circuit theory i

Enginr 124 WS 2002 Part 2: Charge, etc.

Circuit Theory I

Part 2: Charge, Current, Voltage, Energy, Power, Sources

2.1 Major Topics of the Course

2.2 Circuits Quantities

2.3 Charge

2.4 Current

2.5 Voltage

2.6 Energy and Power

2.7 Practical Examples

2.8 Independent & Dependent Sources


2.2 Fundamental Circuits Quantities

q (charge: coulombs, C)

i (current: amperes, A)

v (voltage: volts, V)

w (energy: joules, J)

p (power: watts, W)

S complex power: volt-amps, VA, watts, Wvolt-amps-reactive, VAR

f (frequency: hertz, Hz)


2.3 Charge

When you rub a comb with a woolen cloth, a negative charge is produced on the comb and a positive charge is produced on the cloth. (Benjamin Franklin defined the charge on the comb as negative.)

The comb acquires its negative charge because some of the electrons on the cloth are rubbed off onto the comb.

One coulomb is defined as the amount of charge on identically charged particles, separated by one meter in a vacuum, that repel each other with a force of 10–7 c2 N, where c is the velocity of light (2.997… x 108 m/s).

Recall that a single electron charge is –1.602 x 10–19 C. This is quite small: For example, the charge needed to plate 1 g of copper is 3,036 C.


2.4 Current

Charge in motion results in energy transfer. Of special interest is the case where the motion is confined to a definite path. In this case, current “flows.” Current is the movement of electrical charge—the flow of electrons through an electronic circuit.

Current (i) is the rate at which charge “flows,” or is transferred.

1 amp = 1 coulomb/second

It is convenient to think of current as the flowing motion of positive charge even though we know that current flow in metallic conductors results from electron motion (that is, the charge carriers are negative and move in the opposite direction.)

i = dq/dt amperes

Memorize me!!


Charge/Current Relationship

+ + + ++ + + +++

The charge and the current flow through the conductor, passing from one side of the imaginary surface to the other.

During the time interval from ta to tb the charge

passing through the imaginary surface is

i(t) i(t)

Memorize me!!b

a

t

tidtq


In defining current, both an arrow and a symbol are required. It takes two!

Example:Circuit analysis shows that in the given circuit, i1

has the value 0.5 A. Give the corresponding physical interpretation.

1 V

10

10

5

12 V

i1 = 0.5 A

Solution:Since i1 is 0.5 A, this means that 0.5 A, or 0.5 C/s

is flowing in the direction of the current-reference arrow. This current flows through the 5 , to the + side of the battery, through the battery, exiting from the – side of the battery.

+

_

+

_


1 V

10

10

5

12 Vi2 = ?+

_

Solution:

Here, the circuit is the same, but the direction of the current reference arrow is reversed. 0.5 A is still flowing through the 5 , to the + side of the battery, through the battery, exiting from the – side of the battery. The value of i2 has to be

– 0.50 A.

Example (cont.)Suppose that the current-reference direction and symbol are changed as indicated. What is the value of i2?


5 i2 = –0.5 A

5 – i3

5 – i4

– i3 = – 0.5 A

i3 = 0.5 A

– i4 = 0.5 A

i4 = – 0.5 A

Example (cont.)

For the same circuit show four ways to represent the 0.5 A flowing through the 5 in the direction towards the positive terminal of the 1 V battery.

1 V

10

10

5

12 V+

_

Solution:

5 i1 = 0.5 A


Since i1 = 0.5 A, 0.5 C of charge must be flowing

through the circuit elements each second.

As an equation,

i1 = dq/dt

= 0.5 A

Current is a through variable.Current flows, but does not accumulate.

Charge is also a through variable.Charge flows, and (sometimes) accumulates.

i1 = 0.5 A

+

Current and Charge-Flow

_

5


Types of Current

DC Current

Exponential Current

Damped Sinusoidal Current

Sinusoidal or

AC Current

Transient Current

i

t

DC:Doesn’t Change!

0 5 10 15 20-1

-0.5

0

0.5

1

0 5 10 15 20-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20-3

-2

-1

0

1

2

3

4

5

0 5 10 15 200

0.2

0.4

0.6

0.8

1


How Large Can Currents Get?

Phenomenon Type Value

Lightning bolt transient 106 A

industrial motor AC 103 A

household appliance AC 101 A

ventricular fibrillation transient 10–1 A

IC memory cell transient 10–7 A

brain cell transient 10–13 A


Example 1: DC current.

Find the charge passing through a point in a conductor if the current passing through the conductor has the waveform indicated.

i

t

2 A

Solution:

In any 1-second interval the charge that has flowed past a given point in the conductor will be:

DC:Direct Current.Does not Change!Current does not change with time.

a

a

a

a

t 1

t

t 1t

2dt

[2t] |

2 C

q

b

a

t

tidtq

Memorize me!!


Example 2

For the given current, which is given in graphical form, find the charge passing through a given point in the time interval 0-4 s.

i

t

10 A

10 2 43

Solution:

For the time interval 0-4 s, the charge passing through a given point on the conductor that carries the current is:

q = i dt

= ½ x 1 x 10 + 1 x 10 + ½ x 1 x 10 + ½ x 1 x 10

= 25 C

4

0


2.5 VoltageThe separation of charge creates an electric force, which was recognized by the 18th century Italian physicist Allessandro Guiseppi Antonio Anastasio Volta. Voltage is the energy per unit charge that is created by the separation.

Memorize me!!

a b

The current (and charge) flowing from a to b, through the circuit element, requires an expenditure of energy. Voltage is the amount of energy expended on one unit of charge to move it. The voltage of point b with respect to point a is the work per unit charge required to move that charge from point a to point b:

circuitelement

i i

vba = wba/q

1 V = 1 J/ C

. . . . . .

We say that there is a voltage difference between points a and b, with point b at the assumed higher voltage level. We use the words assumed higher voltage level because if this turns out to be a negative number, then a is actually at the higher voltage level. This will be an important concept throughout our study: Assume a direction for the voltage and solve the required equations for its value. If this value turns out to be a negative number, we can simply reverse the direction of the assumed voltage and remove the negative sign if we choose. However, it would be equally correct to retain the original assumed polarity and the negative sign in its value.


Voltage is an “across variable.” We say that an electrical voltage of potential difference exists between a and b, or that there is a voltage across the circuit element. The voltage across the element is the amount of energy that must be given up to move a unit positive charge from the + terminal to the – terminal. Voltage is the electrical force, or “pressure,” that causes current to flow in a circuit.

A voltage or potential difference can exist between a pair of electrical terminals whether or not a current is flowing. (There could be nothing connected on the left side of a and b above. That is, they could be dangling in midair.)

ab

circuitelementv

–

+i

i

. . .


Implied-Source Convention

When you see the circuit element with an indicated voltage across it, drawn as stand-alone, the implication may be that there is a source present to supply the indicated voltage, even though that source is not explicitly shown. This usage is referred to as the implied-source convention. This convention is used to simplify the drawings.

a

b

v

–

+

a

b

v

–

+

_

+ v


In defining voltage, a plus-minus sign-pair AND a symbol are needed. It takes two!

a

b

v

–

+

a

b

v

a

b–

+

+ –v

v

v +–

–

+

OK! OK!

OK!

OK!

a

b


a

b

v

–

+

Different orientation, same story:

a is v volts above b.

b is v volts below a.

The voltage rise from b to a is v volts.

The voltage rise from a to b is –v volts.

The voltage drop from b to a is –v volts.

The voltage drop from a to b is v volts.

v +–b a

a is v volts above b.

b is v volts below a.

The voltage rise from b to a is v volts.

The voltage rise from a to b is –v volts.

The voltage drop from b to a is –v volts.

The voltage drop from a to b is v volts.


Voltage and Potential Difference

The voltage across two nodes in the circuit is the potential difference between the two nodes. The potential difference may utilize a third node as a point of reference.

The analogy with elevation difference:

0 feet (sea level, the reference )

20,000 feet

29,000 feet

Mount Everest

Sea

= 9,000 feet

–

+


a

b

vab

Double-Subscript Notation

vba is the drop from b to a. (Note

that this voltage is not specifically shown here.)

vab is the drop from a to b.

The voltage at a is higher than that at b by vab.

The plus-minus sign-pair is not needed here.

For the symbol vab, regard terminal a as the +

terminal and terminal b as the – terminal.

For the symbol vba, regard terminal a as the –

terminal and terminal b as the + terminal.


1 V

10

10

5

12 V+

_

VoltmetersThe voltage across two nodes is measured by a voltmeter. The voltmeter measures the voltage drop between its + terminal and its – terminal.

The voltmeters below are measuring branch voltages: the voltages across the branches containing the 5 resistor branch and the 1 V voltage source branch.

+ _1.00+ _–2.50

– 2.5 V +

voltmeter voltmeter


2.6 Energy and Power

q

a

b

In keeping with the principle of conservation of energy, the energy that is utilized in moving the charge through the element is conserved (although it does change form).

Depending on the particular kind of circuit element present, this energy:

-- comes from the circuit element, or

-- goes to the circuit element and is converted irreversible into heat, light, acoustic energy, or some other non-electrical form, or

-- goes to the circuit element and is stored in some form that is still available as electric energy.

circuitelement


Power in Terms of v and i

w = energy in joules, J

p = power in watts, W

= energy/time

= dw/dt

Recall that v dw/dq, so that

dw = vdq so that

p = dw/dt

= vdq/dt

Check using dimensions: J/C x C/s = J/s = W.

Power is the product of the across variable v and the through variable i. In a translational system, p = force x velocity = through x across.

Memorize me!!

p = vi


Example: Energy Calculation (A&S P1.13)

The voltage and current for a device are shown. Find the total energy absorbed by the device during the interval 0-4 s.

v(t)

–

+

i(t)

0 1 2 3 4 t(s)

0 1 2 3 4 t(s)

i(mA)

v(V)

Solution:

t t

0 0w(t) p(t)dt v(t)i(t)dt

By referring to the graph we find that the equations for i(t) and v(t) are:

25t for 0 t 2i(t)

25t 100 for 2 t 4

10t for 0 t 1

v(t) 10 for 1 t 3

10t 40 for 3 t 4

50

10


Solution (cont.):

25t for 0 t 2i(t)

25t 100 for 2 t 4

10t for 0 t 1

v(t) 10 for 1 t 3

10t 40 for 3 t 4

1 2 3 4

0 1 2 3

1 2 3 42 2

0 1 2 3

t 2 t 3 t 4t 32 2 3 2t 2t 1 t 2 t 3 t

w(4) 25t 10tdt 25t 10dt 25t 100 10dt 25t 100 10t 40 dt

250 t dt 250 tdt 10 25t 100 dt 250t 2000t 4000 dt

250 250 25 250t 10 t 100 t t 1000 t

3 2 2 3

t 3

24000

250 125 250125 3 10 100 37 1000 7 4000

3 2 3

916.7 mJ

0 1 2 3 4 t(s)

0 1 2 3 4 t(s)

i(mA)

v(V)

50

10


circuitelementv

–

+i

i

Within the circuit element, the current arrow is pointing in the direction from + to – (in the direction of the voltage drop v within the element).

When this passive sign convention is being used:

If v i > 0, then the circuit element is absorbing energy (getting hot!--if it’s a resistor).

If v i < 0, then the circuit element is supplying energy. (A resistor can’t do this!)

. . .

Passive Sign Convention


power absorbed by the circuit element:

pabsorbed = v i

power supplied by the circuit element:

psupplied = – v i

The power absorbed by a circuit element and the power supplied by that same circuit element are related by

pabsorbed = – psupplied

* We say it this way to make the point that the current and voltage notations may already be set up for you on the circuit and you may not have the flexibility to change them.

When the passive convention is the one that “fits”*:

anycircuitelement: v

–

+i

v

–

+

i

…

…


anycircuitelement: v

–

+

i

power absorbed by the circuit element:

pabsorbed = – v i

power supplied by the circuit element:

psupplied = v i

The power absorbed by a circuit element and the power supplied by that same circuit element are related by

pabsorbed = – psupplied.

When the active convention is the one that “fits”:

v

–

+

i


Example 1

Determine how much power the circuit element is either absorbing or delivering.

v = 2 V–

+3 A

. . .

Solution:

Remember: When the passive sign convention is being used: If the numerical value of the product vi is positive, then power is truly being absorbed by the circuit element. If vi is negative, then the circuit element is absorbing negative power, or delivering it to the external circuit. In this case,

p = vi

= (2) (3)

= 6 W absorbed by the circuit element

circuitelementp = ???

Red (absorbing)3 A


v = – 2 V+

–. . .

Solution:

Here, you need to recognize that the current reference direction is from + to – within the circuit element, so that the passive convention is again the one being used. Thus,

p = vi

= (– 2) (– 3)


–3 A

circuitelementp = ???

Example 2


Red (absorbing)

–3 A


v = 4 V

+. . .

Solution:

Once more, the passive convention is in effect, since the direction of the reference current is from + to – with the circuit element. Thus,

p = vi

= 4 (–5)

= –20 W absorbed , or 20 W delivered

– 5 A

–

circuitelement

p = ???

Example 3


Blue (supplying)– 5 A


Example 4


Solution:

Here, the active convention is in effect, since the direction of the reference current is from – to + within the circuit element. Remember: When the active sign convention is being used: The power absorbed is the negative of the product of the voltage and the current. In this case,

p = –vi

= –(10) (–6)


v = 10 V

+. . .

– 6 A

–

circuitelement

p = ???

Red (absorbing)– 6 A


Source-Load Circuit

A simple model for many circuits consists of one source and one load. The source produces the power and the load consumes the power.

There is a voltage rise at the source and a voltage drop at the load (when proceeding in the direction of the reference current).

Source Load V

+

i

–

Voltage drop in the direction of the current: power consumed.

Voltage rise in the direction of the current: power supplied.

The source get its energy through this connection.


Example 6

Show the proper cable connections for using a car battery in a one car to “jump-start” another car that has a discharged battery.

Solution:

Connect the jumper cables as follows.

+

+

i

i

Normal battery Discharged battery

i

i

+

v

+

v

Delivering vi watts Absorbing vi watts

Voltage sources CAN absorb power!


Source-Load Circuit: Hydraulic Analog

At the source (pump) there is a pressure rise in the direction of the flow.

At the turbine (load) there is a pressure drop in the direction the flow.

Pump Turbine

High pressure

Low pressureThe pump get its energy through this connection.


Power Conservation Theorem

The sum of the powers absorbed by all the elements in a circuit equals zero.

Example

Verify the power conservation theorem for the given circuit.

+300 V –

+60 V –

+60 V –

+ 240 V –

2 A

2 A 1.5 A

0.5 A

Solution:

p1 = – (300) (2) = – 600 W

p2 = (240) (2) = 480 W

p3 = (60) (1.5) = 90 W

p4 = (60) (0.5) = 30 W

pi = – 600 + 480 + 90 + 30

= 0

3 4

Blue

Red

Red

Red

White

1

2


Passive Element

A passive circuit element is one that never supplies energy.

Example

Show that the given element behaves passively for the given conditions.

+ V –

i

i = sin t, v = cos t

Solution:

p = vi = cos t sin t = .5 sin 2t

w = pdt

= .5 sin 2t dt

= – .25 cos 2t |

= 2(1 – cos 2t)

>= 0 Passive!

t

0t

0t

0


2.7 Practical Examples

Example 1.

The energy stored in car batteries is specified in ampere-hours instead of joules. A typical 12 V car battery may contain 200 ampere-hours of energy.

Calculate this energy in megajoules:

Solution:

p = v x i

w = v x i x t (since vx i is constant)

= 12 V x 1 A x 200 hr

= 12 V x 1 A x 200 x 3600 s

= 8.5 x 106 J

= 8.5 MJ

This is the total energy stored in the fully charged battery.

We note that if the headlights of the vehicle are inadvertently left on after the engine is shut down and they consume 50 W, the current drawn by the lights will be 50 W / 12 V = 4.17 A. Hence, the battery should last 200 A-h / 4.17 A = 48 h--an unrealistic answer in that the battery would not be able to supply 4.17 A right up to the point that it is completely dead.


Example 1 (cont.)

We found that the battery contains 8.5 MJ of energy. Suppose the battery sells for $40. Calculate the cost of the equivalent amount of energy if purchased from a typical electrical utility at a cost of 10¢ per kilowatt-hour.

Solution:

1 kWhr = 1000 J/s x 1 hr x 3600 s/hr

= 3.6 M J ( 10¢ worth of energy)

8.5/3.6 x 10 ¢ 24¢

Only a little more than 2 kWhr of energy is stored in the battery. But it is re-chargeable. And there’s no cord attached!


Example 1 (cont.)

Portable batteries (9 V, D cells, C cells, AA cells, etc.) typically have less than 1 MJ of energy stored in them. Their cost can be 1,000 times the cost of the equivalent cost of energy purchased from an electric utility.

Clearly, batteries are used because of their portability, convenience, etc.--not their low price.


Example 2

A smoke alarm draws 6 A from a 9 V battery. Assuming that the battery stores 18.3 kJ of energy, how often should the battery be replaced?

Solution:

w = p t

= v i t

t = w/vi = 18.3 x 103 / 9 x 6 x 10–6

= 338.9 x 106 s

338.9 x 106 s x 1 min/60 s x 1 hr/60 min x 1 day/24 hr x 1 yr/365 day = 10.7 yr

However, the battery’s internal leakage current will discharge it in about 4 years, and energy is consumed each time the smoke alarm is tested.

Play it safe. Replace the battery as recommended by the manufacturer.

9 V

6 A

smokealarm


Noted in passing:

The human brain consumes about 10–17 J per

lowest-level operation (10 W brain with 1016 synapses operating at about 10 nerve pulses per second).

The fastest microprocessors require about 10 –9 J per switching operation.

Comparing these energies we see that the human brain is 10,000,000 times more efficient than the most efficient conventional computers!


2.8 Independent and Dependent Sources

Independent Voltage Source

+–v V

Lower-case v indicates a time-varying voltage.

UPPER-CASE V indicates a constant (DC) voltage.

V

tt

v

For an independent voltage source, the voltage is v, no matter what.

The current is whatever.

i I

+

–


Independent Current Source

+

v

–

Lower-case v indicates a time-varying current.

Upper-case I indicates a constant (DC) current.

I

tt

i

For an independent current source, the current is i, no matter what.

The voltage is whatever.

i I

+

V

–


Dependent Sources

+–

iVCVS: v = vx

i = whatever

vx is somewhere (not shown)

v = v

CCVS: v = ix

i = whatever

ix is somewhere (not shown)

+–

i

v = i

+

v

–

i = v VCCS: i = vx

v = whatever

vx is somewhere (not shown)

x

x

x


+

v

–

i = ixCCCS: i = ix

v = whatever

ix is somewhere (not shown)

dependent source

independent source

Dependent Sources (cont.)

Memorize what these icons signify!

There are three items that must be specified in order to completely specify any dependent source:

1. the polarity of the source (+ and signs or an arrow with the diamond),

2. the source constant , , , , and

3. the controlling variable, by completely specifying the controlling branch and the polarity of the controlling variable on that branch.


Controlled sources are used primarily to model electronic devices. For example, the junction field-effect transistor (JFET) and the bipolar junction transistor (BJT) are modeled as shown below. These devices are used to construct electronic circuits such as amplifiers and digital computers. Without dependent sources we would not be able to model these important electrical components.

G

D

S

GD

S

ro

+

vGS

gmvGS

JFET

B

C

E

BD

S

ro

iB

BJT

ro


Example Using Dependent Sources

Calculate the power absorbed by the dependent source. Assume that we are given that Ix = –2 mA.

10 V+ –

Ix

+ – V=4Ix

*

Solution:

For the power absorbed, we want the product of the voltage across the dependent source and the current through it in the direction of the voltage drop.

p = – (4 Ix) (8 x 10–3)

= – 4 (–2 x 10 –3) ( 8 x 10 –3)

= 64 mW

8 mA

Minus sign needed because the active convention is present at the dependent source.

* Normally the designation is just 4Ix, not

V=4Ix.


Does it matter?

Absolutely.


10 V + – 8 A 8 V+

–

5 A

= 5 V

+ –

= 3

vxvs vx

What happens if you make the connections, anyway?Sparks fly,circuit breakers pop,nonlinear things occur,sometimes, expensive damage results!

+ –

Non-Permissible Connections

10 V + –

“ShortCircuit”

circuit theory i

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