EE6201 Circuit Theory Regulation 2013 Lecture Notes PDF
EE6201 Circuit Theory Regulation 2013 Lecture Notes PDF
EE6201 Circuit Theory Regulation 2013 Lecture Notes PDF
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UNIT I
BASIC CIRCUITS
ANALYSIS
INTRODUCTION
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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.
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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. A transistor is an active circuit element,
meaning that it can amplify power of a signal. On the other hand, transformer is not an
active element because it does not amplify the power level and power remains same
both in primary and secondary sides. Transformer is an example of passive element.
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In many cases, such as in electronic circuits, the chassis is shorted to the earth
itself for safety reasons.
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Voltage Divider
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Let us consider a simple dc network as shown in Figure 4.1 to find the currents
through different branches using Mesh (Loop) current method.
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UNIT II
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NETWORK REDUCTION
AND NETWORK
THEOREMS FOR DC
AND AC CIRCUITS
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INTRODUCTION
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SUPERPOSITION THEOREM
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power calculations
involve either the product of voltage and current, the square of current or the square of the
voltage, they are not linear operations. This statement can be explained with a simple example
as given below.
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Remarks: The Thevenin equivalent circuit is useful in finding the maximum power that a linear circuit
can deliver to a load.
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Norton's Theorem
Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex,
to an equivalent circuit with just a single current source and parallel resistance connected to a
load. Just as with Thevenin's Theorem, the qualification of linear is identical to that found in the
Superposition Theorem: all underlying equations must be linear (no exponents or roots).
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Contrasting our original example circuit against the Norton equivalent: it looks something like
this:
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Remember that a current source is a component whose job is to provide a constant amount of
current, outputting as much or as little voltage necessary to maintain that constant current.
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As with Thevenin's Theorem, everything in the original circuit except the load resistance has
been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's
Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton)
and Norton resistance (RNorton).
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As before, the first step is to identify the load resistance and remove it from the original circuit:
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Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a
direct wire (short) connection between the load points and determine the resultant current. Note
that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced
the load resistor with a break (open circuit):
With zero voltage dropped between the load resistor connection points, the current through R 1 is
strictly a function of B1's voltage and R1's resistance: 7 amps (I=E/R). Likewise, the current
through R3 is now strictly a function of B2's voltage and R3's resistance: 7 amps (I=E/R). The
total current through the short between the load connection points is the sum of these two
currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source
current (INorton) in our equivalent circuit:
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Remember, the arrow notation for a current source points in the direction opposite that of
electron flow. Again, apologies for the confusion. For better or for worse, this is standard
electronic symbol notation. Blame Mr. Franklin again!
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To calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating
Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed),
remove the power sources (in the same style as we did with the Superposition Theorem: voltage
sources replaced with wires and current sources replaced with breaks), and figure total
resistance from one load connection point to the other:
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REVIEW:
Norton's Theorem is a way to reduce a network to an equivalent circuit composed of a
single current source, parallel resistance, and parallel load.
Steps to follow for Norton's Theorem:
(1) Find the Norton source current by removing the load resistor from the original
circuit and calculating current through a short (wire) jumping across the open connection
points where the load resistor used to be.
(2) Find the Norton resistance by removing all power sources in the original circuit
(voltage sources shorted and current sources open) and calculating total resistance
between the open connection points.
(3) Draw the Norton equivalent circuit, with the Norton current source in parallel with
the Norton resistance. The load resistor re-attaches between the two open points of the
equivalent circuit.
(4) Analyze voltage and current for the load resistor following the rules for parallel
circuits
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As with the Thevenin equivalent circuit, the only useful information from this analysis is the
voltage and current values for R2; the rest of the information is irrelevant to the original circuit.
However, the same advantages seen with Thevenin's Theorem apply to Norton's as well: if we
wish to analyze load resistor voltage and current over several different values of load resistance,
we can use the Norton equivalent circuit again and again, applying nothing more complex than
simple parallel circuit analysis to determine what's happening with each trial load.
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Nortons Theorem
In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that Thevenin
reduces his circuit down to a single resistance in series with a single voltage. Norton on the other hand reduces his
circuit down to a single resistance in parallel with a constant current source. Nortons Theorem states that "Any linear
circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in
parallel with a Single Resistor". As far as the load resistance, RL is concerned this single resistance, RS is the value
of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit
current at the output terminals as shown below.
The value of this "constant current" is one which would flow if the two output terminals where shorted together while
the source resistance would be measured looking back into the terminals, (the same as Thevenin).
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For example, consider our now familiar circuit from the previous section.
To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40
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When the terminals A and B are shorted together the two resistors are connected in parallel across their two
respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can
now be calculated as:
If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively
connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the
terminals A and B giving us the following circuit.
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Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following
Nortons equivalent circuit.
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Ok, so far so good, but we now have to solve with the original 40
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The voltage across the terminals A and B with the load resistor connected is given as:
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The basic procedure for solving a circuit using Nortons Theorem is as follows:
1. Remove the load resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
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In a circuit, power supplied to the load is at its maximum when the load resistance is equal to the source resistance.
In the next tutorial we will look at Maximum Power Transfer. The application of the maximum power transfer
theorem can be applied to either simple and complicated linear circuits having a variable load and is used to find the
Example-L.8.5 For the circuit shown in fig.8.10(a), find the current through resistor ( branch) using
Nortons theorem & hence calculate the voltage across the current source (). 21LRR== abIcgV
Solution:
Step-1: Remove the resistor through which the current is to be found and short the terminals a and
b (see fig.8.10(b)).
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Step-2: Any method can be adopted to compute the current flowing through the a-b branch. Here, we
apply mesh current method.
Loop-1
3 R (I I ) = 0, where I = - 2A
4
R I = 3 + R I = 3 2 2 = - 1 I = - 0.5A
4 1
4 2
Loop-3
133323333- RI-R(I-I)=0- 3I-4(I+2)=0- 7I-8=08 I=-=7
N138-7+1I=(I-I)=-0.5+=714
9A14=(current is flowing from a to b)
Step-3: To compute R , all sources are replaced with their internal resistances. The equivalent
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resistance between a and b terminals is same as the value of Thevenins resistance of the circuit
shown in fig.8.3(d).
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Step-4: Replace the original circuit with an equivalent Nortons circuit as shown in fig.8.10(d).
R1.555I=I=0.643=0.39A (a to b)R+R1.555+1
In order to calculate the voltage across the current source the following procedures are adopted.
Redraw
the
original circuit
indicating
the
current
direction in the
load.
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T.2 When a complicated dc circuit is replaced by a Norton equivalent circuit, it consists of ------ in ---- with one -------. [2]
T.3 The dual of a voltage source is a -----------. [1]
T.4 When a Thevenin theorem is applied to a network containing a current source; the current source
is eliminated by --------- it. [1]
T.5 When applying Nortons theorem, the Norton current is determined with the output terminals -------------, but the Norton resistance is found with the output terminals ---------.and subsequently all
the independent sources are replaced -----------. [3]
T.6 For a complicated circuit, the Thevenin resistance is found by the ratio of -------- voltage and ----------- current. [2]
T.7 A network delivers maximum power to the load when its -------- is equal to the -------- of circuit
at the output terminals. [2]
T.8 The maximum power transfer condition is meaningful in ------------ and --------- systems. [2]
T.9 Under maximum power transfer conditions, the efficiency of the system is only --------- %. [1]
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UNIT III
RESONANCE AND
COUPLED CIRCUITS
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Ideal Transformer
In this lesson, we shall study two winding ideal transformer, its properties and working
principle under no load condition as well as under load condition. Induced voltages in primary and
secondary are obtained, clearly identifying the factors on which they depend upon. The ratio between
the primary and secondary voltages are shown to depend on ratio of turns of the two windings. At the
end, how to draw phasor diagram under no load and load conditions, are explained. Importance of
studying such a transformer will be highlighted. At the end, several objective type and numerical
problems have been given for solving.
Key Words: Magnetising current, HV & LV windings, no load phasor diagram, reflected current,
equivalent circuit
23.2 Introduction
Transformers are one of the most important components of any power system. It basically
changes the level of voltages from one value to the other at constant frequency. Being a static
machine the efficiency of a transformer could be as high as 99%.
Big generating stations are located at hundreds or more km away from the load center (where the
power will be actually consumed). Long transmission lines carry the power to the load centre from
the generating stations. Generator is a rotating machines and the level of voltage at which it generates
power is limited to several kilo volts only
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UNIT IV
TRANSIENT RESPONSE
FOR DC CIRCUITS
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UNIT V
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ANALYSING THREE
PHASE CIRCUITS
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As shown earlier, normally the voltage generated, which is also transmitted and then distributed to
the consumer, is the sinusoidal waveform with a frequency of 50 Hz in
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