology:

› b = L + n - 1

How many branches, nodes, loops ?

3 Nodes 3 Loops 5 Branches 5 = 3 + 3 - 1

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› How many branches, nodes and loops are there ?

3 Nodes 3 Loops 5 Branches 5 = 3 + 3 - 1

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› An essential node is a node where three or more circuit elements join.

› An essential branch is a branch path connecting two essential nodes.

› A network with bE branches, nE nodes, and L independent loops will satisfy the fundamental theorem of network topology:

› bE = L + nE - 1

How many essential branches, essential nodes, loops ?

2 Essential Nodes 3 Loops 4 Essential Branches 4 = 3 + 2 - 1

Node 3 is not an essential node.

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KIRCHHOFF’S CURRENT LAW

Ch2. IV. Kirchhoff’s Current Law (KCL)

› Charge is not created nor destroyed in electric circuits. For this physical reason, we have Kirchoff’s Current Law.

› Kirchhoff’s Current Law states that the algebraic sum of currents entering (or leaving) a node (or a closed boundary) is zero.

1 2 3 4 5 0i i i i i− + + − =

i1

i2

i3

i4 i5

1 2 3 4 5 0i i i i i− + + + − =

Sum currents entering node: Sum currents

leaving boundary:

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Ch2. IV. Kirchhoff’s Current Law

B

C

i2

i1 Supernode

F

A

E

D

G

i3

id

ieic

1 2 3i.e. KCL at supernode: i i i= +

1

2

3

c e

c d

d e

i i ii i ii i i

= += ++ =

To verify the claim note that:

(now just add these equations)

KCL can be applied to supernodes. A supernode is obtained by defining a boundary that enclosed more than one node.

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Ch2. IV. Kirchhoff’s Current Law mAI 501 −=

mAmAmAIT 204010 ++=

0410 1 =−− ImAmA 01241 =−+ mAmAI03 12 =−+ ImAI

1I Find TI Find

1I Find21 I and I Find

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KIRCHHOFF’S VOLTAGE LAW

Ch2. V. Kirchhoff’s Voltage Law (KVL)

› Electric circuits obey the principle of conservation of energy.

› Similar to a gravitational field, the electric field is conservative. This means that if a charge moves around a loop and comes back to the same point, the net energy gain or loss must be zero.

› Recall that ΔW = qV

+

AV

BBV

q

CV

−

+ABV −

+BC

V

−+ CAV

ABqVW =∆

BCqVW =∆

CAqVW =∆

Total Energy Gain Around Loop ( ) 0AB BC CDq V V V= + + =

- VCA +

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› Electric circuits obey the principle of conservation of energy, for this physical reason we have Kirchoff’s Voltage Law.

› Kirchhoff’s Voltage Law states that the algebraic sum of the voltages around a closed path, or loop, is zero.

01

=∑=

M

mnv

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Loop 3

A

B

C E

D

va

+

_

+_ vd+ _vb

vc

+

_ve

+

_

0e c dv v v− − =

Note that in obtaining loop equations, we sum the voltage drops around the loop (could also sum voltage rise). In the circuit analysis you are free to choose the direction of the KVL loop. KVL for loop 3 can be obtained by taking the difference of KVL for loop 2 and 1, i.e. it does not contain any new information.

Loop 1Loop 2

KVL for loop 1:

KVL for loop 2:

KVL for loop 3:

0a c bv v v− − =

0a e d bv v v v− + − =

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› Find Vae and Vec?

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10 24 014[V]

ae

ae

VV

+ − ==

GIVEN THE CHOICE USE THE SIMPLEST LOOP

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4 6 010[V]

ec

ec

VV

+ + == −


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KIRCHOFF’S LAWS & OHM’S LAW

Ch2. VI. Kirchhoff’s Laws and Ohm’s Law

Find io ?

KCL at node b: 1 6 0oi i− + =

Ohm’s Law for 10Ω Resistor: 120 10b oV i− =

Ohm’s Law for 50Ω Resistor: 150bV i=

1 6 0oi i− + =

1120 50 10oi i− =

( )120 50 6 10o oi i− + =

180 60oi− =

[ ]3 Aoi = −

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Ch2. VI. Kirchhoff’s Laws, Ohm’s Law, and Dependent Sources

Find vo ? (Note vo = Vb)

KCL at node b: 0 5 0i i i∆ ∆− + =

Ohm’s Law for 5Ω Resistor: 500 5bV i∆− =

Ohm’s Law for 20Ω Resistor: 0 20bV i=

( )6 20 120bV i i∆ ∆= =

500 5120

bb

VV− =

12000 24 b bV V− =

[ ]480 VbV =

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Find vo ?

KCL at Boundary means no current here!

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Find vo ?

Ohm’s Law for 6Ω Resistor: [ ]5 A3si =

Ohm’s Law for 2Ω and 3Ω Resistor: ( )053 2 33

i = +

[ ]0 1 Ai =

[ ]0 3 Vv =

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› How to electrically model the human body:

› Rskin (dry)=15 kΩ, Rskin (wet)=150 Ω

› Rlimb=100 Ω

› Rtrunk=200 Ω

ELECTRICAL SAFETY

Ch2. VII. Electrical Safety

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ELECTRICAL SAFETY


› First, consider electrostatic shock on a dry day:

› Around 20,000 to 40,000 V and 40 A – but only

for a few microseconds.

› Current flow is mainly over the body surface

› Annoying to you, deadly for electrical components

› Consider 50 Hz, 230 V electrical outlets, danger is shown to the right. 0.1 to 0.2 A is fatal causing ventricular fibrillation. 60 Hz current penetrates the body more deeply and electrical outlet sustains the current.

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ELECTRICAL SAFETY


Man working in damp basement using a drill. Insulation on wire nicked and touches casing of drill. Damp concrete relatively good conductor. If the case is not grounded, man could receive fatal shock. If case is grounded, the circuit breaker shorts and lights go out. Which do you prefer ?

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ELECTRICAL SAFETY


Boys in and near a pool. Ground fault in pool lighting. Vinyl lining of pool means water is electrically insulated. One boy in pool and one outside. If they touch and outside boy is grounded, who is more likely to die?

Outside boy, because current more likely to go through the heart.

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ELECTRICAL SAFETY


A crane operator touches high-voltage line at 7200 V. The crane is about 10 m from the pole which is earthed. He jumps out of the crane runs toward the pole and dies – why?

There is a 7200 V/10 m or 720 V/m potential gradient between the crane and the pole. If the man’s running stride is 1 m, his body experiences 720 V. He dies, but a man standing still survives.

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