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ECEN 301 Discussion #5 – Ohm’s Law 1

Date Day Class No. Title Chapters HW Due date Lab Due date Exam 17 Sept Wed 5 Ohm’s Law 2.6 18 Sept Thu LAB 1 19 Sept Fri Recitation HW 2 20 Sept Sat 21 Sept Sun 22 Sept Mon 6 Things Practical 2.6 – 2.8 LAB 2 23 Sept Tue 24 Sept Wed 7 Network Analysis 3.1 – 3.2

Schedule…

ECEN 301 Discussion #5 – Ohm’s Law 2

Divided We Fall

Matthew 12:25-26 25 And Jesus knew their thoughts, and said unto them, Every kingdom divided against itself is brought to desolation; and every city or house divided against itself shall not stand: Nephi 7:2 2 And the people were divided one against another; and they did separate one from another into tribes, every man according to his family and his kindred and friends; and thus they did destroy the government of the land.

ECEN 301 Discussion #5 – Ohm’s Law 3

Lecture 5 – Resistance & Ohm’s Law

ECEN 301 Discussion #5 – Ohm’s Law 4

Open and Short Circuits

Short circuit: a circuit element across which the voltage is zero regardless of the current flowing through it

Resistance approaches zero Flow of current is unimpeded ex: an ideal wire

• In reality there is a small resistance

i R = 0 v = 0 v + –

ECEN 301 Discussion #5 – Ohm’s Law 5

Open and Short Circuits

Undesirable short circuit – accidental connection between two nodes that are meant to be at different voltages. The resulting excessive current causes:

• verheating, fire, or explosion

ECEN 301 Discussion #5 – Ohm’s Law 6

Open and Short Circuits

Open circuit: a circuit element through which zero current flows regardless of the voltage applied to it

Resistance approaches infinity No current flows ex: a break in a circuit

• At sufficiently high voltages arcing occurs

i R ∞ i = 0 v + –

ECEN 301 Discussion #7 – Node and Mesh Methods 7

Open and Short Circuits

i2

+ v2 –

vs

+ _ + v3 – + v1 – R2 R1 R3

i3 i1

Example: What happens to R2 and R3 if R1 is shorted?

• Vs = 2V, R1 = 2Ω, R2 = R3 = 4Ω, i1 = 500mA
• R2 and R3 = ¼ W rating, R1 = ½ W rating

ECEN 301 Discussion #7 – Node and Mesh Methods 8

Open and Short Circuits

Example: What happens to R2 and R3 if R1 is shorted?

• Vs = 2V, R1 = 2Ω, R2 = R3 = 4Ω, i1 = 500mA
• R2 and R3 = ¼ W rating, R1 = ½ W rating

i2

+ v2 –

vs

+ _ + v3 – + v1 – R2 R1 R3

i3 i1 V mA R i v 1 2 500

1 1 1

V v v v v v vs 1

1 2 2 2 1

mA R v i 250 4 1

2 2 2

ECEN 301 Discussion #7 – Node and Mesh Methods 9

Open and Short Circuits

i2

+ v2 –

vs

+ _ + v3 – + v1 – R2 R1 R3

i3 i1 W mW V mA v i P 2 1 500 1 500

1 1 1

Example: What happens to R2 and R3 if R1 is shorted?

• Vs = 2V, R1 = 2Ω, R2 = R3 = 4Ω, i1 = 500mA
• R2 and R3 = ¼ W rating, R1 = ½ W rating

W mW V mA v i P 4 1 250 1 250

2 2 2

ECEN 301 Discussion #7 – Node and Mesh Methods 10

Open and Short Circuits

i2

+ v2 –

vs

+ _ + v3 – R2 R3

i3 i1

Example: What happens to R2 and R3 if R1 is shorted?

• Vs = 2V, R1 = 2Ω, R2 = R3 = 4Ω, i1 = 500mA
• R2 and R3 = ¼ W rating, R1 = ½ W rating

V v v v v

s s

2

2 2

mA R v i 500 4 2

2 2 2

W V mA v i P 1 2 500

2 2 2

BAD! R2 and R3 are damaged

ECEN 301 Discussion #5 – Ohm’s Law 11

Series Resistors

Series Rule: two or more circuit elements are said to be in series if the current from one element exclusively flows into the next element.

From KCL: all series elements have the same current

N n n EQ

R R

1

R1 ∙ ∙ ∙ ∙ ∙ ∙ R2 R3 Rn RN REQ

ECEN 301 Discussion #5 – Ohm’s Law 12

Series Resistors

Demonstration of series rule: apply KVL and Ohm’s law on the circuit

i V R R i R R R i R i R i R i v v v V v v v V

s EQ EQ s s

) ( ) ( ) ( ) (

3 2 1 3 2 1 3 2 1 3 2 1

Vs

+ _ + v1 – R1 – v3 + R3 R2 + v2 – 1.5V i

ECEN 301 Discussion #5 – Ohm’s Law 13

Series Resistors

Voltage divider: the voltage across each resistor in a series circuit is directly proportional to the ratio of its resistance to the total series resistance of the circuit

NB: the ratio will always be <= 1

EQ n

R R

s EQ n s N n n n

V R R V R R R R R R v  

3 2 1

ECEN 301 Discussion #5 – Ohm’s Law 14

Series Resistors

Voltage divider: the voltage across each resistor in a series circuit is directly proportional to the ratio of its resistance to the total series resistance of the circuit

s EQ

V R R v

1 1

Vs

+ _ + v1 – R1 – v3 + R3 R2 + v2 – 1.5V i

s EQ

V R R v

2 2 s EQ

V R R v

3 3

ECEN 301 Discussion #5 – Ohm’s Law 15

Series Resistors

Example1: Determine v3

Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω

Vs

+ _ – v1 + R1 + v3 – R3 R2 + v2 – i

ECEN 301 Discussion #5 – Ohm’s Law 16

Series Resistors

Example1: Determine v3

Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω

V V V

s

1 3 24 8 R R v : Divider

• ltage

Using

EQ 3 3

Vs

+ _ – v1 + R1 + v3 – R3 R2 + v2 – i 24 8 6 10 : Rule Series Using

3 2 1

R R R REQ

ECEN 301 Discussion #5 – Ohm’s Law 17

Parallel Resistors

Parallel Rule: two or more circuit elements are said to be in parallel if the elements share the same terminals

From KVL: the elements will have the same voltage

N EQ N EQ

R R R R R R R R R R 1 1 1 1 1 1 1 1 1 1

3 2 1 3 2 1

 

+ R1 – + R2 – + R3 – + Rn – + RN – + REQ – NB: the parallel combination of two resistors is often written: R1 || R2

ECEN 301 Discussion #5 – Ohm’s Law 18

Parallel Resistors

Demonstration of parallel rule: apply KCL and Ohm’s law on the circuit

i1 i2 i3 + R1 –

is

+ R2 – + R3 – + v – Node a

s EQ EQ s s

i v R R v R R R v R v R v R v i i i i i i i i 1 1 1 1 : a node at KCL

3 2 1 3 2 1 3 2 1 3 2 1

ECEN 301 Discussion #5 – Ohm’s Law 19

Parallel Resistors

Current divider: the current in a parallel circuit divides in proportion to the resistances of the individual parallel elements

s n EQ s EQ n s N n n n

i R R i R R i R R R R R R i 1 1 1 1 1 1 1 1

3 2 1

 

NB: the ratio will always be <= 1

n EQ

R R

ECEN 301 Discussion #5 – Ohm’s Law 20

Parallel Resistors

Current divider: the current in a parallel circuit divides in proportion to the resistances of the individual parallel elements

s EQ s EQ

i R R i R R i

1 1 1

/ 1 / 1

s EQ i

R R i

2 2 s EQ i

R R i

3 3

i1 i2 i3 + R1 –

is

+ R2 – + R3 – NB: it makes sense that the smaller the resistor, the larger the amount of current that will flow.

ECEN 301 Discussion #5 – Ohm’s Law 21

Parallel Resistors

Example2: find i1

R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A

i1 i2 i3 – R1 +

i

s

– R2 + – R3 +

ECEN 301 Discussion #5 – Ohm’s Law 22

Parallel Resistors

Example2: find i1

R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A

13 20 20 1 10 2 1 20 1 2 1 10 1 1 1 1 1 1 terminals share elements all :

3 2 1

R R R REQ NB

i1 i2 i3 – R1 +

i

s

– R2 + – R3 +

A i R R i

s EQ

615 . 65 40 4 20 13 10 1 1 1 : divider current Using

1 1

ECEN 301 Discussion #5 – Ohm’s Law 23

Parallel Resistors

Example2: find i1

R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A

s EQ EQ s EQ s s EQ s EQ s EQ s s

i R R i R R R R i i R R i R R i R R i i i i i i i i / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 / 1 : dividing current with holds KCL t Verify tha

3 2 1 3 2 1 3 2 1 3 2 1

i1 i2 i3 – R1 +

i

s

– R2 + – R3 +

ECEN 301 Discussion #5 – Ohm’s Law 24

Parallel Resistors

 The parallel combination of resistors is often written:

 For two resistors: R1 || R2  For three resistors: R1 || R2 || R3  etc.

 In each case REQ for the parallel combination must be found

2 1 3 1 3 2 3 2 1 3 2 1 2 1 3 1 3 2 3 2 1 3 2 1

1 1 1 1 1 || || R R R R R R R R R R R R R R R R R R R R R R R R

1 2 2 1 2 1 1 2 2 1 2 1

1 1 1 1 || R R R R R R R R R R R R

+ R1 – + R2 – + R3 – + R1 – + R2 – +

R1|| R2|| R3

– +

R1|| R2

–

ECEN 301 Discussion #5 – Ohm’s Law 25

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

+ R2 –

vs

+ _ + R3 – + R1 – + v –

ECEN 301 Discussion #5 – Ohm’s Law 26

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

+ R2 –

vs

+ _ + R3 – + R1 – + v –

is

R2 || R3

vs

+ _ + R1 – + v –

ECEN 301 Discussion #5 – Ohm’s Law 27

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

R2 || R3

vs

+ _ + R1 – + v –

s s s s s s EQ

v R R R R R R R R v R R R R R R R R R R R R v R R R R R R R R R R R R v R R R R R R R R R v R R R R R v R R R v ) ( ) ( ) ( 1 1 ) || ( || || : divider voltage Using

3 2 3 1 2 1 3 2 3 2 3 1 2 1 3 2 3 2 3 2 3 2 3 2 3 1 2 1 3 2 3 2 3 2 3 2 1 3 2 3 2 3 2 1 3 2 3 2

ECEN 301 Discussion #5 – Ohm’s Law 28

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

is

R2 || R3

vs

+ _ + R1 – + v – ) ( 2 10 5 ) 5 ( ) 10 10 10 ( 10 ) ( : divider voltage Using

3 3 3 3 3 3 3 2 6 3 3 3 2 3 1 2 1 3 2

V R R V R R R v R R R R R R R R v

s

NB: notice how R3 controls v

ECEN 301 Discussion #5 – Ohm’s Law 29

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

V V V R R v k R 3 5 ) ( 10 2 10 10 5 ) ( 2 10 5 : 1 If

3 3 3 3 3 3 3

is

R2 || R3

vs

+ _ + R1 – + v – NB: notice how R3 controls v

) ( 12 5 ) ( 10 2 . 10 10 5 . ) ( 2 10 5 : 1 . If

3 3 3 3 3 3 3

V V V R R v k R

ECEN 301 Discussion #5 – Ohm’s Law 30

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

v : As

3

R

NB: notice how R3 controls v

2 3

v : words

• ther

In 2 5 v : As v R

is

+ R2 –

vs

+ _ + R3 – + R1 – + v –

ECEN 301 Discussion #5 – Ohm’s Law 31

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

2 3

R gh throu goes current no

• rds,
• ther w

In v : As R

is

+ R2 –

vs

+ _ + R1 – + v –

ECEN 301 Discussion #5 – Ohm’s Law 32

Parallel Resistors

Example3: find v

vs = 5V, R1 = 1kΩ, R2 = 1kΩ

) R through goes ent curr (no v : words

• ther

In 2 5 v : As

3 2 3

v R

is

+

R2

–

vs

+ _

+ R1 – + v –