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Schedule Date Day Class Title Chapters HW Lab Exam No. Due - - PowerPoint PPT Presentation

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Schedule Date Day Class Title Chapters HW Lab Exam No. Due date Due date Kirchoffs Laws 2.2 2.3 8 Sept Mon 2 NO LAB 9 Sept Tue NO LAB 2.4 2.5 10 Sept Wed 3 Power 11 Sept Thu NO LAB 12 Sept Fri Recitation

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SLIDE 1

ECEN 301 Discussion #2 – Kirchhoff’s Laws 1

Date Day Class No. Title Chapters HW Due date Lab Due date Exam 8 Sept Mon 2 Kirchoff’s Laws 2.2 – 2.3 NO LAB 9 Sept Tue NO LAB 10 Sept Wed 3 Power 2.4 – 2.5 11 Sept Thu NO LAB 12 Sept Fri Recitation HW 1 13 Sept Sat 14 Sept Sun 15 Sept Mon 4 Ohm’s Law 2.5 – 2.6 LAB 1 16 Sept Tue

Schedule…

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SLIDE 2

ECEN 301 Discussion #2 – Kirchhoff’s Laws 2

Divine Source

2 Nephi 25:26 26 And we talk of Christ, we rejoice in Christ, we preach of Christ, we prophesy of Christ, and we write according to our prophecies, that our children may know to what source they may look for a remission of their sins.

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SLIDE 3

ECEN 301 Discussion #2 – Kirchhoff’s Laws 3

Lecture 2 – Kirchhoff’s Current and Voltage Laws

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SLIDE 4

ECEN 301 Discussion #2 – Kirchhoff’s Laws 4

Charge

Elektron: Greek word for amber

~600 B.C. it was discovered that static charge on a piece

  • f amber could attract light objects (feathers)

Charge (q): fundamental electric quantity

Smallest amount of charge is that carried by an electron/proton (elementary charges):

C q q

p e 19

10 602 . 1 / /

Coulomb (C): basic unit of charge.

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SLIDE 5

ECEN 301 Discussion #2 – Kirchhoff’s Laws 5

Electric Current

Electric current (i): the rate of change (in time) of charge passing through a predetermined area (IE the cross-sectional area of a wire).

 Analogous to volume flow rate in hydraulics  Current (i) refers to ∆q (dq) units of charge that flow through a cross- sectional area (Area) in ∆t (dt) units of time i

Area

A dt dq t q i

Ampere (A): electric current unit. 1 ampere = 1 coulomb/second (C/s) Positive current flow is in the direction

  • f positive charges (the opposite direction
  • f the actual electron movement)
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SLIDE 6

ECEN 301 Discussion #2 – Kirchhoff’s Laws 6

Charge and Current Example

For a metal wire, find:

 The total charge (q)  The current flowing in the wire (i)

Given Data:

  • wire length = 1m
  • wire diameter = 2 x 10-3m
  • charge density = n = 1029 carriers/m3
  • charge of an electron = qe = -1.602 x 10-19
  • charge carrier velocity = u = 19.9 x 10-6 m/s
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SLIDE 7

ECEN 301 Discussion #2 – Kirchhoff’s Laws 7

Charge and Current Example

For a metal wire, find:

 The total charge (q)  The current flowing in the wire (i)

3 6 2 2 3 2

10 2 10 2 ) 1 ( area length Volume m m m r L

Given Data:

  • wire length = 1m
  • wire diameter = 2 x 10-3m
  • charge density = n = 1029 carriers/m3
  • charge of an electron = qe = -1.602 x 10-19
  • charge carrier velocity = u = 19.9 x 10-6 m/s

carriers m carriers m n V N

23 3 29 3 6

10 10 10 density carrier volume carriers

  • f

Number

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SLIDE 8

ECEN 301 Discussion #2 – Kirchhoff’s Laws 8

Charge and Current Example

For a metal wire, find:

 The total charge (q)  The current flowing in the wire (i)

Given Data:

  • wire length = 1m
  • wire diameter = 2 x 10-3m
  • charge density = n = 1029 carriers/m3
  • charge of an electron = qe = -1.602 x 10-19
  • charge carrier velocity = u = 19.9 x 10-6 m/s

C carrier C carriers q N q

e 3 19 23

10 33 . 50 / 10 602 . 1 10 rer charge/car carriers

  • f

number Charge

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SLIDE 9

ECEN 301 Discussion #2 – Kirchhoff’s Laws 9

Charge and Current Example

For a metal wire, find:

 The total charge (q)  The current flowing in the wire (i)

Given Data:

  • wire length = 1m
  • wire diameter = 2 x 10-3m
  • charge density = n = 1029 carriers/m3
  • charge of an electron = qe = -1.602 x 10-19
  • charge carrier velocity = u = 19.9 x 10-6 m/s

A s m m C s m u m C L q i 1 / 10 9 . 19 / 10 33 . 50 ) / ( ) / ( locity carrier ve length unit per density charge carrier Current

6 3

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SLIDE 10

ECEN 301 Discussion #2 – Kirchhoff’s Laws 10

Kirchhoff’s Current Law (KCL)

KCL: charge must be conserved – the sum of the currents at a node must equal zero.

N n n

i

1

+ _

1.5 V

i i

i1 i2 i3 Node 1 At Node 1:

  • i + i1 + i2 + i3 = 0

OR: i - i1 - i2 - i3 = 0

NB: a circuit must be CLOSED in order for current to flow

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SLIDE 11

ECEN 301 Discussion #2 – Kirchhoff’s Laws 11

Kirchhoff’s Current Law (KCL)

Potential problem of too many branches on a single node:

not enough current getting to a branch

Suppose:

  • all lights have the same resistance
  • i4 needs 1A

What must the value of i be?

+ _

1.5 V

i i

i1 i2 i6 i3 i4 i5 Node 1

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SLIDE 12

ECEN 301 Discussion #2 – Kirchhoff’s Laws 12

Kirchhoff’s Current Law (KCL)

Potential problem of too many branches on a single node:

not enough current getting to a branch

  • i + i1 + i2 + i3 + i4 + i5 + i6 = 0

BUT: since all resistances are the same:

i1 = i2 = i3 = i4 = i5 = i6 = in

  • i + 6in = 0

6in = i 6(1A) = i i = 6A

+ _

1.5 V

i i

i1 i2 i6 i3 i4 i5 Node 1

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SLIDE 13

ECEN 301 Discussion #2 – Kirchhoff’s Laws 13

Kirchhoff’s Current Law (KCL)

Example1: find i0 and i4

is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A

+ _

Vs

is i0 i1 i2 i3 i4

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SLIDE 14

ECEN 301 Discussion #2 – Kirchhoff’s Laws 14

Kirchhoff’s Current Law (KCL)

Example1: find i0 and i4

is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A

+ _

Vs

is i0 i1 i2 i3 i4 Node a Node c Node b NB: First thing to do – decide on unknown current directions.

  • If you select the wrong direction it won’t

matter

  • a negative current indicates current is

flowing in the opposite direction.

  • Must be consistent
  • Once a current direction is chosen must

keep it

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SLIDE 15

ECEN 301 Discussion #2 – Kirchhoff’s Laws 15

Kirchhoff’s Current Law (KCL)

Example1: find i0 and i4

is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A

A i i i i i i i 1 3 2 : a Node at Find

2 1 2 1

+ _

Vs

is i0 i1 i2 i3 i4 Node a Node c Node b

A i i i i i i i

s s

5 . 3 5 5 . 1 : c Node at Find

3 4 3 4 4

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SLIDE 16

ECEN 301 Discussion #2 – Kirchhoff’s Laws 16

Kirchhoff’s Current Law (KCL)

Example2: using KCL find is1 and is2

i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A

Vs1

+ _

Vs2

+ _

R2 R3 R4 R5

is1 i2 is2 i4 i3 i5

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SLIDE 17

ECEN 301 Discussion #2 – Kirchhoff’s Laws 17

Kirchhoff’s Current Law (KCL)

Example2: using KCL find is1 and is2

i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A

Vs1

+ _

Vs2

+ _

R2 R3 R4 R5

Supernode is1 i2 is2 i4 i3 i5

A i i i i i i

s s

2 2 : rnode supe at KCL

5 3 1 5 3 1

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SLIDE 18

ECEN 301 Discussion #2 – Kirchhoff’s Laws 18

Kirchhoff’s Current Law (KCL)

A i i i i i i

s s s s

1 2 3 : a e Nod at KCL

1 2 2 2 1 2

Vs1

+ _

Vs2

+ _

R2 R3 R4 R5

is1 i2 is2 i4 i3 i5 Node a

Example2: using KCL find is1 and is2

i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A

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SLIDE 19

ECEN 301 Discussion #2 – Kirchhoff’s Laws 19

Voltage

Moving charges in order to produce a current requires work Voltage: the work (energy) required to move a unit charge between two points Volt (V): the basic unit of voltage (named after Alessandro Volta)

Volt (V): voltage unit. 1 Volt = 1 joule/coulomb (J/C)

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SLIDE 20

ECEN 301 Discussion #2 – Kirchhoff’s Laws 20

Voltage

Voltage is also called potential difference

Very similar to gravitational potential energy Voltages are relative

  • voltage at one node is measured relative to the voltage at another

node

  • Convenient to set the reference voltage to be zero

+

vab _

a b

_

+

vba

vab => the work required to move a positive charge from terminal a to terminal b vba => the work required to move a positive charge from terminal b to terminal a vba = - vab vab = va - vb

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SLIDE 21

ECEN 301 Discussion #2 – Kirchhoff’s Laws 21

Voltage

Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied +

vab

_

a b i

vba

_

a b i

+

  • Since i is going from + to – energy

is being absorbed by the element (passive element)

  • Since i is going from – to + energy is

being supplied by the element (active element)

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SLIDE 22

ECEN 301 Discussion #2 – Kirchhoff’s Laws 22

Voltage

Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied +

vab

_

a b i

vba

_

a b i

+ + _

1.5 V

i i

v1 a b vab + + _ _ Supplying energy (source) (active element) NEGATIVE voltage Absorbing energy (load) (passive element) POSITIVE voltage

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SLIDE 23

ECEN 301 Discussion #2 – Kirchhoff’s Laws 24

Voltage

Ground: represents a specific reference voltage

Most often ground is physically connected to the earth (the ground) Convenient to assign a voltage of 0V to ground

The ground symbol we’ll use (earth ground) Another ground symbol (chasis ground)

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SLIDE 24

ECEN 301 Discussion #2 – Kirchhoff’s Laws 25

Kirchhoff’s Voltage Law (KVL)

KVL: energy must be conserved – the sum of the voltages in a closed circuit must equal zero. V v v v v v

ab a b a ab

5 . 1 5 . 1

N n n

v

1

+ _

1.5 V

i i

v1 a b vab + + _ _ Use Node b as the reference voltage (ground): vb = 0

V v v v v

ab ab

5 . 1

1 1

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SLIDE 25

ECEN 301 Discussion #2 – Kirchhoff’s Laws 26

Kirchhoff’s Voltage Law (KVL)

Example3: using KVL, find v2

vs1 = 12V, v1 = 6V, v3 = 1V

Vs1

+ _ i + v1 – + v3 – + v2 – Source: loop travels from – to + terminals

  • Sources have negative voltage

Load: loop travels from + to – terminals

  • Loads have positive voltage
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SLIDE 26

ECEN 301 Discussion #2 – Kirchhoff’s Laws 27

Kirchhoff’s Voltage Law (KVL)

Example3: using KVL, find v2

vs1 = 12V, v1 = 6V, v3 = 1V

Vs1

+ _ i + v1 – + v3 – + v2 – Source: loop travels from – to + terminals

  • Sources have negative voltage

Load: loop travels from + to – terminals

  • Loads have positive voltage
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SLIDE 27

ECEN 301 Discussion #2 – Kirchhoff’s Laws 28

Kirchhoff’s Voltage Law (KVL)

Example3: using KVL, find v2

vs1 = 12V, v1 = 6V, v3 = 1V

Vs1

+ _ i + v1 – + v3 – + v2 –

V v v v v v v v v

s s

5 1 6 12

3 1 1 2 3 2 1 1

NB: v2 is the voltage across two elements in parallel branches. The voltage across both elements is the same: v2

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SLIDE 28

ECEN 301 Discussion #2 – Kirchhoff’s Laws 29

Kirchhoff’s Voltage Law (KVL)

Example4: using KVL find v1 and v4

vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V

Vs1

+ _

Vs2

+ _ + v3 – + v4 – + v1 – + v5 – + v2 –

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SLIDE 29

ECEN 301 Discussion #2 – Kirchhoff’s Laws 30

Kirchhoff’s Voltage Law (KVL)

Example4: using KVL find v1 and v4

vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V

Vs1

+ _

Vs2

+ _ + v3 – + v4 – + v1 – + v5 – + v2 – Loop1 Loop2 Loop3

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SLIDE 30

ECEN 301 Discussion #2 – Kirchhoff’s Laws 31

Kirchhoff’s Voltage Law (KVL)

Example4: using KVL find v1 and v4

vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V

Vs1

+ _

Vs2

+ _ + v3 – + v4 – + v1 – + v5 – + v2 –

V v v v v v v v v

s s

4 6 2 12 : 1 Loop

3 2 1 1 3 2 1 1

Loop1 Loop2 Loop3

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SLIDE 31

ECEN 301 Discussion #2 – Kirchhoff’s Laws 32

Kirchhoff’s Voltage Law (KVL)

Example4: using KVL find v1 and v4

vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V

Vs1

+ _

Vs2

+ _ + v3 – + v4 – + v1 – + v5 – + v2 –

V v v v v v v

s s

6 2 4 : 2 Loop

2 2 4 2 4 2

Loop1 Loop2 Loop3