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CHAPTER 1
DC CIRCUITS
DEFINATIONS
Linear elements :
In an electric circuit, a linear element is an electrical element with a linear
relationship between current and voltage. Resistors are the most common
example of a linear element; other examples include capacitors, inductors,
and transformers.
Nonlinear Elements :
A nonlinear element is one which does not have a linear input/output
relation. In a diode, for example, the current is a non-linear function of the
voltage.Most semiconductor devices have non-linear characteristics.
Active Elements :
The elements which generates or produces electrical energy are called active
elements. Some of the examples are batteries,
generators,transistors,operational amplifiers,vacuum tubes etc.
Passive Elements :
All elements which consume rather than produce energy are called passive
elements, like resistors,inductors and capacitors.
 In unilateral element, voltage – current relation is not same for
both the direction. Example: Diode, Transistors.
 In bilateral element, voltage – current relation is same for both the
direction. Example: Resistor
 The voltage generated by the source does not vary with any circuit
quantity. It is only a function of time. Such a source is called an
ideal voltage Source.
 The current generated by the source does not vary with any circuit
quantity. It is only a function of time. Such a source is called as an
ideal current source.
 Resistance : It is the property of a substance which opposes the
flow of current through it. The resistance of element is denoted by
the symbol “R”. It is measured in Ohms. R = PL / A Ω
Ohm’s Law:
The current flowing through the electric circuit is directly proportional to
the potential difference across the circuit and inversely proportional to the
resistance of the circuit, provided the temperature remains constant.
+ _
v(t)
i(t)
R
v(t) = Ri(t)
+
_ v(t)
i(t)
R
v(t) = Ri(t)
_
(2.1)
(2.2)
Basic Laws of Circuits
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)
Basic Laws of Circuits
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)
Basic Laws of Circuits
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)
Basic Laws of Circuits
Ohm’s Law: Ohm’s Law: Example 2.1.
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.
Circuit Definitions
 Node – any point where 2 or more circuit elements
are connected together
 Wires usually have negligible resistance
 Each node has one voltage (w.r.t. ground)
 Branch – a circuit element between two nodes
 Loop – a collection of branches that form a closed
path returning to the same node without going
through any other nodes or branches twice
Example
How many nodes, branches & loops?
Three nodes
FEE Unit 2.ppt
FEE Unit 2.ppt
Basic Laws of Circuits
Kirchhoff’s Current Law
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
Basic Laws of Circuits
Kirchhoff’s Current Law
•The algebraic sum of the currents entering a node equal to zero.
Ia
Ib
Ic
Id
Ia +Ib Ic + Id = 0
Ia, Ib, Ic, and Id can each be either a positive
or negative number.
Basic Laws of Circuits
Kirchhoff’s Current Law
• The algebraic sum of the currents leaving a node equal to zero.
Ia
Ib
Ic
Id
Ia - Ib + Ic + Id = 0
Ia, Ib, Ic, and Id can each be either a positive
or negative number.
Basic Laws of Circuits
Kirchhoff’s Current Law: Example 2.2.
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.
-8 A
-3 A
-5 A
-2 A
Basic Laws of Circuits
Kirchhoff’s Current Law: Example 2.3
Find the currents IW, I X, IY, IZ.

 
12 A
IX IY
IZ
6 A
9 A
2 A
IW
IW =
IX =
IY =
IZ =
Basic Laws of Circuits
Kirchhoff’s Current Law
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
Kirchoff’s Voltage Law (KVL)
The algebraic sum of voltages around each loop is
zero
Beginning with one node, add voltages across each
branch in the loop
(if you encounter a + sign first) and subtract voltages
(if you encounter a – sign first)
Σ voltage drops - Σ voltage rises = 0
Or Σ voltage drops = Σ voltage rises
Circuit Analysis
 When given a circuit with sources and resistors
having fixed values, you can use Kirchoff’s two laws
and Ohm’s law to determine all branch voltages
and currents
+
12 v
-
I
7Ω
3Ω
A
B
C
+ VAB -
+
VBC
-
Circuit Analysis
 By Ohm’s law: VAB = I·7Ω and VBC = I·3Ω
 By KVL: VAB + VBC – 12 v = 0
 Substituting: I·7Ω + I·3Ω -12 v = 0
 Solving: I = 1.2 A
+
12 v
-
I
7Ω
3Ω
A
B
C
+ VAB -
+
VBC
-
Example Circuit
Solve for the currents through each resistor
And the voltages across each resistor
Example Circuit
Using Ohm’s law, add polarities and
expressions for each resistor voltage
+ I1∙10Ω -
+
I2∙8Ω
-
+ I3∙6Ω -
+
I3∙4Ω
-
Example Circuit
Write 1st Kirchoff’s voltage law equation
-50 v + I1∙10Ω + I2∙8Ω = 0
+ I1∙10Ω -
+
I2∙8Ω
-
+ I3∙6Ω -
+
I3∙4Ω
-
Example Circuit
Write 2nd Kirchoff’s voltage law equation
-I2∙8Ω + I3∙6Ω + I3∙4Ω = 0
or I2 = I3 ∙(6+4)/8 = 1.25 ∙ I3
+ I1∙10Ω -
+
I2∙8Ω
-
+ I3∙6Ω -
+
I3∙4Ω
-
Example Circuit
 We now have 3 equations in 3 unknowns, so we can
solve for the currents through each resistor, that are
used to find the voltage across each resistor
 Since I1 - I2 - I3 = 0, I1 = I2 + I3
 Substituting into the 1st KVL equation
-50 v + (I2 + I3)∙10Ω + I2∙8Ω = 0
or I2∙18 Ω + I3∙ 10 Ω = 50 volts
Example Circuit
 But from the 2nd KVL equation, I2 = 1.25∙I3
 Substituting into 1st KVL equation:
(1.25 ∙ I3)∙18 Ω + I3 ∙ 10 Ω = 50 volts
Or: I3 ∙ 22.5 Ω + I3 ∙ 10 Ω = 50 volts
Or: I3∙ 32.5 Ω = 50 volts
Or: I3 = 50 volts/32.5 Ω
Or: I3 = 1.538 amps
Example Circuit
 Since I3 = 1.538 amps
I2 = 1.25∙I3 = 1.923 amps
 Since I1 = I2 + I3, I1 = 3.461 amps
 The voltages across the resistors:
I1∙10Ω = 34.61 volts
I2∙8Ω = 15.38 volts
I3∙6Ω = 9.23 volts
I3∙4Ω = 6.15 volts

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FEE Unit 2.ppt

  • 2. DEFINATIONS Linear elements : In an electric circuit, a linear element is an electrical element with a linear relationship between current and voltage. Resistors are the most common example of a linear element; other examples include capacitors, inductors, and transformers. Nonlinear Elements : A nonlinear element is one which does not have a linear input/output relation. In a diode, for example, the current is a non-linear function of the voltage.Most semiconductor devices have non-linear characteristics. Active Elements : The elements which generates or produces electrical energy are called active elements. Some of the examples are batteries, generators,transistors,operational amplifiers,vacuum tubes etc. Passive Elements : All elements which consume rather than produce energy are called passive elements, like resistors,inductors and capacitors.
  • 3.  In unilateral element, voltage – current relation is not same for both the direction. Example: Diode, Transistors.  In bilateral element, voltage – current relation is same for both the direction. Example: Resistor  The voltage generated by the source does not vary with any circuit quantity. It is only a function of time. Such a source is called an ideal voltage Source.  The current generated by the source does not vary with any circuit quantity. It is only a function of time. Such a source is called as an ideal current source.  Resistance : It is the property of a substance which opposes the flow of current through it. The resistance of element is denoted by the symbol “R”. It is measured in Ohms. R = PL / A Ω
  • 4. Ohm’s Law: The current flowing through the electric circuit is directly proportional to the potential difference across the circuit and inversely proportional to the resistance of the circuit, provided the temperature remains constant. + _ v(t) i(t) R v(t) = Ri(t) + _ v(t) i(t) R v(t) = Ri(t) _ (2.1) (2.2)
  • 5. Basic Laws of Circuits 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)
  • 6. Basic Laws of Circuits 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)
  • 7. Basic Laws of Circuits 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)
  • 8. Basic Laws of Circuits Ohm’s Law: Ohm’s Law: Example 2.1. 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.
  • 9. Circuit Definitions  Node – any point where 2 or more circuit elements are connected together  Wires usually have negligible resistance  Each node has one voltage (w.r.t. ground)  Branch – a circuit element between two nodes  Loop – a collection of branches that form a closed path returning to the same node without going through any other nodes or branches twice
  • 10. Example How many nodes, branches & loops?
  • 14. Basic Laws of Circuits Kirchhoff’s Current Law 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
  • 15. Basic Laws of Circuits Kirchhoff’s Current Law •The algebraic sum of the currents entering a node equal to zero. Ia Ib Ic Id Ia +Ib Ic + Id = 0 Ia, Ib, Ic, and Id can each be either a positive or negative number.
  • 16. Basic Laws of Circuits Kirchhoff’s Current Law • The algebraic sum of the currents leaving a node equal to zero. Ia Ib Ic Id Ia - Ib + Ic + Id = 0 Ia, Ib, Ic, and Id can each be either a positive or negative number.
  • 17. Basic Laws of Circuits Kirchhoff’s Current Law: Example 2.2. 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.
  • 18. -8 A -3 A -5 A -2 A Basic Laws of Circuits Kirchhoff’s Current Law: Example 2.3 Find the currents IW, I X, IY, IZ.    12 A IX IY IZ 6 A 9 A 2 A IW IW = IX = IY = IZ =
  • 19. Basic Laws of Circuits Kirchhoff’s Current Law 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
  • 20. Kirchoff’s Voltage Law (KVL) The algebraic sum of voltages around each loop is zero Beginning with one node, add voltages across each branch in the loop (if you encounter a + sign first) and subtract voltages (if you encounter a – sign first) Σ voltage drops - Σ voltage rises = 0 Or Σ voltage drops = Σ voltage rises
  • 21. Circuit Analysis  When given a circuit with sources and resistors having fixed values, you can use Kirchoff’s two laws and Ohm’s law to determine all branch voltages and currents + 12 v - I 7Ω 3Ω A B C + VAB - + VBC -
  • 22. Circuit Analysis  By Ohm’s law: VAB = I·7Ω and VBC = I·3Ω  By KVL: VAB + VBC – 12 v = 0  Substituting: I·7Ω + I·3Ω -12 v = 0  Solving: I = 1.2 A + 12 v - I 7Ω 3Ω A B C + VAB - + VBC -
  • 23. Example Circuit Solve for the currents through each resistor And the voltages across each resistor
  • 24. Example Circuit Using Ohm’s law, add polarities and expressions for each resistor voltage + I1∙10Ω - + I2∙8Ω - + I3∙6Ω - + I3∙4Ω -
  • 25. Example Circuit Write 1st Kirchoff’s voltage law equation -50 v + I1∙10Ω + I2∙8Ω = 0 + I1∙10Ω - + I2∙8Ω - + I3∙6Ω - + I3∙4Ω -
  • 26. Example Circuit Write 2nd Kirchoff’s voltage law equation -I2∙8Ω + I3∙6Ω + I3∙4Ω = 0 or I2 = I3 ∙(6+4)/8 = 1.25 ∙ I3 + I1∙10Ω - + I2∙8Ω - + I3∙6Ω - + I3∙4Ω -
  • 27. Example Circuit  We now have 3 equations in 3 unknowns, so we can solve for the currents through each resistor, that are used to find the voltage across each resistor  Since I1 - I2 - I3 = 0, I1 = I2 + I3  Substituting into the 1st KVL equation -50 v + (I2 + I3)∙10Ω + I2∙8Ω = 0 or I2∙18 Ω + I3∙ 10 Ω = 50 volts
  • 28. Example Circuit  But from the 2nd KVL equation, I2 = 1.25∙I3  Substituting into 1st KVL equation: (1.25 ∙ I3)∙18 Ω + I3 ∙ 10 Ω = 50 volts Or: I3 ∙ 22.5 Ω + I3 ∙ 10 Ω = 50 volts Or: I3∙ 32.5 Ω = 50 volts Or: I3 = 50 volts/32.5 Ω Or: I3 = 1.538 amps
  • 29. Example Circuit  Since I3 = 1.538 amps I2 = 1.25∙I3 = 1.923 amps  Since I1 = I2 + I3, I1 = 3.461 amps  The voltages across the resistors: I1∙10Ω = 34.61 volts I2∙8Ω = 15.38 volts I3∙6Ω = 9.23 volts I3∙4Ω = 6.15 volts