Fall 2001Circuits I1 Chapter 1 - Basic Concepts 1.1System of Units 1.2Basic Quantities 1.3Circuit Elements

Fall 2001 Circuits I 1

Chapter 1 - Basic Concepts

• 1.1 System of Units

• 1.2 Basic Quantities

• 1.3 Circuit Elements


Reading Assignment:

Chapter 1, Pages 1-11

HomeworkEnd-of-Chapter Problems: P1.13, P1.18, P1.21

(PLUS WebCT Problems)


Electrotechnology pervades our day-to-day lives:

electrical appliances

Refrigerators, stoves, stereos, hairdryers, ...

communication devices

telephones, faxes, computer and data networks

computers

main frames, personal (PCs), workstations, embedded systems, PLCs

instrumentation

meters, measurement systems, controls

“Today we live in a predominately electrical world.”


System of Units

Unit of Measure Symbol Measuresmeter m distance

kilogram kg mass (weight)second s timeampere A electric current

degree Kelvin K temperaturecandela cd luminous intensity

The international system of units (SI) is composed of the following basic units of measure:


Standard SI PrefixesStandard prefixes are employed with SI units to represent powers of ten. These prefixes, sometimes referred to as engineering notation, are used extensively in representing the values of various circuit parameters.

Prefix Power Prefix Powerpico (p) 10-12 kilo (k) 10+3

nano (n) 10-9 mega (M) 10+6

micro () 10-6 giga (G) 10+9

milli (m) 10-3 tera (T) 10+12

1 103 106 109 1012

kilo(k)

mega(M)

giga(G)

tera(T)

10-12 10-9 10-6 10-3

pico(p)

nano(n)

micro( or u)

milli(m)


TI-85 MODE Function

To Select Engineering Notation


Example - Using Engineering Notation

If the local power company charges $0.12 per kWh (kilowatt-hour) of energy used, how much would it cost to illuminate a 100-Watt lamp for an entire year?

Solution:

12.105$Cost

kWhper12.0$kWh876Cost

kWh876W

day

hours24days365W100W

)kWhorhourskW(tPW

TimePowerEnergy

TI-85 Screen

(Wh)

(kWh)


Basic Quantities

• What is Electricity? - A natural force (like wind or water flow) which can be harnessed for some beneficial purpose.

• What causes electricity? Just as wind is due to the movement of air molecules, electricity is caused by the movement of charged particles (i.e., electrons).

• The study of electricity has been going on since about 600 B.C. when it was noticed that amber, when rubbed, acquired the ability to pick up light objects (we call this static electricity).


DefinitionsCharge is an intrinsic property of matter (like mass and size).

Symbol1: q or Q Unit: coulomb (C)

The elemental unit of charge is the charge of one electron, e:

e = -1.6022 x 10-19 C (1 proton = +1.6022 x 10-19 C)

Current is a measure of the amount of charge flowing past a point in a given amount of time.

Symbol1: i or I Unit: ampere (A)

1A 1C/s (coulomb per second)

1Lowercase is used to indicate a variable (time-dependent) quantity while upper case indicates a constant value.


Current has magnitude and direction.

Electrical Conductor Segment

2A

Conventional Current Flow

- - --

Electron Flow

(The flow of electrons is analogous to the flow of water molecules in a pipe or hose.)

Note:2A = 2C per second = (2/1.6022E-19) electrons per second2A 12,500,000,000,000,000,000 electrons per second


Typical Current Magnitudes106

Lightning Bolt104

Large Industrial Motor Current102

Typical Household Appliance Current100

Causes Ventricular Fibrillation in Humans10-2

10-4

10-6

IC Memory Cell Current10-8

10-10

10-12

Synaptic Current (Brain Cell)10-14

Cu

rren

t in

Am

per

es (

A)


Direct Current (dc) Versus Alternating Current (ac)

DC signals are constant with respect to time:

t

Ix

2A

AC signals vary with respect to time:

t

i(t)

2A

-1A

periodic ac signal

t

i(t) = 2e-at

2A

nonperiodic ac signal


Charge - Current Relationships

t

0t

t

0t

t

0t

t

0t

t

0t

t

0t

d)(i)0

t(q)t(q

d)(i)0

t(q)t(q

d)(i)(q

d)(i)(dq

dt)t(i)t(dqdt

)t(dq)t(iCurrent is proportional to the rate of change of charge.

Charge is increased or decreaseddepending on the area under thecurrent function.


Example 1

Given: q(t) = 12t CFind: i(t)

Solution: Since current is proportional to the rate of change of charge, we take the derivative of q(t) to find i(t):

i(t) = q(t)i(t) = 12 A

Since the slope (rate of change) of q(t) is constant at 12 Coulombs per second, the current is constant.


Example 2

Given: The current entering an element is described by thefunction in the graph shown below.

t

(seconds)

i(t)

1 2 3

1

2

3

1tt1t01)t(i

Find: The amount of change in the charge of the elementover the interval 0 to 3 seconds.


Example 2 - Solution

To find the change in the charge of the element, we need to find the area under the function i(t) over the interval 0 to 3 s. This can be done two ways for this example: by integration or by using area formulas for rectangles and triangles.

t

i(t)

1 2 3

1

2

3

C5)5.05.4()01(3

122t1

0

3

1

10

ttdtdt1Q

C5A2s221s3A1AreaQ

Note: 5C is the increase in the charge of the element. We must know the initial charge to determine the total charge of the element.


Definition - VoltageAs stated earlier, current is the flow of charge through a conductor. Voltage is the force required to move the charge through the conductor. Hence, voltage is also referred to as electromotive force and potential difference.

+

-

BatteryLamp

i (current flow)

+

Vab

-

+

Vab

-

i

b

a


BatteryLamp

i (current flow)

+

Vab

-

+

Vab

-

The current from the battery flows through a voltage rise since the battery is supplying energy. There is an equal voltage drop across the lamp since the lamp is absorbing the energy supplied by the battery. This is similar to the counter force created by friction when pushing an object up an incline.

Fab

Fab

b

a


Electromotive force is measured in volts. If one joule of energy is required to move one coulomb of charge through an element, then there is one volt (V) of potential difference across the element.

coulombjoule1volt1

Like current, the voltage across an element may be positive or negative. The assignment of a sign to a voltage is somewhat arbitrary, but usually represents whether potential difference is above or below some reference potential.


As we traverse the circuit in a clockwise direction, there is a voltage rise across the battery; we could assign this a “positive” sign. Since we experience a voltage drop across the lamp, this would be assigned the opposite sign or a “negative” value. We could just as correctly assign all voltage drops a positive sign and, therefore, all voltage rises a negative sign.

+

BatteryLamp

i+

6 V

-

+

6 V

-

“0” volts (ground)

(6 V above ground)


Power and EnergyEnergy is the capacity to do work. For example, energy is stored in a battery and, when connected to a “load” such as a lamp, there is a continuous exchange of energy from the battery to the load. (There is also a transformation from energy stored in a chemical form to energy emitted in the form of heat and light.) Typically, electrical energy is measure in joules (J) or watt-hours (W-h). (1W = 1J/s)

BatteryLamp

i

+

Vab

-

+

Vab

-


dtdwp

Battery Lamp

power flows from the source to the load

+

V

-

+

V

-

i i

p = -vi

Supplies Power

p = +vi

Absorbs Power

(Passive sign convention)

Power is a measure of how much work is done per unit time, or, in other words, a measure of the rate at which energy is transferred (supplied or absorbed). There is always a conservation of energy (power). That is, the battery supplies exactly the amount of power consumed by the load, no more no less. Typically, electrical power is measured in watts (W).


Mathematical Relationships

In electrical terms:

velocityforcepower In general:

ondsecjoule1va1ampvolt1watt1

wattsivp

Using previous definitions:

dqdw

dtdq

dtdw

v

,sodtdqiand

dtdwpbut

ipv


The total energy absorbed or supplied is equal to the area under the power function in the interval t0 to t. If the area is positive, then energy is absorbed by an element; if the area is negative, the element is supplying power.

t

t

t

t

t

t

0

0

00

pdt)t(w)t(w

pdtdw

pdtdwdtdwp


Example 3 - The current in a typical lightening bolt is 20KA, and the bolt lasts typically for 0.1 seconds. If the voltage between the earth and clouds is 500MV, find (a) the total charge transmitted by the bolt and (b) the total energy released by the bolt.

Solution: C2000]t10x2[dt10x2dt)t(iQ 1.00

41.0

0

1.0

0

4 (a)

TJ1J10x1VC10x10w

)V10x5()C2000(VQw1211

8

(b) ( )dw

v tdq

WV

Q

W V Q

Note,

If V is constant,

Then,A typical automobile battery is rated at 12V and 100 amp-hours (Ah). 100Ah = (100 C/s)(3600s) = 3.6 105C 12V3.6 105C = 4.32 106J (MJ)The lightening strike has as much energy as 230,000 batteries!


Example 4 - (a) Find the power absorbed by the element shown and (b) the total energy absorbed over a 10-second interval.

10

0

J400dt40w(b)

10A + 4v -

Solution: p = vi =4v x 10A = 40w(a)


Example 5 - Find the power absorbed or supplied by the elements shown below

Solution: (a) p = -vi = -2v x 4A = -8w(element-a is supplying 8w of power)

4A - 2v +(a)

+ 2v -(b)

-2A

(b) p = -vi = -2v x (-2A) = +4w(element-b is absorbing 4w of power)


Example 6 - Determine the unknown voltage or current.

Solution:(a) -20w = -V x 5A

V = 4v

5A - V=? +(a)

- 5v +(b)

I=?

P = -20w P = 40w

(b) 40w = -(5v x I)I = -8A


Circuit Elements• Circuits (electrical systems) consist interconnected devices.

• Electrical devices can be categorized as either a source or a load.

• A source can be either independent or dependent, a voltage source or a current source.

• Independent voltage sources provide power at a constant voltage

• Independent current sources provide power at a constant current.

• The voltage level of a voltage source is not affected by the amount of power demanded from it.

• Similarly, the current level of a current source is not affected by the amount of power demanded from it.


Independent Voltage Sources:

+V-

Constant (battery)

LOAD

I

+v(t)

-

+-

Time-varying

LOAD

i(t)

p

v

V

Since p = vi, as the demand for power increases, the current, i, increases because v is constant.


Independent Current Sources:

p

i

I

Since p = vi, as the demand for power increases, the voltage,v, increases because i is constant.

+v(t)orV-

LOADi(t)orI


Dependent Source (Controlled) Sources)An independent source provides power to a circuit at a constant voltage or current, and the level of that voltage or current is unaffected by the circuit to which it is attached. Generally, independent sources are used to represent devices such as batteries or generators.

A voltage or current associated with a dependent source is controlled or affected by the circuit to which it is attached. Dependent sources, also called controlled sources, are used to model electronic devices, sensors and other devices which are influenced by some physical or electrical parameter. For example, a device called a thermocouple generates a small voltage (in the millivolt range) which is proportional to temperature. So, a thermocouple can be thought of as a temperature-controlled voltage source. That is, v(t) = KT, where T is the temperature and k is a constant,


Dependent SourcesThere are four types of controlled sources:

voltage-controlled voltage sources (VCVS)current-controlled voltage sources (CCVS)voltage-controlled current sources (VCCS)current-controlled current sources (CCCS)

+vo

-

+-

+v = vo

-

VCVS

io

+-

+v = rio

-

CCVS


Dependent Sources

VCCS+vo

-

i = gvo

CCCSi =vo

io


Example 7 - Given the two networks shown below we want to determine the outputs.

Solution:

(a) Vo = 20 2V = 40V

(b) Io = 50 1mA = 50mA

+Vs = 2V

-

+-

+Vo= 20Vs

-

(a)

(b)

50Is

Is = 1mA Io


Example 8 - Compute the power that is supplied or absorbed by each element in the network below.

Solution:

P36V = -(36V)(4A) = -144W (supplies 144W)P1 = +(12V)(4A) = +48W (absorbs 48W)P2 = +(24V)(2A) = +48W (absorbs 48W)PDS = -(4V)(2A) = -8W (supplies 8W)P1 = +(28V)(2A) = +56W (absorbs 56W)Total = 0W

1Ix

+

-

1

2 3

- +

Ix = 4A2A

2A

+

24V

-

+

36V

-

+

28V

-

+ 12V -


Example 9 - Determine the value of Io.

Solution:

P2A = -(6V)(2A) = -12W P1 = +(6V)(Io) = 6IoW = ? P2 = -(12V)(9A) = -108W P3 = -(10V)(3A) = -30WP4V = -(4V)(8A) = -32W PDS = +(16V)(11A) = 176WSince PTotal = 0W = -182 + 6Io + 176 6Io = 6W Io = 1A

- 6V +

+

-

1 2

3+

-8Ix

Ix = 2A

2A

+

10V

-

+

4V

-

+ 6V - - 12V +9AIo

8A3A

11A

Documents

Fall 2001Circuits I1 Chapter 1 - Basic Concepts 1.1System of Units 1.2Basic Quantities 1.3Circuit Elements