Electronics Electric Bolystad

8/17/2019 Electronics Electric Bolystad

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Electrical circuits – 2010/2011 Control and Systems Eng. A.L. Taghreed M.

DC Series Circuits

A circuit consists of any number of elements joined at terminal points, providing at least oneclosed path through which charge can flow:

Two elements are in series if:

1. They have only one terminal in common (i.e., one lead of one is connected to only one lead of

the other).

2. The common point between the two elements is not connected to another current-carrying

element.

As shown in the circuit diagram above, since all the elements are in series, the network is called a

series circuit.

The current is the same through series elements.

Notes:

1.

A rise of potential occurs in an active element (source) when going from (-) to (+)

through the source, in the direction of current.


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2.

A drop of potential occurs in a passive element (e.g. resistor) when going from (+) to (-)

through the element in the direction of current.

KIRCHHOFF’S VOLTAGE LAW KVL):

States that the algebraic sum of the potential rises and drops around a closed loop (or path) is zero.

For the following circuit with Clock Wise (CW )

direction:

Where,

is the voltage rise. are the voltage drops.

Notes:

Clock wise direction will be used for all applications of KVL .

Plus sign (+) is used for potential rise.

Minus sign (-) is assigned for potential drop.

Series Circuit relations:

We are now ready to summarize current, voltage, resistance and power relations for series circuit:


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Since:

From KVL:

Dividing by I:

This forms Ohm's law:

Finally:

Example1: Determine the unknown voltages for the following networks:


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Solution:

Applying KVL for network a in CW direction yields:

Example2: Find V 1 and V 2 for the network:

Solution:

For path 1, starting at point a in a clockwise direction:

For path 2, starting at point a in a clockwise direction:

The minus sign simply indicates that the actual polarities of the potential difference are opposite the

assumed polarity indicated.

Example3: for the following network find:

a. Find R T.

b. Find I.

c. Find V 1 and V 2.

d. Find the power to the 4 Ω and 6 Ω

resistors.


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e. Find the power delivered by the battery, and compare it to that dissipated

f. By the 4 Ω and 6 Ω resistors combined.

g. Verify KVL.

Solution: a. b.

c.

d.

e.

E

W W W

VOLTAGE DIVIDER RULE

In a series circuit,

The voltage across the resistive elements will divide as the magnitude of the resistance levels

For the following series circuit:


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Then Voltage Divider Rule (VDR) will be:

Where ,

Vx: Voltage across Rx.

Example4: a KΩ and a KΩ are connected in series to 33V Source, without

finding current, calculate:

a. V 1 , V 2 .

b. P 1 , P 2 .

Solution: a.

b.

Home Work: Three resistors R 1 R 2 and R 3 are connected in series to 220V supply. The

combined voltage drop across R 1 and R 2 is 140 V and that across R 2 and R 3 is 180V. if the total

resistance of the circuit is 11 K Ω . Calculate R 1, R 2 and R 3.


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Voltage Sources and Ground

Except for a few special cases, electrical and electronic systems are grounded for reference

and safety purposes. The symbol for ground connection is:

The following figures show grounded supplies:


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Example4:for the given circuit find the voltage at point x:

Solution:

By KVL:

By VDR:

Example5: for the given circuit find the voltage at

point x:


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Solution:

By KVL for loop 1:

Or by KVL for loop 2:

INTERNAL RESISTANCE OF VOLTAGE SOURCES

Every source of voltage, whether a generator, battery, or laboratory supply as shown below, will have

some internal resistance


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Case a: for Ideal voltage source, which has no internal resistance the output voltage = E with load and

without load.

Case b: for not ideal voltage source which has internal resistance the output voltage = E with no load.

Case c: for not ideal voltage supply the output voltage will decrease due to the voltage drop across

the internal resistance.

For case c, by applying KVL around the indicated loop we obtain:


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To find the power delivered to the load:

Example6: a Lamp having a resistance of 5.6 Ω is connected across the battery

in the following circuit, find:

a. The current drawn from the battery.

b. The internal voltage drop across its

internal resistance.

c. The internal voltage of the battery.

d. The power dissipated internally

within the battery.

e.

The total power generated by the battery.

f.

The efficiency of the battery.


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Voltage regulation:

A measure of how close a supply voltage will come to ideal conditions, where the ideal

conditions dictate that for all range of load demand IL), the terminal voltage remain fixed in

magnitude.

For ideal conditions, V FL = V NL and VR% = 0 . Therefore, the smaller the voltage regulation, the

less the variation in terminal voltage with change in load.

It can be shown that:


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~ 1 ~

DC Parallel Circuits

Two elements, branches, or networks are inparallel

if they have two points in

common.

Different ways in which three parallel elements may appear as shown in the following figure:

d

)

In figure (d) elements 1 and 2 are in parallel since they have two common points [a, b].

The parallel combination of 1 and 2 is then in series with element 3 because they are

connected by point b only.

1.

TOTAL CONDUCTANCE AND RESISTANCE


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~ 2 ~

For parallel elements, the total conductance is the sum of the individual conductances.

That is for the following network:

… (1)

G: is the conductance in Siemens (S).

Since increasing levels of conductance will establish higher current levels , the more terms

appearing in Eq. 1. In other words, as the number of resistors in parallel increases, theinput current level will increase for the same applied voltage —the opposite effect of

increasing the number of resistors in series.

…. (2)

Notes:


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~ 3 ~

The total resistance of parallel resistors is always less than the value of the

smallest resistor.

For equal parallel resistors the total resistance will be:

The total resistance of two parallel resistors is:

I t can be shown that for 3 parallel resistors:

parallel elements can be interchanged without changing the total resistance

or input current.

2.

Current and Power of Parallel Circuits:


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~ 4 ~

For the shown parallel network The total resistance is determined by R T = R 1 R 2 /

(R 1 + R 2 ) , and the source current by I s = E / R T .

Since the terminals of the battery are connected directly across the resistors R1 and R2, the

following should be obvious:

The voltage across parallel elements is the same.

Using this fact will result in:

And

Then:

Multiplying both sides by E we have:


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~ 5 ~

For single-source parallel networks, the source current Is) is equal to the sum of the

individual branch currents.

The power dissipated by the resistors and delivered by the source (in Watts ) can be

determined from:

Example1: Six resistors are connected in parallel have the following values:

120, 60, 40, 5, 4 and 2 K Ω . Calculate the:

a. Equivalent conductance of the parallel combination.

b.

Equivalent resistance of the parallel combination.c.

Circuit voltage if 20mW is dissipated by the 5K Ω resistor.

d.

Current in the 40K Ω resistor.

e. Total circuit current.

f. Total power drawn from the supply

Solution:

a.


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~ 6 ~

b.

c. d.

e.

f.

Example2: calculate the total circuit resistance and battery current for the

following circuit. Prove that the power delivered to the circuit equal to the

power dissipated in each resistor.

Solution: the circuit can be redrawn in the following manner:


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~ 7 ~

Checked

4.KIRCHHOFF’S CURRENT LAW

KCL

States that the algebraic sum of the currents entering and leaving a node, system, or

junction is zero.

Example3: Determine I1, I3, I4, and I5 for the following network:

Solution:


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At a :

At b: ∑ ∑

Since R 1 and R 3 are in series and the current is the same in series elements.

At c :

At d:

We can conclude that the current entering is I = 5 A; the net current leavingfrom the far right is I5 =5 A.

5. CURRENT DIVIDER RULE

(CDR):

For the following network:


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~ 9 ~

The voltage drop across any of the resistances (Rx) is:

Since

And the general form of current divider rule (CDL)is:

Particularly for two resistors CDL will be:

Example4:

Determine the


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~ 10 ~

magnitude of the currents I1, I2, and I3 for the network:

Solution:


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6. VOLTAGE SOURCES IN PARALLEL

Voltage sources are placed in parallel as shown in the following figure only if they have the

same voltage rating.

The primary reason for placing two or more batteries in parallel of the same terminal

voltage would be to increase the current rating (and, therefore, the power rating) of the

source.

If two batteries of different terminal voltages were placed in parallel, both would be left

ineffective or damaged because the terminal voltage of the larger battery would try to drop

rapidly to that of the lower supply. Consider two lead-acid car batteries of different terminal

voltage placed in parallel:

The relatively small internal resistances of the batteries are the only current-limiting

elements of the resulting series circuit. The current is:


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~ 12 ~

Which far exceeds the continuous drain rating of the larger supply, resulting in a rapid

discharge of E1 and a destructive impact on the smaller supply.

7. OPEN AND SHORT CIRCUITS

An open circuit is simply two isolated terminals not connected by an element of any kind.

While ashort circuit

is a very low resistance, direct connection between two terminals of a

network.

An open circuit can have a potential difference voltage) across its terminals, but the current is

always zero amperes.

An open circuit exists between terminals a and b .As shown above, the voltage across the

open-circuit terminals is the supply voltage, but the current is zero due to the absence of a

complete circuit.


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~ 13 ~

A short circuit can carry a current of a level determined by the external circuit, but the

potential difference voltage) across its terminals is always zero volts.

Consider the circuit shown below; the current through the 2 Ω resistor is 5 A.

If a short circuit should develop across the 2 Ω resistor,

The total resistance of the parallel combination of the 2 Ω resistor and the short (of essentially

zero ohms) will be

And the current will rise to very high levels, as determined by Ohm’s law:


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~ 14 ~

The effect of the 2 Ω resistor has effectively been “shorted out” by the low-resistance

connection. The maximum current is now limited only by the circuit breaker or fuse in series

with the source.

Example5: Calculate the current I and the voltage V for the network:

Solution:

The 10-k Ω resistor has been effectively shorted out by the jumper, resulting in the equivalentnetwork below, using Ohm's law:


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1

1. MESH ANALYSIS (GENERAL APPROACH)

The second method of analysis to be described is called mesh analysis. The

term mesh is derived from the similarities in appearance between the closed loops of a

network and a wire mesh fence. Applying this method the below steps should be

followed:

1. Assign the current in the clock wise direction (CW) to each closed loop in the network.

2. Show the voltage polarities across each resistor produced by CW mesh currents.

3. Apply KVL around each closed loop.

a) If the resistor has two or more currents, the total current is the current of

the loop in which KVL is being applied plus the currents of other loops in the

same direction, or minus the currents in the opposite direction.

b) The polarity of the voltage source is unaffected.

4. Solve the resulting simultaneous linear equations for the assumed loop currents.

Example1: for the following circuit find the complete solution showing all currents and

voltages by mesh analysis.


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2

Solution:

Two loop currents ( I 1 and I 2) are assigned in the clockwise direction in the windows of

the network. A third loop ( I 3) could have been included around the entire network, but

the information carried by this loop is already included in the other two. Now after steps 1 and 2 are done move on applying step 3 (KVL) around loop I 1 and I 2:

The minus signs indicate that the currents have a direction opposite to that indicated by theassumed loop current.


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3

()() ()()

() ()()

Notes:

*if a current source exist when applying mesh analysis with a resistor in parallel

with it the first step will be to convert all current sources to voltage sources.

*if the current source has no resistance in parallel with it start as before and

assign a mesh current to each independent loop, including the current sources, as

if they were resistors or voltage sources. Then mentally (redraw the network ifnecessary) remove the current sources (replace with open-circuit equivalents),

and apply Kirchhoff’s voltage law to all the remaining independent paths of the

network using the mesh currents just defined.

Example2: determine the currents of the following network using mesh analysis.

Solution: First, the mesh currents for the network are defined, as shown below.


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4

Then the current source is mentally removed, as shown in the network below, and

Kirchhoff’s voltage law is applied to the resulting network. The single path now includingthe effects of two mesh currents is referred to as the path of a super mesh current.


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5

2. MESH ANALYSIS (FORMAT APPROACH)

We will now examine a technique for writing the mesh equations more rapidly and

usually with fewer errors. Before solving the following example using mesh format

approach consider the following:

* The format approach can be applied only to networks in which all current sources have

been converted to their equivalent voltage source.


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6

Example 3: Write the mesh equations for the network below, and find the current

through the 7Ω resistor.

Solution: As indicated in the network, each assigned loop current has a clockwise

direction.


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7

Example4: write mesh equations for the following network.

Solution:


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3. NODAL ANALYSIS (GENERAL APPROACH)

Recall from the development of loop analysis that the general network equations were

obtained by applying Kirchhoff’s voltage law around each closed loop. We will now

employ Kirchhoff’s current law to develop a method referred to as nodal analysis.

A node is defined as a junction of two or more branches. The nodal analysis method is

applied as follows:

1. Determine the number of nodes within the network.

2. Pick a reference node, and label each remaining node with a subscripted value of

voltage: V1, V2, and so on.

3. Apply Kirchhoff’s current law at each node except the reference. Assume that all

unknown currents leave the node for each application of Kirchhoff’s current law.

4. Solve the resulting equations for the nodal voltages.

Example5: Apply nodal analysis to the network shown below.

Solution: step 1&2: the network has 2 nodes. The lower node is defined as the referencenode at ground potential (zero volts) and the

other node V1, the voltage from node 1 to the

ground.

Step3: applying KCL:


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Example6: determine nodal voltages for the following network.

Solution: step1, 2 as indicated in the figure below:

step3: applying KCL for the node V1:


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For node V2 as illustrated in the figure below:

Resulting in two equations and two unknowns:


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12


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13

Notes:

In applying nodal analysis if there exists voltage source in the network with a resistor in

series then convert them to current sources.

If the voltage source exists without a resistor in series then start as before and assign a

nodal voltage to each independent node of the network, including each independent

voltage source as if it were a resistor or current source. Then mentally replace the

independent voltage sources with short-circuit equivalents, and apply KCL to the

defined nodes of the network.

Example7: Determine the nodal voltages V1 and V2 using nodal analysis.

Solution:Replacing the independent voltage source of 12 V with a short-circuit equivalent

will result in the network:


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15

4. NODAL ANALYSIS (FORMAT APPROACH) It will allow us to write nodal equations rapidly and in a form that is convenient

for the use of determinants. A major requirement, however, is that all voltage sources

must first be converted to current sources before the procedure is applied . This approach

will be applied and explained by the following example.

Example8: write nodal equations for the following network.

Solution:


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16


Solution:

In this network 4 nodes can be taken to solve it

i.e.V4 can be included if desired

But three nodes are sufficient to solve this

network:


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18

Note the symmetry about the major diagonal above. This is a check for your

solution.


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1. Superposition Theorm

The superposition theorem states the following:

The cur rent through, or voltage across, an element in a l inear bil ateral network is equalto the algebraic sum of the cur rents or voltages produced independently by each sour ce.

The most obvious advantage of this method is that it does not require the use of a

mathematical technique such as determinants to find the required voltages or currents.Instead, each source is treated independently, and the algebraic sum is found to determine a

particular unknown quantity of the network.

When applying this theorem, the difference in potential between the terminals of the

voltage source must be set to zero (short circuit); removing a current source requires thatits terminals be opened (open circuit). Any internal resistance or conductance associatedwith the displaced sources is not eliminated but must still be considered.

The total current through any portion of the network is equal to the algebraic sum of the

currents produced independently by each source.

*Note:The total power delivered to a resistive element must be determinedusing the total current through or the total voltage across the element

and cannot be determined by a simple sum of the power levels

established by each source.


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Example1: Using superposition,

a. find the current through the 6Ω resistor of thenetwork.

b. Demonstrate that superposition is not applicable

to power levels.

Solution:

a. * Considering the effect of 36V source:

* Considering the effect of 9A source:

The total current through the 6Ω resistor is:


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As mentioned previously, the superposition principle is not applicable to power effects

since power is proportional to the square of the current or voltage ( I 2 R or V 2/ R).

Obviously, x + y ≠ z, or 24 W + 216 W ≠ 384 W, and superposition does not hold.

However, for a linear relationship, such as that between the voltage and current of the

fixed-type 6Ω resistor, superposition can be applied, as demonstrated by the above

graph.


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22

Example2: Using the principle of superposition, find the current I 2 through the 12Ωk

resistor.

Solution:

* Considering the effect of 6mA source:

* Considering the effect of 9V voltage source:


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23

2. THÉVENIN’S THEOREM

Thévenin’s theorem states the following:Any two-terminal , linear bil ateral dc network can be replaced by

an equivalent cir cuit consisting of a voltage source and a series

resistor .

The following sequence of steps will lead to the proper value of RTh and E th:

1. Remove that portion of the network across which the Thévenin equivalent circuit is to be

found and Mark the terminals of the remaining two-terminal network.

R Th:

2. Calculate R Th by first setting all sources to zero (voltage sources are replaced by short

circuits and current sources by open circuits) and then finding the resultant resistance

between the two marked terminals. ( If the internal resistance of the voltage and/or current

sources is included in the original network, it must remain when the sources are set to

zero.)

ETh:

3. Calculate ETh by first returning all sources to their original position and finding the open-

circuit voltage between the marked terminals.

Conclusion:

4. Draw the Thévenin equivalent circuit with the portion of the circuit previously removed

replaced between the terminals of the equivalent circuit.


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Example3: Find the Thévenin equivalent circuit for the network in the shaded area ofthe network shown below.

Solution: step1 is applied as shown below:


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Step 2 is applied to find RTh:

Step3:

Step 4 :


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Example4: Find the Thévenin equivalent circuit for the network in the shaded area of the

bridge network .

Solution:

Step1:

Step2:


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Step3:

step4:


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Example5: find Thévenin circuit for the network illustrated below.

Solution:

Step1:

Step2:


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Step3: applying Superposition and *taking the effect of E1.

* For the source E2:


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Step4:


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1

1. MESH ANALYSIS GENERAL APPROACH)

The second method of analysis to be described is called mesh analysis. The

term mesh is derived from the similarities in appearance between the closed loops of a

network and a wire mesh fence. Applying this method the below steps should be followed:

1. Assign the current in the clock wise direction (CW) to each closed loop in the network.

2. Show the voltage polarities across each resistor produced by CW mesh currents.

3. Apply KVL around each closed loop.

a

If the resistor has two or more currents, the total current is the current of the

loop in which KVL is being applied plus the currents of other loops in the

same direction, or minus the currents in the opposite direction.

b

The polarity of the voltage source is unaffected.

4. Solve the resulting simultaneous linear equations for the assumed loop currents.

Example1: for the following circuit find the complete solution showing all currents and

voltages by mesh analysis.

Solution:

Two loop currents (I 1 and I 2) are assigned in the clockwise direction in the windows of the

network. A third loop (I 3) could have been included around the entire network, but the

information carried by this loop is already included in the other two.


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2

Now after steps 1 and 2 are done move on applying step 3 (KVL) around loop I 1 and I 2 :

The minus signs indicate that the currents have a direction opposite to that indicated by the

assumed loop current.

()() ()()

() ()()


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3

Notes:

*if a current source exist when applying mesh analysis with a resistor in parallel

with it the first step will be to convert all current sources to voltage sources.

*if the current source has no resistance in parallel with it start as before and assign

a mesh current to each independent loop, including the current sources, as if they

were resistors or voltage sources. Then mentally (redraw the network if necessary)

remove the current sources (replace with open-circuit equivalents), and apply

Kirchhoff’s voltage law to all the remaining independent paths of the network using

the mesh currents just defined.

Example

2

: determine the currents of the following network using mesh analysis.

Solution: First, the mesh currents for the network are defined, as shown below.

Then the current source is mentally removed, as shown in the network below, and

Kirchhoff’s voltage law is applied to the resulting network. The single path now including the

effects of two mesh currents is referred to as the path of a super mesh current.


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4


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5

2. MESH ANALYSIS FORMAT APPROACH)

We will now examine a technique for writing the mesh equations more rapidly and usually

with fewer errors. Before solving the folloing example using mesh format approach

consider the following:

* The format approach can be applied only to networks in which all current sources have

been converted to their equivalent voltage source.

Example 3: Write the mesh equations for the network below, and find the current

through the 7 Ω resistor.

Solution: As indicated in the network, each assigned loop current has a clockwise

direction.


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6

Example4: write mesh equations for the following network.

Solution:


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7

3. NODAL ANALYSIS GENERAL APPROACH)

Recall from the development of loop analysis that the general network equations were

obtained by applying Kirchhoff’s voltage law around each closed loop. We will now

employ Kirchhoff’s current law to develop a method referred to as nodal analysis.

A node is defined as a junction of two or more branches. The nodal analysis method is

applied as follows:

1. Determine the number of nodes within the network.

2. Pick a reference node, and label each remaining node with a subscripted value of

voltage: V 1, V 2, and so on.3. Apply Kirchhoff’s current law at each node except the reference. Assume that all

unknown currents leave the node for each application of Kirchhoff’s current law.

4. Solve the resulting equations for the nodal voltages.

Example5: Apply nodal analysis to the network shown below.

Solution: step 1&2 : the network has 2 nodes. Thelower node is defined as the reference node at groundpotential (zero volts) and the other node V1, the

voltage from node 1 to the ground.

Step3 : applying KCL:


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8

Example

6

:determine nodal voltages for the following network.


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9

Solution: step1, 2 as indicated in the figure below:

step3: applying KCL for the node V 1:

For node V2 as illustrated in the figure below:


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Resulting in two equations and two unknowns:


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11

Notes:

In applying nodal analysis if there exists voltage source in the network with a resistor inseries then convert them to current sources.

If the voltage source exists without a resistor in series then start as before and assign a

nodal voltage to each independent node of the network, including each independent

voltage source as if it were a resistor or current source. Then mentally replace the

independent voltage sources with short-circuit equivalents, and apply KCL to the

defined nodes of the network.

Example 7: Determine the nodal voltages V1 and V2 using nodal analysis.

Solution:


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12

Replacing the independent voltage source of 12 V with a short-circuit equivalent

will result in the network:


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13

4. NODAL ANALYSIS FORMAT APPROACH) It will allow us to write nodal equations rapidly and in a form that is convenient

for the use of determinants. A major requirement, however, is that all voltage sources must

first be converted to current sources before the procedure is applied . This approach will be

applied and explained by the following example.


Solution:


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14


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15


Solution:

In this network 4 nodes can be taken to solve it

i.e.V4 can be included if desired

But three nodes are sufficient to solve this

network:


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16

Note the symmetry about the major diagonal above. This is a check for your

solution.


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17

1. Superposition Theorm

The superposition theorem states the following:

The current through, or voltage across, an element in a linear bilateral network is equal to the

algebraic sum of the currents or voltages produced independently by each source.

The most obvious advantage of this method is that it does not require the use of a

mathematical technique such as determinants to find the required voltages or currents.

Instead, each source is treated independently , and the algebraic sum is found to determine a

particular unknown quantity of the network.

When applying this theorem, the difference in potential between the terminals of thevoltage source must be set to zero (short circuit); removing a current source requires that its

terminals be opened (open circuit) . Any internal resistance or conductance associated with

the displaced sources is not eliminated but must still be considered.

The total current through any portion of the network is equal to the algebraic sum of the

currents produced independently by each source.

*Note:The total power delivered to a resistive element must be determined

using the total current through or the total voltage across the element

and cannot be determined by a simple sum of

the power levels established by each source.

Example1: Using superposition,


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18

a. find the current through the 6 Ω resistor of the network.b. Demonstrate that superposition is not applicable to power levels.

Solution:

a. * Considering the effect of 36V source:

* Considering the effect of 9A source:

The total current through the 6Ω resistor is:


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19

As mentioned previously, the superposition principle is not applicable to power effects

since power is proportional to the square of the current or voltage (I 2R or V 2 / R ).

Obviously, x + y ≠ z, or 24 W + 216 W ≠ 384 W, and superposition does not hold.

However, for a linear relationship, such as that between the voltage and current of the

fixed-type 6 Ω resistor, superposition can be applied, as demonstrated by the above

graph.


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20

Example2: Using the principle of superposition, find the current I 2 through the 12 Ω kresistor.

Solution:

* Considering the effect of 6mA source:

* Considering the effect of 9V voltage source:


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21

2.THÉV NIN’S

THEOREM

Thévenin’s theorem states the following:

Any two-terminal, linear bilateral dc network can be replaced by an

equivalent circuit consisting of a voltage source and a series resistor

.

The following sequence of steps will lead to the proper value of R Th

and E th :

1. Remove that portion of the network across which the Théveninequivalent circuit is to be found and Mark the terminals of the remaining two-terminal

network.

RTh:

2. Calculate R Th by first setting all sources to zero (voltage sources are replaced by short

circuits and current sources by open circuits ) and then finding the resultant resistance

between the two marked terminals. (If the internal resistance of the voltage and/or current

sources is included in the original network, it must remain when the sources are set to zero.)

ETh:

3. Calculate ETh by first returning all sources to their original position and finding the open-

circuit voltage between the marked terminals.

Conclusion:

4. Draw the Thévenin equivalent circuit with the portion of the circuit previously removed

replaced between the terminals of the equivalent circuit.


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22

Example3: Find the Thévenin equivalent circuit for the network in the shaded area ofthe network shown below.

Solution: step1 is applied as shown below:

Step 2 is applied to find R Th :


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Step3:

Step 4 :


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24

Example4: Find the Thévenin equivalent circuit for the network in the shaded area of the

bridge network .

Solution:

Step1:

Step2:


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Step3:


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step4:

Example5: find

Thévenin circuit for the

network illustrated below.

Solution:

Step1:


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Step2:

Step3: applying Superposition and *taking

the effect of E1.

* For the source E2:


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Step4:


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3 NORTON’S THEOREM

States the following:

Any two-terminal linear bilateral dc network can be replaced by an equivalent

circuit consisting of a current source and a parallel resistor.

The steps leading to the proper values of IN and RN are now listed.

1. Remove that portion of the network across which the Norton equivalent circuit is

found.

2.

Mark the terminals of the remaining two-terminal network. RN:

3. Calculate RN by first setting all sources to zero (voltage sources are replaced with

short circuits and current sources with open circuits) and then finding the resultant

resistance between the two marked terminals. (If the internal resistance of the

voltage and/or current sources is included in the original network, it must remain

when the sources are set to zero.) Since RN _ RTh, the procedure and value obtained

using the approach described for Thévenin’s theorem will determine the proper valueof RN. IN:

4.

Calculate IN by first returning all sources to their original position and then finding

the short-circuit current between the marked terminals. It is the same current that

would be measured by an ammeter placed between the marked terminals

Conclusion:

5. Draw the Norton equivalent circuit with the portion of the circuit previously

removed replaced between the terminals of the equivalent circuit


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The Norton and Thévenin equivalent circuits can also be found from each other by

using the source transformation discussed earlier and reproduced:

Example 1Find the Norton equivalent circuit for the network external to the 9Ω resistor


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Solution:Steps 1 and 2:

Step 3:

Step 4: As shown below, the Nortoncurrent is the same as the current through the 4Ω resistor. Applying the current divider

rule,


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Step 5:

Example2:- Find the Norton equivalent circuit for the portion of the network to the left of a-b.

Solution:Steps 1 and 2:

Step 3:


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Step 4: (Using superposition) For the 7-V battery :

For the 8-A source we find that both R1 and

R2 have been“short circuited” by the direct

connection between a and b, and

The result is


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Step 5:

4 MAXIMUM POWER TRANSFER THEOREM

The maximum power transfer theorem states the following:

A load will receive maximum power from a linear bilateral dc network

when its total resistive value is exactly equal to the Thévenin resistance of the

network as “seen” by the load.

For the network below, maximum power will be delivered to the load when


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From past discussions, we realize that a Thévenin equivalent circuit can be found

across any element or group of elements in a linear bilateral dc network. Therefore,

if we consider the case of the Thévenin equivalent circuit with respect to the

maximum power transfer theorem, we are, in essence, considering the total effects

of any network across a resistor RL. For the Norton equivalent circuit of the

following network, maximum power will be delivered to the load when

This result will be used to its fullest advantage in the analysis of transistor networks, where

the most frequently applied transistor circuit model employs a current source rather than a voltage

source.

For Thévenin circuit we have:

Let us now consider an example where ETh = 60 V and RTh = 9Ω, as shown

below:


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The power to the load is determined by

Note, in particular, that PL is, in fact, a maximum when RL = RTh = 9Ω .

The power curve increases more rapidly toward its maximum value than it

decreases after the maximum point, clearly revealing that a small change in load

resistance for levels of RL

below RTh

will have a more dramatic effect on the power delivered than similar changes in RL above the RTh level.

If we plot VL and IL versus the same resistance scale, we find that both

change nonlinearly, with the terminal voltage increasing with an increase in load

resistance as the current decreases. Note again that the most dramatic changes in

VL and IL occur for levels of RL less than RTh. As pointed out on the plot, when

RL = RTh, VL = ETh/2 and

IL = I max/2, with I max = ETh/ RTh.


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The dc operating efficiency of a system is defined by the ratio of the power

delivered to the load to the power supplied by the source; that is,

Also

The power delivered to RL under maximum power conditions ( RL = RTh) is


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Example3 A dc generator, battery, and laboratory supply are connected to aresistive load RL

a. For each, determine the value of RL for maximum power transfer to RL.

b. Determine RL for 75% efficiency.

Solutions:

a. For the dc generator, RL = RTh = Rint= 2.5 Ω,

For the battery, RL = RTh = Rint = 0.5 Ω

For the laboratory supply, RL = RTh = Rint = 40 Ω

b. For the dc generator,


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Example 4:- For the network below, determine the value of R formaximum power to R, and calculate the power delivered under theseconditions.

Solution:


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5 MILLMAN’S THEOREM

Through the application of Millman’s theorem, any number of parallel

voltage sources can be reduced to one. The three voltage sources in the following

figure can be reduced to one. This would permit finding the current through or

voltage across RL without having to apply a method such as mesh analysis, nodal

analysis, superposition, and so on. The theorem can best be described by applying

it to the network. Basically, three steps are included in its application.

Step 1: Convert all voltage sources to current sources This is performed below:


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Step 2: Combine parallel current sources, where

I T = I 1 + I 2 + I 3 and GT = G1 + G2 + G3

Step 3: Convert the resulting current source to a voltage source, and the desiredsingle-source network is obtained:


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In general, Millman’s theorem states that for any number of parallel voltage

sources,

The plus-and-minus signs include those cases where the sources may not be

supplying energy in the same direction.

The equivalent resistance is


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Example 5 :- Using Millman’s theorem, find the current through and voltage

across the resistor RL

Solution:

The minus sign is used for E 2 / R2 because that supply has the opposite polarity of

the other two. The chosen reference direction is therefore that of E 1 and E 3. The

total conductance is unaffected by the direction, and


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6. SUBSTITUTION THEOREM

The substitution theorem states the following:

If the voltage across and the current through any branch of a dc bilateral

network are known, this branch can be replaced by any combination of elements

that will maintain the same voltage across and current through the chosen

branch.

More simply, the theorem states that for branch equivalence, the terminal voltage

and current must be the same. Consider the circuit:

In which the voltage across and current through the branch a b are

determined. Through the use of the substitution theorem, a number of equivalent a-

b branches are shown :

Note that for each equivalent, the terminal voltage and current are the same.

Also consider that the response of the remainder of the circuit is unchanged by


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You will also recall from the discussion of bridge networks that V= 0 and I = 0

were replaced by a short circuit and an open circuit, respectively. This substitution

is a very specific application of the substitution theorem.

7. RECIPROCITY THEOREM:

The reciprocity theorem is applicable only to single-source networks. It is,

therefore, not a theorem employed in the analysis of multisource networks

described thus far. The theorem states the following:

The current I in any branch of a network, due to a single voltage source E

anywhere else in the network, will equal the current through the branch in

which the source was originally located if the source is placed in the branch in

which the current I was originally measured.

In other words, the location of the voltage source and the resulting current

may be interchanged without a change in current. The theorem requires that the polarity of the voltage source have the same correspondence with the direction of

the branch current in each position.


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In the representative network below, the current I due to the voltage source E

was determined. If the position of each is interchanged as shown in (b), the current

I will be the same value as indicated.

To demonstrate the validity of this statement and the theorem, consider the

network, in which values for the elements of (a) above have been assigned.

The total resistance is


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For the following network that corresponds to the above:

We have:

The uniqueness and power of such a theorem can best be demonstrated byconsidering a complex, single-source network such as the one shown:


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~ 1 ~

1. The electric field:

Electric field exists in the region around any charged body. It is

represented by electr ic flux li nes, which are drawn to indicate the strength of the

electric field at any point around the charged body; that is, the denser the lines of flux,the stronger the electric field.

The flux density is represented by:

The larger the charge Q in coulombs, the greater the number of flux lines since:

The electri c field strength at a point is the force acting on a unit positive charge at that point; that is,

The force exerted on a unit positive charge , by a charge , r meters away asdetermined by Coulomb's Law:


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Then:

Electri c fl ux l ines always extend from a positively charged body to a negatively

charged body, always extend or terminate perpendicular to the charged sur faces, and

never intersect.

2. Capacitors

Simply two parallel conducting plates separated by an insulating material (like,air), is called a capacitor.

Capacitance is a measure of a capacitor’s ability to store charge on its plates —

in other words, its storage capacity.

A capacitor has a capacitance of 1 farad i f 1 coulomb of charge is deposited on the

plates by a potential difference of 1 volt across the plates.


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The farad is named after Michael Faraday a nineteenth century English chemist and

physicist. The farad, however, is generally too large a measure of capacitance for most

practical applications, so the microfarad () or picofarad () is morecommonly used.

If a potential difference of V volts is applied across the two plates separated by adistance of d , the electric field strength between the plates is

The electrons within the insulator are unable to leave the parent atom and travel to the

positive plate. The positive components (protons) and negative components (electrons)

of each atom do shift , however [as shown in Fig. above], to form dipoles . When the

dipoles align themselves as shown in figure below, the material is polarized.


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The purpose of the dielectric (the insulating material), therefore, is to create an

electric field to oppose the electric field set up by free charges on the parallel plates. For

this reason, the insulating material is referred to as a dielectric, di for “opposing” and

electric for “electric field.”

The ratio of the flux density to the electric field intensity in the dielectric is called

the permittivity of the dielectric:

It is a measure of how easily the dielectric will “permit” the establishment of flux lines

within the dielectric.

For a vacuum, the value of e (denoted by ) is F/m. The ratio ofthe permittivity of any dielectric to that of a vacuum is called the relative permittivity;

It simply compares the permittivity of the dielectric to that of air. In equation form,

Also prove that:


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~ 5 ~

Example1: For the following capacitor:a. Determine the capacitance.

b. Determine the electric field strength between the plates if 450 V are

applied across the plates.

c. Find the resulting charge on each plate.

Solution:


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~ 6 ~

Example2: A sheet of mica 1.5 mm thick having the same area as the plates is inserted

between the plates of example1

a. Find the electric field strength between the plates.

b. Find the charge on each plate.

c. Find the capacitance.

Solution:


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~ 7 ~

3. TRANSIENTS IN CAPACITIVE NETWORKS: a. CHARGING PHASE

Consider the following network:

When the switch is thrown into position 1 (t=0) the capacitor behaves as a short circuit

and the current:

()

As C is charged the flow of charge will stop, the current i will be zero, and the voltage

will cease to change in magnitude:

For all future analysis:

A capacitor can be replaced by an open-cir cuit equivalent once the charging phase in

a dc network has passed.


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~ 8 ~

The following figure shows the behavior of the circuit current;

The mathematical expression of the current will be:

The factor RC in is called the time constant of the system and has the units of time Its

symbol is the Greek letter (tau).

Below is a time chart that explains the change of with during the charging phase.


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~ 9 ~

Concerning the charging voltage across the capacitor we have:

Then the mathematical expression of Vc is:

After 5 time constants the voltage across the capacitor will be .


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Example3

: a. Find the mathematical expressions for the transient behavior of v i , andv

R

for the circuit of the following network when the switch is moved to position 1. Plot the

curves of v

i

and vR.

b. How much time must pass before it can be assumed, for all practical purposes, that

Solution:


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~ 11 ~


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b. Discharge phase:

For the network below if the capacitor is fully charged and the switch is thrown

at position 2 the capacitor will begin to discharge at a rate sensitive to the same time

constant .

The equation for the decaying voltage across the capacitor would be the

following:

In essence, the capacitor functions like a battery with a decreasing terminal

voltage. Note in particular that the current iC has reversed direction, changing the

polarity of the voltage across R.

The resulting curve will have the same shape as the curve for and in thelast section. During the discharge phase, the current

will also decrease with time, as

defined by the following equation:


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~ 13 ~

The complete discharge will occur, for all practical purposes, in five time

constants. If the switch is moved between terminals 1 and 2 every five time constants,

the wave shapes of figure below will result for vC, iC, and vR .

Since the polarity of vC is the same for both the charging and the discharging

phases, the entire curve lies above the axis. The current iC reverses direction during the

charging and discharging phases, producing a negative pulse for both the current and the

voltage vR . Note that the voltage vC never changes magnitude instantaneously but that

the current iC has the ability to change instantaneously, as demonstrated by its vertical

rises and drops to maximum values.


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~ 14 ~

Example4: Find the mathematical expressions for the transient behavior of

v C , i C , and v R for the following circuit after the closing of the switch. Plot the

curves for v C , i C , and v R .

Solution:


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~ 15 ~

If the charging phase is disrupted before reaching the supply voltage, the

capacitive voltage will be less, and the equation for the discharging voltage vC will take

on the form

Where V i is the starting or initial voltage for the discharge phase. The equation

for the decaying current is also modified by simply substituting V i for E; that is,

Example5:a. Find the mathematical expression for the transient behavior of the voltage

across the capacitor of the network below if the switch is thrown into

position 1 at t = 0 s .

b. Repeat part (a) for iC.

c. Find the mathematical expressions for the response of v C and i C if theswitch is thrown into position 2 at 30 ms.

d. Find the mathematical expressions for the voltage v C and current i C if the

switch is thrown into position 3 at t = 48 ms.

e. Plot the waveforms obtained in parts (a) through (d) on the same time axis

for the voltage v C and the current i C using the defined polarity and current

direction.

Solution:


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~ 16 ~


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Example6:a. Find the mathematical expression for the transient behavior of the voltage across the

capacitor of the following network if the switch is thrown into position 1 at t = 0 s. b. Repeat part (a) for iC.c. Find the mathematical expression for the response of vC and iC if the switch is thrown

into position 2 at t = 1t of the charging phase.

d. Plot the waveforms obtained in parts (a) through (c) on the same time axis for the

voltage vC and the current iC using the defined polarity and current direction.


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~ 18 ~

Solution:

a. Charging phase: Converting the current source to a voltage source will result in


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~ 19 ~


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2. Initial values:

If the capacitor has an initial value of charge (i.e. at t=0 Q≠0) and has a initial voltage

before the charging face Vi:

3. Instantaneous values:

To determine the voltage or current at a particular instant of time that is not an

integral multiple of , as in the previous sections. There are also occasions when thetime to reach a particular voltage or current is required. There are two forms that require

some development. First, consider the following sequence:


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~ 21 ~


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~ 22 ~

4. THÉVENIN EQUIVALENT: :Occasions will arise in which the network does not have the simple series form. It

will then be necessary first to find the Thévenin equivalent circuit for the network

external to the capacitive element . E Th will then be the source voltage E and RTh will be

the resistance R. The time constant is then t = RThC.

Example7: for the following network

a. Find the mathematical expression for the transient behavior of the voltage

vC and the current iC following the closing of the switch (position 1 at t = 0 s).

b. Find the mathematical expression for the voltage vC and current iC as a

function of time if the switch is thrown into position 2 at t = 9 ms.

c. Draw the resultant waveforms of parts (a) and (b) on the same time axis.


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~ 23 ~

Solution:

The resultant Thévenin equivalent circuit with the capacitor replaced is shown

below:


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~ 24 ~

with Vf = E th and V i = 0 V, we find that:


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~ 25 ~

c. The figures are:


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~ 26 ~

Example8: the following capacitor is initially charged to 40 V . Find the mathematical

expression for vC after the closing of the switch.

Solution: The network is redrawn in:


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~ 27 ~

1. THE CURRENT iC The current i C associated with a capacitance C is related to the voltage across thecapacitor by

Where is a measure of the change in in a vanishingly small period of time.The function is called the derivative of the voltage with respect to time t.If the voltage fails to change at a particular instant, then

The average current is defined by the equation

Where ∆ indicates a finite (measurable) change in charge, voltage, or time.

Note:If the voltage increases with time, the average current is the change involtage divided by the change in time, with a positive sign. If the voltagedecreases with time, the average current is again the change in voltage

divided by the change in time, but with a negative sign.


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~ 28 ~

Example1: Find the waveform for the average current if the voltage across a

capacitor is as shown below

Solution:

a. From 0 ms to 2 ms, the voltage increases linearly from 0 V to 4 V , the change in

voltage (with a positive sign since the voltage increases withtime). The change in time , and


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~ 29 ~

d. From 11 ms on, the voltage remains constant at 0 and , so . Thewaveform for the average current for the impressed voltage is as shown

Notes In general, the steeper the slope, the greater the current, and when the

voltage fails to change, the current is zero. It is not the magnitude of the voltage across a capacitor that determines

the current but rather how quickly the voltage changes across the

capacitor. An applied steady dc voltage of 10,000 V would (ideally) not

create any flow of charge (current), but a change in voltage of 1 V in avery brief period of time could create a significant current.

The method described above is only for waveforms with straight-line

(linear) segments. For nonlinear (curved) waveforms, a method of

calculus (differentiation) must be employed.

Homework:

1.

Find the waveform for the average current if the voltage across a capacitor is as shown


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~ 30 ~

2. CAPACITORS IN SERIES AND PARALLEL Increasing levels of capacitance can be obtained by placing capacitors in parallel,

while decreasing levels can be obtained by placing capacitors in series.

For capacitors in series, the charge is the same on each capacitor:

Since Q1=Q2=Q3=QT then:


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~ 31 ~

The voltage across each capacitor of the previous figure is

A similar equation will result for each capacitor of the network.

For capacitors in parallel, as shown above, the voltage is the same across each

capacitor, and the total charge is the sum of that on each capacitor:


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~ 32 ~

Example2

: Find the voltage across and charge on each capacitor for the network

Solution:


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~ 33 ~

Example3: Find the voltage across and charge on each capacitor of thenetwork after each has charged up to its final value.

Solution:


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~ 34 ~


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1

1. MAGNETIC FIELDS

In the region surrounding a permanent magnet there exists a magnetic field, which can be

represented by magnetic flux lines similar to electric flux lines. Magnetic flux lines, however, do

not have origins or terminating points as do electric flux lines but exist in continuous loops, as

shown

The symbol for magnetic flux is the Greek letter (phi). The magnetic flux lines radiatefrom the north pole to the south pole, returning to the north pole through the metallic bar.

If a nonmagnetic material, such as glass or copper, is placed in the flux paths surrounding

a permanent magnet, there will be an almost unnoticeable change in the flux distribution.

However, if a magnetic material, such as soft iron, is placed in the flux path, the flux lines will

pass through the soft iron rather than the surrounding air because flux lines pass with greater ease

through magnetic materials than through air.

This principle is put to use in the shielding of sensitive

electrical elements and instruments that can be affected by

stray magnetic fields.


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A magnetic field is present around every wire that carries an electric current. The direction of the

magnetic flux lines can be found simply by placing the thumb of the right hand in the direction of

conventional current flow and noting the direction of the

fingers. (This method is commonly called the right-hand rule.)

If the conductor is wound in a single-turn coil, the resulting flux will flow in

a common direction through the center of the coil.

A coil of more than one turn would produce a magnetic

field that would exist in a continuous path through and

around the coil.

The flux distribution of the coil is quite similar to that of the permanent magnet. The flux

lines leaving the coil from the left and entering to the right simulate a north and a south pole,

respectively. The principal difference between the two flux distributions is that the flux lines are

more concentrated for the permanent magnet than for the coil. Also, since the strength of a

magnetic field is determined by the density of the flux lines, the coil has weaker field strength.

The field strength of the coil can be effectively increased by placing certain materials,

such as iron, steel, or cobalt, within the coil to increase the flux density within the coil. By

increasing the field strength with the addition of the core, we have devised an electromagnet.

Applications for electromagnetic effects are shown


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2. FLUX DENSITY

The number of flux lines per unit area is calledthe flux density

An instrument designed to measure flux density in gauss (CGS system) 1 T = 104 gauss.

3. PERMEABILITY

If cores of different materials with the same physical

dimensions are used in the electromagnet, the strength of the magnet will vary in accordance with

the core used. Materials in which flux lines can readily be set up are said to be magnetic and to

have high permeability. The permeability (µ) of a material, therefore, is a measure of the ease

with which magnetic flux lines can be established in the material.* The permeability of free space

µo

(vacuum) is


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* the permeability of all nonmagnetic materials, such as copper, aluminum, wood, glass, and air,

is the same as that for free space.

* Materials that have permeabilities slightly less than that of free space are said to bediamagnetic

,

and those with permeabilities slightly greater than that of free space are said to beparamagnetic

.

* Magnetic materials, such as iron, nickel, steel, cobalt, and alloys of these metals, have

permeabilities hundreds and even thousands of times that of free space. Materials with these very

high permeabilities are referred to asferromagnetic

.

The ratio of the permeability of a material to that of free space is called its relative permeability;

that is,

In general, for ferromagnetic materials,µ r

≥ 100, and for nonmagnetic materials,µ r

= 1


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5

Inductors

1. F R D Y’S L W OF ELECTROMAGNETIC INDUCTION

If a conductor is moved through a magnetic field so that it cuts magnetic lines of flux, a

voltage will be induced across the conductor, as shown

The greater the number of flux lines cut per unit time (by increasing the speed

with which the conductor passes through the field), or the stronger the magnetic fieldstrength (for the same traversing speed) the greater will be the induced voltage across the

conductor.

If the conductor is held fixed and the magnetic field is moved so that its flux lines cut the

conductor, the same effect will be produced.

If a coil of N turns is placed in the region of a changing flux, as illustrated, a voltage will

be induced across the coil as determined by

Faraday’s law:


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2.

LENZ’S L W

It was shown that the magnetic flux linking a coil of N turns with a current I has the

distribution of:

A changing flux linking a coil induces a voltage across the coil. For this coil,

therefore, an induced voltage is developed across the coil due to the change in current

through the coil. The polarity of this induced voltage tends to establish a current in the coil

that produces a flux that will oppose any change in the original flux. In other words, the

induced effect (e ind ) is a result of the increasing current through the coil. However, the

resulting induced voltage will tend to establish a current that will oppose the increasing

change in current through the coil.


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The instant the current begins to increase in magnitude, there will be an opposing

effect trying to limit the change. It is “choking” the change in current through the coil.

Hence, the termchoke

is often applied to the inductor or coil.

This effect is an example of

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