Schedule…
Date
Day
17 Sept
Wed
18 Sept
Thu
19 Sept
Fri
20 Sept
Sat
21 Sept
Sun
22 Sept
Mon
Class
No.
5
Title
Ohm’s Law
Chapters
HW
Due date
Lab
Due date
Exam
2.6
LAB 1
Recitation
6
Things Practical
HW 2
2.6 – 2.8
LAB 2
23 Sept
Tue
24 Sept
Wed
ECEN 301
7
Network Analysis
3.1 – 3.2
Discussion #5 – Ohm’s Law
1
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
2
Lecture 5 – Resistance & Ohm’s Law
ECEN 301
Discussion #5 – Ohm’s Law
3
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
+
v
i
R=0
v=0
–
ECEN 301
Discussion #5 – Ohm’s Law
4
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:
overheating, fire, or explosion
ECEN 301
Discussion #5 – Ohm’s Law
5
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
+
v
i
R ∞
i=0
–
ECEN 301
Discussion #5 – Ohm’s Law
6
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
+ v1 –
R1
vs
+
_
ECEN 301
i1
+
v2
–
i2
R2
i3
+
v3 R3
–
Discussion #7 – Node and Mesh Methods
7
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
v1
+ v1 –
R1
vs
+
_
i1
+
v2
–
i2
R2
500 mA 2
1V
i3
+
v3 R3
–
i1 R1
vs
v1 v2
v2
0
v2 v1
1V
ECEN 301
Discussion #7 – Node and Mesh Methods
i2
v2
R2
1
4
250 mA
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
+ v1 –
R1
vs
+
_
ECEN 301
i1
+
v2
–
i2
R2
i3
+
v3 R3
–
P1
i1 v1
500 mA 1V
P2
i2 v2
500 mW
250 mA 1V
250 mW
1
W
2
1
W
4
Discussion #7 – Node and Mesh Methods
9
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
vs
i1
vs
+
_
ECEN 301
+
v2
–
i2
R2
v2
0
v2
vs
2V
i3
+
v3 R3
–
i2
v2
R2
2
4
500 mA
Discussion #7 – Node and Mesh Methods
P2
i2 v2
500 mA 2V
1W
BAD! R2 and
R3 are damaged
10
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
REQ
Rn
n 1
R1
R2
R3
Rn
∙∙∙
RN
∙∙∙
REQ
ECEN 301
Discussion #5 – Ohm’s Law
11
Series Resistors
Demonstration of series rule: apply KVL and
Ohm’s law on the circuit
i
+ v1 –
Vs
Vs
R1
Vs
+
_
1.5V
– v3 +
R3
ECEN 301
v1 v2 v3
+
v2
–
0
v1 v2 v3
(i R1 ) (i R2 ) (i R3 )
R2
i ( R1 R2
R3 )
i REQ
REQ
Discussion #5 – Ohm’s Law
Vs
i
12
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
vn
R1 R2
Rn
Vs
R3  Rn  RN
Rn
Vs
REQ
Rn
NB: the ratio R
will always be <= 1
EQ
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
i
+ v1 –
R1
Vs
+
_
1.5V
– v3 +
+
v2
–
R2
v1
R1
Vs
REQ
v2
R2
Vs
REQ
v3
R3
Vs
REQ
R3
ECEN 301
Discussion #5 – Ohm’s Law
14
Series Resistors
Example1: Determine v3
Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω
– v1 +
i
R1
R2
+
v2
–
+
_
Vs
+ v3 –
R3
ECEN 301
Discussion #5 – Ohm’s Law
15
Series Resistors
Example1: Determine v3
Vs = 3V, R1 = 10Ω, R2 = 6Ω, R3 = 8Ω
– v1 +
i
Using Series Rule :
REQ R1 R2 R3
R1
R2
+
v2
–
+
_
Vs
10 6 8
24
+ v3 –
Using Voltage Divider :
R3
v3
Vs
R EQ
8
3
24
1V
R3
ECEN 301
Discussion #5 – Ohm’s Law
16
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
1
REQ
+
R1
–
ECEN 301
+
R2
–
1
R1
+
R3
–
1 1
1

R2 R3
RN
+
Rn
–
REQ
1
R1
1
R2
+
RN
–
Discussion #5 – Ohm’s Law
1
1
1

R3
RN
+
REQ
–
NB: the parallel
combination of two
resistors is often
written: R1 || R2
17
Parallel Resistors
Demonstration of parallel rule: apply KCL and
Ohm’s law on the circuit
KCL at node a :
is i1 i2 i3 0
Node a
is i1 i2 i 3
+
is
i1
i2
+
R1
–
+
R2
–
v
R1
i3
+
R3
–
v
–
REQ
ECEN 301
Discussion #5 – Ohm’s Law
v
R2
v
1
R1
v
1
REQ
1
R2
v
R3
1
R3
v
is
18
Parallel Resistors
 Current divider: the current in a parallel circuit divides in
proportion to the resistances of the individual parallel elements
in
1
Rn
1
R1
1
R2
1

R3
1
Rn

1
RN
is
1
Rn
i
1 s
REQ
REQ
Rn
ECEN 301
REQ
is
NB: the ratio
Discussion #5 – Ohm’s Law
Rn
will always be <= 1
19
Parallel Resistors
Current divider: the current in a parallel circuit
divides in proportion to the resistances of the
individual parallel elements
i1
is
i1
i2
+
R1
–
+
R2
–
1 / R1
is
1 / REQ
REQ
i3
+
R3
–
R1
is
i2
i3
REQ
R2
REQ
R3
is
is
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
20
Parallel Resistors
Example2: find i1
R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A
i1
i2
–
R1
+
–
R2
+
i
s
ECEN 301
i3
–
R3
+
Discussion #5 – Ohm’s Law
21
Parallel Resistors
Example2: find i1
R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A
NB : all elements share terminals
1
REQ
1 1
1
R1 R2 R3
i1
i2
–
R1
+
–
R2
+
i
s
ECEN 301
i3
–
R3
+
1
1 1 1
10 2 20
1
2 10 1
20
20
13
Discussion #5 – Ohm’s Law
Using current divider :
i1
1
R1
i
1 s
REQ
1
10 4
13
20
40
65
0.615 A
22
Parallel Resistors
Example2: find i1
R1 = 10Ω, R2 = 2 Ω, R3 = 20 Ω, Is = 4A
Verify tha t KCL holds with current dividing :
is
i1
i2
–
R1
+
–
R2
+
i
s
i1 i2
i3
0
is
i1 i2
i3
–
R3
+
i3
1 / R1
is
1 / REQ
is
is
1 / R2
is
1 / REQ
1 / R3
is
1 / REQ
1 / R1 1 / R2 1 / R3
1 / REQ
1 / REQ
1 / REQ
is
ECEN 301
Discussion #5 – Ohm’s Law
23
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
R1 || R2
+
R2
–
+
R1
–
+
R1|| R2
–
ECEN 301
1
1
R1
1
R2
1
R2 R1
R1 R2
R1 R2
R2 R1
R1 || R2 || R3
+
R1
–
+
R2
–
+
R3
–
+
R1|| R2|| R3
–
Discussion #5 – Ohm’s Law
1
R1
1
1
R2
1
R3
1
R2 R3 R1 R3 R1 R2
R1 R2 R3
R1 R2 R3
R2 R3 R1 R3 R1 R2
24
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
+
vs
+
_
is
ECEN 301
+
R2
–
+
R3
–
v
–
Discussion #5 – Ohm’s Law
25
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
+ R1 –
+
vs
+
_
is
ECEN 301
+
R2
–
+
R3
–
v
+
vs
–
Discussion #5 – Ohm’s Law
+
_
is
R2 || R3
v
–
26
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
Using voltage divider :
R2 || R3
v
vs
REQ
R2 || R3
vs
( R1 R2 || R3 )
+ R1 –
R2 R3
R2 R3
+
vs
+
_
is
R2 || R3
v
R2 R3
R2 R3
–
1
R1
R2 R3
R2 R3
R1 R2
vs
1
vs
R1 R3 R2 R3
R2 R3
R2 R3
R2 R3
vs
( R2 R3 ) ( R1 R2 R1 R3 R2 R3 )
( R1 R2
ECEN 301
Discussion #5 – Ohm’s Law
R2 R3
vs
R1 R3 R2 R3 )
27
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
+
vs
+
_
is
R2 || R3
v
Using voltage divider :
R2 R3
v
vs
( R1 R2 R1 R3 R2 R3 )
(10
–
6
2
R3 10 3
R3 10 3
5 R3
10
3
2 R3
R3 10
3
)
(5V )
(V )
NB: notice how R3 controls v
ECEN 301
Discussion #5 – Ohm’s Law
28
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
If R3 1k :
+
vs
+
_
is
R2 || R3
v
–
v
5R3
10 3
2 R3
(V )
5 10 3
10 3
2 10 3
5
V
3
(V )
If R3
0.1k :
v
5R3
10
3
2 R3
(V )
0.5 10 3
10 3
0.2 10 3
5
(V )
12
(V )
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
+ R1 –
As R3
+
vs
+
_
is
+
R2
–
+
R3
–
v
v
–
0:
0
As R3
:
5
2
In other words :
v
v
v2
NB: notice how R3 controls v
ECEN 301
Discussion #5 – Ohm’s Law
30
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
+
vs
+
_
is
+
R2
–
v
–
As R3
0:
v 0
In other w ords, no current
goes throu gh R 2
ECEN 301
Discussion #5 – Ohm’s Law
31
Parallel Resistors
Example3: find v
vs = 5V, R1 = 1kΩ, R2 = 1kΩ
+ R1 –
As R3
+
vs
+
+
_
is
R2
–
v
:
5
v
2
In other words :
–
v
v2
(no curr ent goes through R 3 )
ECEN 301
Discussion #5 – Ohm’s Law
32