Unit 1 Dc Circuit Analysis PDF 1 8 Meg

UNIT UNIT-- I DC CIRCUIT ANALYSIS

Sources-Transformation Sources-Transformation and manipulation, manipulation, Network Network theorems theorems - Superposition theorem, Thevenin’s theorem, Norton’s theorem, theorem, Reciprocity theorem, Millman’s theorem, theorem, Compensation Compensation theorem, theorem, Maximum Maximum power transfer transfer theorem theorem and Tellegen Tellegen’s ’s theorem theorem – Applica Application tion to DC circui circuitt analys analysis. is.

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 uncontrolled source of energy, resistors, capacitors, capacitors, inductors, etc.

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Analysis of electric circuits refers to computations computations required to determine determine the unknown quantities such as voltage, current and power associated with one or more elements in the circuit.

BASIC ELEMENTS & INTRODUCTORY 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 elements in two categories, 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 Non-linear system is that whose parameters change with voltage or current. More specifically, non-linear circuit OHM’S LAW The potential difference (voltage) across an ideal conductor is proportional to the current t hrough 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 Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Engineering College , Madagadipet, Puducherry

Sri Manakula Manakul a Vinayagar Vinayaga 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 ‘R 1’, ‘R 2’ & ‘R 3’ are connected connected in series with a voltage source source ‘v’. So, the Individual voltages ‘V1’, ‘V2’ & ‘V3’ are given by,

In general,

Note:- voltage division division rule is is only applicable applicable for that that circuit circuit in which resistances resistances are in series with a battery source. source. CURRENT DIVISION RULE Current division rule is applied on a circuit for finding individual current current of an element or a resistance.

Formula:Suppose that, two resistances ‘R 1’ & ‘R 2’ are connected in parallel with a current source ‘I’. So, the Individual current ‘i 1’ & ‘i2’ are given by,

Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Engineering College , Madagadipet, Puducherry


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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 ‘R 1’, ‘R 2’ & ‘R 3’ are connected connected in series with a voltage source source ‘v’. So, the Individual voltages ‘V1’, ‘V2’ & ‘V3’ are given by,

In general,

Note:- voltage division division rule is is only applicable applicable for that that circuit circuit in which resistances resistances are in series with a battery source. source. CURRENT DIVISION RULE Current division rule is applied on a circuit for finding individual current current of an element or a resistance.

Formula:Suppose that, two resistances ‘R 1’ & ‘R 2’ are connected in parallel with a current source ‘I’. So, the Individual current ‘i 1’ & ‘i2’ are given by,



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In general,

Note:- Current division division rule is only applicable applicable for that circuit circuit in which the resistances (or any passive passive elements are in parallel) parallel) are in parallel with a current source. Current 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. problem. To make easy, easy, we convert our given source. Means that, 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 source A 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=I∙R V=I∙R 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. source. The circuit will be as,



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Kirchhoff’s 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:

 V  0

for a closed loop

An alternate way of stating Kirchhoff’s 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 V 1. Continuing through resistors R2 and R3, we have additional drops of V 2 and V 3 respectively. By applying Kirchhoff’s voltage law around the closed loop, we arrive at the following mathematical statement for the given circuit:

Kirchhoff’s Current Law

The summation of currents entering a node is equal to the summation of currents leaving the node. In mathematical form, Kirchhoff’s current law is stated as follows:

Figure 6–5 is an illustration of Kirchhoff’s current law. Here we see that the node has two currents entering, I 1 = 5 A and I 5 = 3 A, and three currents leaving, I 2 = 2 A, I 3 =4 A, and I 4 =2 A. Now we can see that Equation applies in the illustration, namely.


Sri Manakula Vinayagar

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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 Kirchhoff’s 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 both I 2 and I 3, 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 Kirchhoff’s 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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 

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 through R L 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 through R L is determined from Ohm’s law as

Next, we determine the current through R L 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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The resulting current through R L is found with the current divider rule as



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EXAMPLE Determine the voltage drop across the resistor R2 of the circuit shown in Figure.

Solution Since this circuit has three separate sources, it is necessary to determine the voltage across R2 due to each individual source. First, we consider the voltage across R2 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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Finally, the voltage due to the 32-V source is found by analyzing the circuit of Figure.

Thevenin’s Theorem Thévenin’s 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, Thévenin’s 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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The following steps provide a technique which converts any circuit into its Thévenin 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 Thévenin 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 Thévenin voltage, E Th. 6. Draw the Thévenin 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 Thévenin equivalent circuit external to the resistor R L for the circuit of Figure. Use the Thévenin equivalent circuit to calculate the current through RL

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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Step 3: Setting the sources to zero, we have the circuit.

Step 4: The Thevenin resistance between the terminals is RTh = 24Ω . Step 5: From Figure, the open-circuit voltage between terminals a and b is found as

V ab = 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 when R L = 0K Ω, 2KΩ and 5KΩ,



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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 Thévenin resistance of the circuit is

Step 5: Although several methods are possible, we will use the superposition theorem to find the opencircuit voltage V ab. Figure shows the circuit for determining the contribution due to the 15-V source.



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Figure shows the circuit for determining the contribution due to the 5-mA source.

Step 6: The resulting Thévenin 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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NORTON’S THEOREM Norton’s theorem is a circuit analysis technique which is similar to Thévenin’s theorem. By using this theorem the circuit is reduced to a single current source and one parallel resistor. As with the Thévenin equivalent circuit, the resulting two-terminal circuit is equivalent to the original circuit when connected to any external branch or component. In summary, Norton’s 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, R N , 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 current I N . 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 R L for the circuit of Figure. Use the Norton equivalent circuit to calculate the current through R L

Solution Steps 1 and 2: Remove load resistor R L 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 through R L

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 b but 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 Thévenin (Norton) resistance looking back at the circuit .

From Figure we see that once the network has been simplified using either Thévenin’s or Norton’s theorem, maximum power will occur when

The following equations determine the power delivered to the load:

Under maximum power conditions ( R L = RTh=R N ), 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 Thévenin resistance of the circuit is

Step 5: Although several methods are possible, we will use the superposition theorem to find the opencircuit voltage V ab. 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 Thévenin equivalent circuit

Maximum power will be transferred to the load when R L = 1.5k Ω Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Engineering College , Madagadipet, Puducherry


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Let R L = 1.5k Ω, 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 R 2 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 Ω- 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 is I= 200 mA. In order for Kirchhoff’s 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

MILLMAN’S THEOREM Millman’s 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, Millman’s 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) Lecture notes prepared by P. RAJA, Associate Professor, Department of ECE, Engineering College , Madagadipet, Puducherry

(b) Sri Manakula Vinayagar

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To find the values of the equivalent voltage source E eq and series resistance Req , 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 Ohm’s law (i.e., I 1=E1/R 1, I2=E2/R 2, 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 Millman’s Theorem to simplify the circuit and Use the simplified circuit to find the current in the load resistor, R L



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Solution we express the equivalent voltage source as

The equivalent circuit using Millman’s 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 sources E 1 and E 3 have magnitudes which overcome the polarity and magnitude of the source E 2. 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.


a. Calculate the current I b. Remove voltage source E and place it into the branch with R3 . Show that the current through the branch which formerly had E is now the same as the current I .



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Solution

For the circuit, we determine the current I as follows


a. Determine the voltage V across resistor R3 b. Remove the current source I 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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Unit 1 Dc Circuit Analysis PDF 1 8 Meg

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