Chapter_2_Basic_Circuit_Laws.ppt

Chapter 2
Basic Laws
of Electric Circuits
Ohms Law
Kirchhoff’s Current Law
Kirchhoff’s Voltage Law

Chapter Motivation
There are too many people praying
for mountains of difficulty to be
removed, when what they really need
is the courage to climb them!

Chapter Outcomes
An ability to design and conduct
experiments, as well as to analyze and
interpret data

Basic Laws of Circuits
Ohm’s Law:
The voltage across a resistor is directly proportional to the current
moving through the resistor.
+ _
v(t)
i(t)
R
v(t) = Ri(t)
+
_ v(t)
i(t)
R
v(t) = Ri(t)
_
(2.1)
(2.2)
1

Ohm’s Law:
Directly proportional means a straight line relationship.
v(t)
i(t)
R
The resistor is a model and will not produce a straight line
for all conditions of operation.
v(t) = Ri(t)
2

Ohm’s Law: About Resistors:
The unit of resistance is ohms( ).
A mathematical expression for resistance is
l
R
A


 
2
: ( )
: ( )
:
l Thelengthof theconductor meters
A Thecross sectional area meters
Theresistivity m



(2.3)
3

Ohm’s Law: About Resistors:
We remember that resistance has units of ohms. The reciprocal of
resistance is conductance. At one time, conductance commonly had units
of mhos (resistance spelled backwards).
In recent years the units of conductance has been established as seimans (S).
Thus, we express the relationship between conductance and resistance as
1
G
R
 (2.4)
We will see later than when resistors are in parallel, it is convenient
to use Equation (2.4) to calculate the equivalent resistance.
(S)
4

Ohm’s Law: Ohm’s Law: Example
Consider the following circuit.
+
_
115V RMS
(ac)
R
(100 Watt light bulb)
V
Determine the resistance of the 100 Watt bulb.
2
2
2
2
115
132.25
100
V
P VI I R
R
V
R ohms
P
  
  
(2.5)
A suggested assignment is to measure the resistance of a 100 watt light
bulb with an ohmmeter. Debate the two answers.
5

Ohm’s Law: Property of Resistance:
Resistivities of some basic materials
Material Resistivity (ohm meters) Common Use
silver 1.6x10-8 conductor
copper 1.7x10-8 conductor
aluminum 2.8x10-8 conductor
gold 2.5x10-8 conductor
carbon 4.1x10-5 semiconductor
germanium 47x10-2 semiconductor
silicon 6.4x102 semiconductor
paper 1x1010 insulator
mica 5x1011 insulator
glass 1x1012 insulator
teflon 3x1012 insulator
6

Nodes, Branches & Loops
• Page 35

As a consequence of the Law of the conservation of charge, we have:
•The sum of the current entering a
node (junction point) equal to the
sum of the currents leaving.
Ia
Ib
Ic
Id
Ia, Ib, Ic, and Id can each be either a positive
or negative number.
Ia + Ib = Ic + Id
11

•The algebraic sum of the currents entering a node equal to zero.
Ia
Ib
Ic
Id
Ia +Ib Ic + Id = 0
or negative number.
12

• The algebraic sum of the currents leaving a node equal to zero.
Ia
Ib
Ic
Id
Ia - Ib + Ic + Id = 0
or negative number.
13

Kirchhoff’s Current Law: Example
Find the current I x.
4 A
2 A
-1 A 6 A
IX
9 A
Ans: IX = 22 A
14
Highlight the box
then use bring to
front to see answer.

Kirchhoff’s current law can be generalized to include a surface.
We assume the elements within the surface are interconnected.
A closed 3D surface
We can now apply Kirchhoff’s current law in the 3 forms we discussed
with a node. The appearance might be as follows:
Currents entering and
leaving a closed surface
that contains interconnected
circuit elements
16

Circuit
Circuit
Circuit
1 1
j
j
j N k M
k
k
i i

 


 
1
0
r Q
r
r
i




1
0
m Q
m
m
i




As a consequence of the Law of the conservation of charge, we have:
17

Kirchhoff’s Current Law:
surface
1
surface
2
-2A
4A
IB IC
9A
2A IA
Find the currents IA, IB, and IC in the circuit below.
18

Kirchhoff’s Current Law: Solution of Example
surface
1
surface
2
-2A
4A
IB IC
9A
2A IA


node 1 node 2
At surface 1: IB = 2A: At node 1, Ic = 0A: At node 2, IA = 9A
19

Basic Electric Circuit Concepts
End of Lecture 2
circuits
Ohm’s Law

Kirchhoff’s Voltage Law:
Kirchhoff’s voltage law tells us how to handle voltages in an
electric circuit.
Kirchhoff’s voltage law basically states that the algebraic sum of
the voltages around any closed path (electric circuit) equal zero.
The secret here, as in Kirchhoff’s current law, is the word
algebraic.


 There are three ways we can interrupt that the algebraic sum of
the voltages around a closed path equal zero. This is similar to
what we encountered with Kirchhoff’s current law.
1

Consideration 1: Sum of the voltage drops around a circuit
equal zero. We first define a drop.
We assume a circuit of the following configuration. Notice that
no current has been assumed for this case, at this point.
+
+
+
+
_
_
_
_
v1
v2
v4
v3
Figure 3.1
2

Kirchhoff’s Voltage Law: Consideration 1.
We define a voltage drop as positive if we enter the positive terminal
and leave the negative terminal.
+ _
v1
The drop moving from left to right above is + v1.
+
_ v1
The drop moving from left to right above is – v1.
Figure 3.2
Figure 3.3
3

+
+
+
+
_
_
_
_
v1
v2
v4
v3
Figure 3.4
Consider the circuit of Figure 3.4 once
again. If we sum the voltage drops in the clockwise direction around the
circuit starting at point “a” we write:
- v1 – v2 + v4 + v3 = 0
- v3 – v4 + v2 + v1 = 0
•
“a”
 drops in CW direction starting at “a”
 drops in CCW direction starting at “a”
4

Consideration 2: Sum of the voltage rises around a circuit
equal zero. We first define a drop.
We define a voltage rise in the following diagrams:
+
_ v1
Figure 3.5
+ _
v1
Figure 3.6
The voltage rise in moving from left to right above is + v1.
The voltage rise in moving from left to right above is - v1.
5

Kirchhoff’s Voltage Law: Consider the circuit of Figure 3.7 once
again. If we sum the voltage rises in the clockwise direction around the
circuit starting at point “a” we write:
+
+
+
+
_ _
_
v1
v2
v4
v3
Figure 3.7
•
“a”
+ v1 + v2 - v4 – v3 = 0
+ v3 + v4 – v2 – v1 = 0
rises in the CW direction starting at “a”
rises in the CCW direction starting at “a”
6
_

Consideration 3: Sum of the voltage rises around a circuit
equal the sum of the voltage drops.
Again consider the circuit of Figure 3.1 in which we start at
point “a” and move in the CW direction. As we cross elements
1 & 2 we use voltage rise: as we cross elements 4 & 3 we use
voltage drops. This gives the equation,
+
+
+
+
_
_
_
_
v1
v2
v4
v3
v1 + v2 = v4 + v3
1
2
3
4
p
7

Kirchhoff’s Voltage Law: Comments.
• We note that a positive voltage drop = a negative voltage rise.
• We note that a positive voltage rise = a negative voltage drop.
• We do not need to dwell on the above tongue twisting statements.
• There are similarities in the way we state Kirchhoff’s voltage
and Kirchhoff’s current laws: algebraic sums …
However, one would never say that the sum of the voltages
entering a junction point in a circuit equal to zero.
Likewise, one would never say that the sum of the currents
around a closed path in an electric circuit equal zero.
8

Kirchhoff’s Voltage Law: Further details.
For the circuit of Figure 3.8 there are a number of closed paths.
Three have been selected for discussion.
+
+
+
+ +
+
+
+
+
+
+
-
- -
-
-
-
-
-
-
-
-
v1
v2
v4
v3
v12
v11 v9
v8
v6
v5
v7
v10
+
-
Figure 3.8
Multi-path
Circuit.
Path 1
Path 2
Path 3
9

Kirchhoff’s Voltage Law: Further details.
For any given circuit, there are a fixed number of closed paths
that can be taken in writing Kirchhoff’s voltage law and still
have linearly independent equations. We discuss this more, later.
Both the starting point and the direction in which we go around a closed
path in a circuit to write Kirchhoff’s voltage law are arbitrary. However,
one must end the path at the same point from which one started.
Conventionally, in most text, the sum of the voltage drops equal to zero is
normally used in applying Kirchhoff’s voltage law.
10

Kirchhoff’s Voltage Law: Illustration from Figure 3.8.
+
+
+
+ +
+
+
+
+
+
+
-
- -
-
-
-
-
-
-
-
-
v1
v2
v4
v3
v12
v11 v9
v8
v6
v5
v7
v10
+
-
“a”
•
Blue path, starting at “a”
- v7 + v10 – v9 + v8 = 0
•
“b”
Red path, starting at “b”
+v2 – v5 – v6 – v8 + v9 – v11
– v12 + v1 = 0
Yellow path, starting at “b”
+ v2 – v5 – v6 – v7 + v10 – v11
- v12 + v1 = 0
Using sum of the drops = 0
11

Kirchhoff’s Voltage Law: Double subscript notation.
Voltages in circuits are often described using double subscript notation.
• •
a b
Consider the following:
Figure 3.9: Illustrating double subscript notation.
Vab means the potential of point a with respect to point b with
point a assumed to be at the highest (+) potential and point b
at the lower (-) potential.
12

Kirchhoff’s Voltage Law: Double subscript notation.
Task: Write Kirchhoff’s voltage law going in the clockwise
direction for the diagram in Figure 3.10.
• •
•
•
b a
x
y
Going in the clockwise direction, starting at “b”, using rises;
vab + vxa + vyx + vby = 0
Figure 3.10: Circuit for illustrating double subscript notation.
13

Kirchhoff’s Voltage Law: Equivalences in voltage notations
The following are equivalent in denoting polarity.
+
-
v1 v1
•
•
a
b
+ -
v2
= =
v2 = - 9 volts means the right hand side
of the element is actually positive.
vab = v1
Assumes the upper terminal is positive in all 3 cases
14

Kirchhoff’s Voltage Law: Application.
Given the circuit of Figure 3.11. Find Vad and Vfc.
5 V
8 V
15 V
12 V
20 V 10 V
30 V
a b c
d
e
f
+ _
+
+
_
_
+
+
+
+
_
_
_
_




Using drops = 0; Vad + 30 – 15 – 5 = 0 Vab = - 10 V
Vfc – 12 + 30 – 15 = 0 Vfc = - 3 V
Figure 3.11: Circuit for illustrating KVL.
15

Kirchhoff’s Voltage Law: Single-loop circuits.
We are now in a position to combine Kirchhoff’s voltage and current
Laws to the solution of single loop circuits. We start by developing the
Voltage Divider Rule. Consider the circuit of Figure 3.12.
R2
R1
v2
v1
+ +
+
_
_
_
v i1
v = v1 + v2
v1 = i1R1, v2 = i1R2
Figure 3.12: Circuit for developing
voltage divider rule.
then,
v = i1(R1 + R2) , and i1 =
v
(R1 + R2)
so,
v1 =
vR1
(R1 + R2)
* You will be surprised by how much you use this in circuits.
*
16

Find V1 in the circuit shown in Figure 3.13.
V
R3 R2
R1
V1
I
+
_
1 2 3
1 1
( )
, ,
V
I
R R R
V IR so wehave

 

1
1
1 2 3
( )
VR
V
R R R

 
Figure 3.13
17

V
R3 R2
R1
I
+
_
Example 3.1: For the circuit of Figure 3.14, the following is known:
R1 = 4 ohms, R2 = 11 ohms, V = 50 volts, P1 = 16 watts
Find R3.
Figure 3.14: Circuit for example 3.1.
Solution:
P1 = 16 watts = I2R1
I = 2 amps
V = I(R1 + R2 + R3), giving,
R1 + R2 + R3 = 25, then solve for R3,
R3 = 25 – 15 = 10 ohms
, thus,
18

Example 3.2: For the circuit in Figure 3.15 find I, V1, V2, V3, V4 and the
power supplied by the 10 volt source.
+ +
+
+
+
+
+
_ _
_
_
_
_
_
V1
V4
V3 V2
30 V 10 V
15  40 
5 
20 
20 V
I
"a"

Figure 3.15: Circuit for example 3.2.
For convenience, we start at point “a” and sum voltage drops =0 in the
direction of the current I.
+10 – V1 – 30 – V3 + V4 – 20 + V2 = 0 Eq. 3.1
19

Kirchhoff’s Voltage Law: Single-loop circuits. Ex. 3.2 cont.
We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I Eq. 3.2
We substitute the above into Eq. 3.1 to obtain Eq. 3.3 below.
10 + 20I – 30 + 15I + 5I – 20 + 40I = 0 Eq. 3.3
Solving this equation gives, I = 0.5 A.
Using this value of I in Eq. 3.2 gives;
V1 = - 10 V
V2 = 20 V
V3 = - 7.5 V
V4 = 2.5 V
P10(supplied) = -10I = - 5 W
(We use the minus sign in –10I because the current is entering the + terminal)
In this case, power is being absorbed by the 10 volt supply.
20

Kirchhoff’s Voltage Law: Single-loop circuits, Equivalent Resistance.
+ +
+
+
_
_
+
_ _
_
+
+
_
_
V1
V4
V3 V2
VS1 VS3
R2 R4
R3
R1
VS2
I
"a"

Given the circuit of Figure 3.16. We desire to develop an equivalent circuit
as shown in Figure 3.17. Find Vs and Req.
Figure 3.16: Initial circuit for
development.
VS
Req
+
_
I Figure 3.17: Equivalent circuit
for Figure 3.16
21

+ +
+
+
_
_
+
_ _
_
+
+
_
_
V1
V4
V3 V2
VS1 VS3
R2 R4
R3
R1
VS2
I
"a"

Figure 3.16: Initial circuit.
Starting at point “a”, apply KVL going clockwise, using drops = 0, we have
VS1 + V1 – VS3 + V2 + VS2 + V4 + V3 = 0
or
- VS1 - VS2 + VS3 = I(R1 + R2 + R3 + R4) Eq. 3.4
22

Consider again, the circuit of Figure 3.17.
VS
Req
+
_ I
Figure 3.17: Equivalent circuit
of Figure 3.16.
Writing KVL for this circuit gives;
VS = IReq compared to - VS1 - VS2 + VS3 = I(R1 + R2 + R3 + R4)
Therefore;
VS = - VS1 - VS2 + VS3 ; Req = R1 + R2 + R3 + R4 Eq. 3.5
23

We make the following important observations from Eq. 3.5:
• The equivalent source of a single loop circuit can be
obtained by summing the rises around the loop of
the individual sources.
• The equivalent resistance of resistors in series is equal
to the sum of the individual resistors.
24

Example 3.3: Find the current I in the circuit of Figure 3.18.
+ +
_
_ _
+
10 V 40 V
15  10 
5 
20 
20 V
I
Figure 3.18: Circuit for
example 3.3.
From the previous discussion we have the following circuit.
50 V 50 
+
_ I Therefore, I = 1 A
25

End of Lesson 3
circuits
Kirchhoff’s Voltage Law

Chapter_2_Basic_Circuit_Laws.ppt

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