unit 1 dc circuit analysis pdf 1 8 meg
TRANSCRIPT
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Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Sri Manakula Vinayagar
Engineering College , Madagadipet, Puducherry Page 1
UNIT- I
DC CIRCUIT ANALYSIS
Sources-Transformation and manipulation, Network theorems - Superposition theorem, Theveninstheorem, Nortons theorem, Reciprocity theorem, Millmans theorem, Compensation theorem,
Maximum power transfer theorem and Tellegens theorem Application to DC circuit analysis.
The interconnection of various electric elements in a prescribed manner comprises as an electric circuit in
order to perform a desired function.
The electric elements include controlled and uncontrolled source of energy, resistors, capacitors,
inductors, etc.
Analysis of electric circuits refers to computations required to determine the unknown quantitiessuch as voltage, current and power associated with one or more elements in the circuit.
BASIC ELEMENTS & INTRODUCTORY CONCEPTS
Electrical Network: A combination of various electric elements (Resistor, Inductor, Capacitor, Voltage
source, Current source) connected in any manner what so ever is called an electrical network. We may
classify circuit elements in two categories, passive and active elements.
Passive Element: The element which receives energy (or absorbs energy) and then either converts it into
heat (R) or stored it in an electric (C) or magnetic (L ) field is called passive element.
Active Element: The elements that supply energy to the circuit is called active element. Examples of
active elements include voltage and current sources, generators, and electronic devices that require power
supplies.
Bilateral Element: Conduction of current in both directions in an element (example: Resistance;
Inductance; Capacitance) with same magnitude is termed as bilateral element.
Non-Linear Circuit: Non-linear system is that whose parameters change with voltage or current. More
specifically, non-linear circuit
OHMS LAW
The potential difference (voltage) across an ideal conductor is proportional to the current through
it. The constant of proportionality is called the "resistance", R.
Ohm's Law is given by:
V = I R
where V is the potential difference between two points which include a resistance R
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VOLTAGE DIVISION RULE
Voltage division rule is applied on a circuit for finding individual voltage of an element or a resistance.
Formula:-
Suppose that, three resistances R1, R2 & R3 are connected in series with a voltage source v. So, the
Individual voltages V1, V2 & V3 are given by,
In general,
Note:- voltage division rule is only applicable for that circuit in which resistances are in series with a
battery source.
CURRENT DIVISION RULE
Current division rule is applied on a circuit for finding individual current of an element or a resistance.
Formula:-
Suppose that, two resistances R1 & R2 are connected in parallel with a current source I. So, the
Individual current i1 & i2 are given by,
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Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Sri Manakula Vinayagar
Engineering College , Madagadipet, Puducherry Page 3
In general,
Note:- Current division rule is only applicable for that circuit in which the resistances (or any passive
elements are in parallel) are in parallel with a current source. Current division rule can be used with a
current source as well as voltage source.
SOURCE TRANSFORMATION
Many times when we are solving a circuit or a problem it is quite difficult to solve it by using same
source which is given in the problem. To make easy, we convert our given source. Means that, a current
source can be changed by a voltage source and vice-verse. But have to fulfill it necessary condition.
(1) Current source to voltage sourceA current source can be converted into a voltage source if and only if a resistance is parallel to this current
source. The circuit will be as,
Current source is converted into a voltage source in which a parallel resistance R becomes in series with
this voltage source. And the value of this voltage source is equals to,
V=IR
Note:-The direction of voltage source is depends upon the direction of current source.
(2) Voltage source to Current source
A voltage source can be converted into a current source if and only if a resistance is in series with this
voltage source. The circuit will be as,
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Kirchhoffs Voltage Law
The summation of voltage rises and voltage drops around a closed loop is equal to zero. Symbolically,
this may be stated as follows:
0V for a closed loop
An alternate way of stating Kirchhoffs voltage law is as follows: The summation of voltage rises is equal
to the summation of voltage drops around a closed loop.
If we consider the circuit
By arbitrarily following the direction of the current, I, we move through the voltage source, which
represents a rise in potential from point a to point b.Next, in moving from point b to point c, we pass
through resistor R1, which presents a potential drop of V1. Continuing through resistors R2 and R3, we
have additional drops of V2 and V3 respectively. By applying Kirchhoffs voltage law around the closed
loop, we arrive at the following mathematical statement for the given circuit:
Kirchhoffs Current Law
The summation of currents entering a node is equal to the summation of currents leaving the node. In
mathematical form, Kirchhoffs current law is stated as follows:
Figure 65 is an illustration of Kirchhoffs current law. Here we see that the node has two currents
entering,I1 = 5 A andI5 = 3 A, and three currents leaving,I2 = 2 A,I3 =4 A, andI4 =2 A. Now we can
see that Equation applies in the illustration, namely.
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Engineering College , Madagadipet, Puducherry Page 5
Voltage Sources are in series
If the rises in one direction were equal to the rises in the opposite direction, then the resultant voltage
source would be equal to zero.
Note : Voltage sources of different potentials should never be connected in parallel, since to do so
would contradict Kirchhoffs voltage law. However, when two equal potential sources are connected in
parallel, each source will deliver half the required circuit current. For this reason automobile batteries are
sometimes connected in parallel to assist in starting a car with a weak battery.
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Current Sources are in parallel
When several current sources are placed in parallel, the circuit may be simplified by combining the
current sources into a single current source. The magnitude and direction of this resultant source is
determined by adding the currents in one direction and then subtracting the currents in the opposite
direction.
Since all of the current sources are in parallel, they can be replaced by a single current source. The
equivalent current source will have a direction which is the same as bothI2 andI3, since the magnitude of
current in the downward direction is greater than the current in the upward direction. The equivalent
current source has a magnitude of
I=2 A+6 A -3 A =5 A
Note : Current sources should never be placed in series. If a node is chosen between the current sources,
it becomes immediately apparent that the current entering the node is not the same as the current leaving
the node. Clearly, this cannot occur since there would then be a violation of Kirchhoffs current law.
Superposition Theorem
The superposition theorem is a method which allows us to determine the current through or the voltage
across any resistor or branch in a network. The advantage of using this approach instead of mesh analysis
or nodal analysis is that it is not necessary to use determinants or matrix algebra to analyze a given circuit.
The theorem states the following:
The total current through or voltage across a resistor or branch may be determined by summing the
effects due to each independent source.
In order to apply the superposition theorem it is necessary to remove all sources other than the
one being examined.
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Engineering College , Madagadipet, Puducherry Page 7
In order to zero a voltage source, we replace it with a short circuit, since the voltage across a
short circuit is zero volts.
A current source is zeroed by replacing it with an open circuit, since the current through an
open circuit is zero amps.
Example
Consider the circuit of Figure
Determine the current in the load resistor,R
Verify that the superposition theorem does not apply to power.
Solution
We first determine the current throughRL due to the voltage source by removing the current source
and replacing in with an open circuit (zero amps) as shown in Figure .
The resulting current throughRL is determined from Ohms law as
Next, we determine the current throughRL due to the current source by removing the voltage source and
replacing it with a short circuit (zero volts) as shown in Figure.
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Engineering College , Madagadipet, Puducherry Page 8
The resulting current throughRL is found with the current divider rule as
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Engineering College , Madagadipet, Puducherry Page 9
EXAMPLE
Determine the voltage drop across the resistorR2 of the circuit shown in Figure.
Solution
Since this circuit has three separate sources, it is necessary to determine the voltage acrossR2 due to each
individual source. First, we consider the voltage acrossR2 due to the 16-V source as shown in Figure.
Next, we consider the current source. The resulting circuit is shown in Figure .
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Engineering College , Madagadipet, Puducherry Page 10
Finally, the voltage due to the 32-V source is found by analyzing the circuit of Figure.
Thevenins Theorem
Thvenins theorem is a circuit analysis technique which reduces any linear bilateral network to an
equivalent circuit having only one voltage source and one series resistor. The resulting two-terminal
circuit is equivalent to the original circuit when connected to any external branch or component. In
summary, Thvenins theorem is simplified as follows:
Any linear bilateral network may be reduced to a simplified two-terminal circuit consisting of a single
voltage source in series with a single resistor.
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Engineering College , Madagadipet, Puducherry Page 11
The following steps provide a technique which converts any circuit into its Thvenin equivalent:
1. Remove the load from the circuit.
2. Label the resulting two terminals. We will label them as a and b, although any notation may be
used.
3. Set all sources in the circuit to zero.
Voltage sources are set to zero by replacing them with short circuits (zero volts). Current sources are set to zero by replacing them with open circuits (zero amps).
4. Determine the Thvenin equivalent resistance,RTh, by calculating the resistance seen between
terminals a and b. It may be necessary to redraw the circuit to simplify this step.
5. Replace the sources removed in Step 3, and determine the open-circuit voltage between the
terminals. If the circuit has more than one source, it may be necessary to use the superposition
theorem. In that case, it will be necessary to determine the open-circuit voltage due to each source
separately and then determine the combined effect. The resulting open-circuit voltage will be the
value of the Thvenin voltage,ETh.
6. Draw the Thvenin equivalent circuit using the resistance determined in Step 4 and the voltage
calculated in Step 5. As part of the resulting circuit, include that portion of the network removed
in Step 1.
Example
Determine the Thvenin equivalent circuit external to the resistorRL for the circuit of Figure. Use the
Thvenin equivalent circuit to calculate the current throughRL
Solution
Steps 1 and 2: Removing the load resistor from the circuit and labelling the remaining terminals, we
obtain the circuit
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Engineering College , Madagadipet, Puducherry Page 12
Step 3: Setting the sources to zero, we have the circuit.
Step 4: The Thevenin resistance between the terminals isRTh = 24 .
Step 5: From Figure, the open-circuit voltage between terminals a and b
is found as
Vab = 20 V - (24 )(2 A) = 28.0 V
Step 6: The resulting Thevenin equivalent circuit is
Example
Find the Thevenin equivalent circuit. Using the equivalent circuit, determine the current through the load
resistor whenRL = 0K, 2K and 5K,
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Engineering College , Madagadipet, Puducherry Page 13
Solution
Steps 1, 2, and 3: After removing the load, labelling the terminals, and setting the sources to zero, we
have the circuit
Step 4: The Thvenin resistance of the circuit is
Step 5: Although several methods are possible, we will use the superposition theorem to find the open-
circuit voltage Vab. Figure shows the circuit for determining the contribution due to the 15-V source.
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Engineering College , Madagadipet, Puducherry Page 14
Figure shows the circuit for determining the contribution due to the 5-mA source.
Step 6: The resulting Thvenin equivalent circuit
From this circuit, it is now an easy matter to determine the current for any value of load resistor:
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Engineering College , Madagadipet, Puducherry Page 15
NORTONS THEOREM
Nortons theorem is a circuit analysis technique which is similar to Thvenins theorem. By using this
theorem the circuit is reduced to a single current source and one parallel resistor. As with the Thvenin
equivalent circuit, the resulting two-terminal circuit is equivalent to the original circuit when connected to
any external branch or component. In summary, Nortons theorem may be simplified as follows:
Any linear bilateral network may be reduced to a simplified two-terminal circuit consisting of a single
current source and a single shunt resistor.
The following steps provide a technique which allows the conversion of any circuit into its Norton
equivalent:1. Remove the load from the circuit.
2. Label the resulting two terminals. We will label them as a and b, although any notation may be
used.
3. Set all sources to zero. As before, voltage sources are set to zero by replacing them with short
circuits and current sources are set to zero by replacing them with open circuits.
4. Determine the Norton equivalent resistance, RN , by calculating the resistance seen between
terminals a and b. It may be necessary to redraw the circuit to simplify this step.
5. Replace the sources removed in Step 3, and determine the current which would occur in a short if
the short were connected between terminals a and b. If the original circuit has more than one
source, it may be necessary to use the superposition theorem. In this case, it will be necessary to
determine the short-circuit current due to each source separately and then determine the
combined effect. The resulting short-circuit current will be the value of the Norton currentIN.
6. Sketch the Norton equivalent circuit using the resistance determined in Step 4 and the current
calculated in Step 5. As part of the resulting circuit, include that portion of the network removed
in Step 1.
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Example
Determine the Norton equivalent circuit external to the resistor RL for the circuit of Figure. Use the
Norton equivalent circuit to calculate the current throughRL
Solution
Steps 1 and 2: Remove load resistorRL from the circuit and label the remaining terminals as a and b. The
resulting circuit is
Step 3: Zero the voltage and current sources as shown in the circuit
Step 4: The resulting Norton resistance between the terminals is
Step 5: The short-circuit current is determined by first calculating the current through the short due to
each source. The circuit for each calculation is illustrated in Figure
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Voltage Source, E: The current in the short between terminals a and b
Current Source, I: By examining the circuit for the current source [Figure (b)] we see that the short circuit
between terminals a and b effectively removes R1 from the circuit. Therefore, the current through the
short will be
The negative sign indicates that the short-circuit current is actually from terminal b toward terminal a.
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Example
Consider the circuit
Find the Norton equivalent circuit external to terminals a and b.
Determine the current throughRL
Solution
Steps 1 and 2: After removing the load (which consists of a current source in parallel with a resistor), we
have the circuit
Step 3: After zeroing the sources, we have the network shown in Figure
Step 4: The Norton equivalent resistance is found as
Step 5: In order to determine the Norton current we must again determine the short-circuit current due to
each source separately and then combine the results using the superposition theorem.
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Voltage Source, E: Referring to Figure (a), notice that the resistor R2 is shorted by the short circuit
between terminals a and b and so the current in the short circuit is
Figure (a)
Current Source, I: Referring to Figure (b), the short circuit between terminals a and b will now eliminate
both resistors.
The current through the short will simply be the source current. However, since the current will not be
from a to bbut rather in the opposite direction, we write
Now the Norton current is found as the summation of the short-circuit currents due to each source:
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MAXIMUM POWER TRANSFER THEOREM
In amplifiers and in most communication circuits such as radio receivers and transmitters, it is often
desired that the load receive the maximum amount of power from a source.
The maximum power transfer theorem states the following:A load resistance will receive maximum
power from a circuit when the resistance of the load is exactly the same as the Thvenin (Norton)resistance looking back at the circuit.
From Figure we see that once the network has been simplified using either Thvenins or Nortons
theorem, maximum power will occur when
The following equations determine the power delivered to the load:
Under maximum power conditions (RL = RTh=RN), the above equations may be used to determine the
maximum power delivered to the load
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EXAMPLE
Consider the circuit
Determine the value of load resistance required to ensure that maximum power is transferred to
the load.
Find VL,IL and PL when the maximum power is delivered to the load
Solution
First we have to simplify the given network using thevnins theorem
Steps 1, 2, and 3: After removing the load, labelling the terminals, and setting the sources to zero, we
have the circuit
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Step 4: The Thvenin resistance of the circuit is
Step 5: Although several methods are possible, we will use the superposition theorem to find the open-
circuit voltage Vab. Figure shows the circuit for determining the contribution due to the 15-V source.
Figure shows the circuit for determining the contribution due to the 5-mA source.
Step 6: The resulting Thvenin equivalent circuit
Maximum power will be transferred to the load when RL = 1.5k
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Let RL = 1.5k, we see that half of the Thevenins voltage will appear across the load resistor and half will
appear across the Thevenins resistance. So at maximum power,
Note :
By using the maximum power transfer theorem, we see that under the condition of maximum power the
efficiency of the circuit is
SUBSTITUTION THEOREM
The substitution theorem states the following: Any branch within a circuit may be replaced by an
equivalent branch, provided the replacement branch has the same current through it and voltage across it
as the original branch.
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This theorem is best illustrated by examining the operation of a circuit. Consider the circuit
The voltage Vab and the current I in the circuit
The resistor R2 may be replaced with any combination of components, provided that the resulting
components maintain the above conditions
EXAMPLE
If the indicated portion in the circuit of Figure is to be replaced with a current source and a 240- shunt
resistor, determine the magnitude and direction of the required current source.
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Solution
The voltage across the branch in the original circuit is
which results in a current of
Finally, we know that the current entering terminal a isI= 200 mA. In order for Kirchhoffs current law
to be satisfied at this node, the current source must have a magnitude of 150 mA and the direction must be
downward, as shown in Figure
MILLMANS THEOREM
Millmans theorem is used to simplify circuits having several parallel voltage sources as illustrated in
Figure (a). Although any of the other theorems developed in this chapter will work in this case, Millmans
theorem provides a much simpler and more direct equivalent. In circuits of the type shown in Figure(a),
the voltage sources may be replaced with a single equivalent source as shown in Figure (b).
(a) (b)
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To find the values of the equivalent voltage sourceEeq and series resistanceReq , we need to convert each
of the voltage sources of Figure (a) into its equivalent current source using the technique (source
transformation). The value of each current source would be determined by using Ohms law (i.e.,
I1=E1/R1, I2=E2/R2, etc). After the source conversions are completed, the circuit appears as shown in
Figure (c)
(c)
It is now possible to replace the n current sources with a single current source having a magnitude given
as
The general expression for the equivalent voltage is
Example
Use Millmans Theorem to simplify the circuit and Use the simplified circuit to find the current in the
load resistor,RL
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Solution
we express the equivalent voltage source as
The equivalent circuit using Millmans theorem is shown in Figure (c).
Notice that the equivalent voltage source has a polarity which is opposite to the originally assumed
polarity. This is because the voltage sourcesE1 andE3 have magnitudes which overcome the polarity and
magnitude of the sourceE2.
From the equivalent circuit, the current through the load resistor:
RECIPROCITY THEOREM
The reciprocity theorem is a theorem which can only be used with single source circuits. This theorem,
however, may be applied to either voltage sources or current sources. The theorem states the following:
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Voltage Sources
A voltage source causing a current I in any branch of a circuit may be removed from the original location
and placed into that branch having the current I. The voltage source in the new location will produce a
current in the original source location which is exactly equal to the originally calculated current, I.
When applying the reciprocity theorem for a voltage source, the following steps must be followed:1. The voltage source is replaced by a short circuit in the original location.
2. The polarity of the source in the new location is such that the current direction in that branch
remains unchanged.
Current Sources
A current source causing a voltage V at any node of a circuit may be removed from the original location
and connected to that node. The current source in the new location will produce a voltage in the original
source location which is exactly equal to the originally calculated voltage, V.
When applying the reciprocity theorem for a current source, the following conditions must be met:
1. The current source is replaced by an open circuit in the original location.
2. The direction of the source in the new location is such that the polarity of the voltage at the node
to which the current source is now connected remains unchanged.
Example
Consider the circuit
a. Calculate the currentI
b. Remove voltage sourceE and place it into the branch withR3 . Show that the current through the
branch which formerly hadE is now the same as the currentI.
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Solution
For the circuit, we determine the currentI as follows
Example
Consider the circuit
a. Determine the voltage V across resistorR3
b. Remove the current sourceI and place it between node b
and the reference node. Show that the voltage across the
former location of the current source (node a) is now the
same as the voltage V.
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Solution
a. The node voltages for the circuit
b. After relocating the current source from the original location, and connecting it between node b and
ground, we obtain the circuit shown in Figure.
Assignment
1. Given the circuit, use superposition to calculate the current
through each of the resistors.
2. Use superposition to determine the voltage drop across each
of the resistors of the circuit in Figure.
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3. Find the Thvenin equivalent external to RL