EE1151-CIRCUITS THEORY

UNIT I : CIRCUIT ANALYSIS TECHNIQUES

Fundamentals

Ohm's Law states the voltage across a resistor, R (or impedance, Z) is directly proportional to the current passing through it (the resistance/impedance is the proportionality constant)

Kirchhoff's Voltage Law (KVL): the algebraic sum of the voltages around any loop of N elements is zero (like pressure drops through a closed pipe loop)

Kirchhoff's Current Law (KCL): the algebraic sum of the currents entering any node is zero, i.e., sum of currents entering equals sum of currents leaving (like mass flow at a junction in a pipe)

Kirchhoff's Current Law (KCL)

KCL states that the algebraic sum of the currents in all the branches which converge in a common node is equal to zero

Iin = Iout

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law states that the algebraic sum of the voltages between successive nodes in a closed path in the network is equal to zero.

E = IR

Solution using Kirchhoff’s Voltage and current laws

Steps to solve circuit by Kirchhoff’s Laws.

1. Construct circuit with circuit magic schematics editor.

Circuit sample from circuit magic.

2. Construct loops. (See “creating loop” section in user guide) Number of loops (and number of Kirhhoff’s Voltage laws equations) can be determined using following formula. Loop can not include branches with current sources. Due current sources resistance equal infinity.

Loop Number = Branch Number –(Nodes Number –1) – Current sources Number

3. Select Analyze->Solve by Kirhhoff’s laws menu item

4. In dialog box press OK button. if no warning shown. 5. Read solution.

Writing Kirchhoff current law for 3-1 nodes

(Note number of Kirchhoff current laws equations equal Nodes Number –1)

(Node 1)J1+I3+I4+I7=0

(Node 2)-J1+I2-I4=0

Wrining Kirchoff voltage law for 5-1-(3-1) loops

(Loop1) I3·R3-I7·R5=-E2

(Loop2) I2·R2-I7·R5+I4·R4=E1-E2

Linear equations

I3+I4+I7=-2

I2-I4=2

10I3-10I7=-10

11I2+10I4-10I7=-7

Equations solution

I1=2

I2=0,692

I3=-0,846

I4=-1,308

I7=0,154

Ohm’s law

Ohm's law is the main basic electrical law and defines the resistance of a device to the flow of electrons.

There are three different notations of Ohm’s law

1. Unknown current

2. Unknown voltage

3. Unknown resistance

Resistors in Series & Resistors in Parallel:

A series circuit is one with all the loads in a row. Like links in a chain. There is only one path for the electricity to flow.

A parallel circuit is one that has two or more paths for the electricity to flow. In other words, the loads are parallel to each other.

Star-Delta Transformation:

 

Delta - Star Transformation:

Nodal AnalysisNodal analysis is generally best in the case of several voltage sources. In nodal analysis, the variables (unknowns) are the "node voltages."

Nodal Analysis Procedure:

1. Label the N node voltages. The node voltages are defined positive with respect to a common point (i.e., the reference node) in the circuit generally designated as the ground (V = 0).

2. Apply KCL at each node in terms of node voltages. a. Use KCL to write a current balance at N-1 of the N nodes of the circuit

using assumed current directions, as necessary. This will create N-1 linearly independent equations.

b. Take advantage of supernodes, which create constraint equations. For circuits containing independent voltage sources, a supernode is generally used when two nodes of interest are separated by a voltage source instead of a resistor or current source. Since the current (i) is unknown through the voltage source, this extra constraint equation is needed.

c. Compute the currents based on voltage differences between nodes. Each resistive element in the circuit is connected between two nodes; the current in this branch is obtained via Ohms law where Vm is the positive side and current flows from node m to n (that is, I is m --> n).

3. Determine the unknown node voltages; that is, solve the N-1 simultaneous equations for the unknowns, for example using Gaussian elimination or matrix solution methods.

Nodal Analysis Example 1. Label the nodal voltages. 2. Apply KCL.

a. KCL at top node gives IS = IL + IC

b. Supernode constraint eq. of VL = VS

c.

3. Solve for VT for instance.

Loop or Mesh AnalysisMesh (loop) analysis is generally best in the case of several current sources. In loop analysis, the unknowns are the loop currents. Mesh analysis means that we choose loops that have no loops inside them.

Loop Analysis Procedure:

1. Label each of the loop/mesh currents. 2. Apply KVL to loops/meshes to form equations with current variables.

a. For N independent loops, we may write N total equations using KVL around each loop. Loop currents are those currents flowing in a loop; they are used to define branch currents.

b. Current sources provide constraint equations. 3. Solve the equations to determine the user defined loop currents.

Mesh Analysis Example:

1. Label mesh currents. 2. Apply KVL.

a. Left loop KVL: VS = R1I1 + R2(I1-I2)

b. Constraint equation I2 = -IS.

3. Solve for I1 and I2. Note: Branch current from mesh currents: IM = I1 - I2

SuperpositionIn any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as the algebraic sum of the individual contributions of each source acting alone.

Procedure:

1. For each independent voltage and current source (repeat the following): a. Replace the other independent voltage sources with a short circuit (i.e., v

= 0). b. Replace the other independent current sources with an open circuit (i.e., i

= 0).

Note: Dependent sources are not changed!

c. Calculate the contribution of this particular voltage or current source to the desired output parameter.

2. Algebraically sum the individual contributions (current and/or voltage) from each independent source.

Source TransformationAn ac voltage source V in series with an impedance Z can be replaced with an ac current source of value I=V/Z in parallel with the impedance Z.An ac current source I in parallel with an impedance Z can be replaced with an ac voltage source of value V=IZ in series with the impedance Z.

Likewise, a dc voltage source V in series with a resistor R can be replaced with a dc current source of value i = v/R in parallel with the resistor R; and vice versa.

Thévenin's and Norton's TheoremsThévenin's Theorem states that we can replace entire network, exclusive of the load, by an equivalent circuit that contains only an independent voltage source in series with an impedance (resistance) such that the current-voltage relationship at the load is unchanged.

Norton's Thereom is identical to Thévenin's Theorem except that the equivalent circuit is an independent current source in parallel with an impedance (resistor). Hence, the Norton equivalent circuit is a source transformation of the Thévenin equivalent circuit.

Thévenin Equivalent Circuit Norton Equivalent Circuit

Procedure:

1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable).

2. Thevenin: Compute the open circuit voltage, VOC.Norton: Compute the short circuit current, ISC.

3. Compute the Thevenin equivalent resistance, RTh (or impedance, ZTh). a. If there are only independent sources, then short circuit all the voltage

sources and open circuit the current sources (just like superposition). b. If there are only dependent sources, then must use a test voltage or current

source in order to calculate RTh= vTest/iTest (or ZTh=VTest/ITest). c. If there are both independent and dependent sources, then compute RTh (or

ZTh) from RTh= vOC/iSC (or ZTh=VOC/ISC). 4. Replace circuit with Thevenin/Norton equivalent.

Thevenin: VOC in series with RTh (or ZTh).Norton: ISC in parallel with RTh (or ZTh).

Note: for 3(b) the equivalent network is merely RTh (or ZTh), that is, no current or voltage sources.MORE EXAMPLES:

Kirchhoff's Voltage LawKirchhoff's Voltage Law states:

The sum of the voltages around a closed circuit path must be zero.

Notice that a closed circuit path insists that if one circuit element is chosen as a starting point, then one must be able to traverse the circuit elements in that loop and return to the element in the beginning.

Mathematically, The Kirchhoff's Voltage Law is given by

∑ vn = 0n

For reference, this law is sometimes called Kirchhoff's Second Law, Kirchhoff's Loop Rule, and Kirchhoff's Second Rule.

Part 2:Kirchhoff's Voltage Law (Cont...)

Figure 5.1: v4 + v1 + v2 + v3 = 0

We observe five voltages in Figure 5.1:v4 across a voltage source, and the four voltages v1, v2, v3 and v5 across the resistors R1, R2, R3 and R5, respectively.The voltage source and resistors R1, R2 and R3 comprise a closed circuit path, thus the sum of the voltages v4, v1, v2 and v3 must

be zero:

∑ vn = v4 + v1 + v2 + v3 = 0n

The resistor R5 is outside the closed path in question, and thus plays no role in the calculation of Kirchhoff's Voltage Law for this path. (Note that alternate closed paths can be defined which include the resistor R5. In these cases, the voltage v5 across R5 must be considered in calculating Kirchhoff's Voltage Law.)

Part 3Now, if we take the point d in the image as our reference point and arbitrarily set its voltage to zero, we can observe how the voltage changes as we traverse the circuit clockwise.

Going from point d to point a across the voltage source, we experience a voltage increase of v4 volts (as the symbol for the voltage source in the image indicates that point a is at a positive voltage with respect to point d).

On traveling from point a to point b, we cross a resistor. We see clearly from the diagram that, since there is only a single voltage source, current must flow from it's positive terminal to its negative terminal--clockwise around the circuit path. Thus from Ohm's Law, we observe that the voltage drops from

Part 4 : Example

Figure 5.2: Example 1

Consider Figure 5.2 with the following Parameters:

V1 = 15V V2 = 7V R1 = 20Ω R2 = 5Ω R3 = 10Ω

Find current through R3 using Kirchhoff's Voltage Law.

point a to point b across resistor R1.

Likewise, the voltage drops across resistors R2 and R3. Having crossed R2 and R3, we arrive back at point d, where our voltage is zero (just as we defined). So we experienced one increase in voltage and three decreases in voltages as we traversed the circuit.

The implication from Kirchhoff's Voltage Law is that, in a simple circuit with only one voltage source and any number of resistors, the voltage drop across the resistors is equal to the voltage applied by the voltage source:

v4 = v1 + v2 + v3

Kirchhoff's Voltage Law can easily be extended to circuitry that contains capacitors.

Solution:

Figure 5.3: Example 1 loops

We can see that there are two closed paths (loops) where we can apply KVL in, Loop 1 and 2 as shown in figure 5.3

From Loop 1 we get:

V1 − VR3 − VR1 = 0

From Loop 2 we get:

V2 − VR3 − VR2 = 0

A bit confused? well look at the explanation in Part 3 of this lesson and Review Passive sign convention.

Part 5 : Example (Continued)The above results can further be simplified as follows:

V1 − (I1 − I2) * R3 − I1 * R1 = 0

Part 6 : Example (Continued)It is clear that: from (3)

..... (1)

and

V2 + (I1 − I2) * R3 − I2 * R2 = 0

..... (2)

By equating above (1) and (2) we can eliminate I2 and hence get the following:

..... (3)

We end up with the above three equations and now substitute the Values given in the above equations and solve the variables.

If you feel lost up to this point do go back to the beginning of the example. Think of this as just another mathematical problem requiring solving by use of simultaneous equations with two unknowns!

Notice that we work with Variables only and try to solve the equation to its simplest form. Only after we have arrived at a simplified equation then that we can substitute in values of Resistors, Voltages and current. This can save you a lot of trouble because, if you go wrong you can easily trace your work to the problem.

.

Substitute the Above Result into (2)

.

The Positive sign for I2 only tells us that Current I2 flows in the same direction to our initial assumed direction. Thus now we can calculate Current through R3 as follows:

.

The Negative sign for IR3 only tells us that Current IR3 flows in the same direction to I2 direction.

Figure 5.4: Exercise 5

Consider Figure 5.4 with the following Parameters:

V1 = 15V V2 = 7V R1 = 20Ω R2 = 5Ω R3 = 10Ω

Find current through R3 using Kirchhoff's Voltage Law.

Part 1: Kirchhoff's Current LawKirchhoff's Current Law states:

The sum of the currents entering a particular point must be zero.

We can now define the electrical point physically connecting two or more electric circuit components, as a NODE. Note that a positive current leaving a point is considered to be a negative current entering that point.

Mathematically, Kirchhoff's Current

Part 2:Kirchhoff's Current Law (Cont...)

Figure 6.1: i1 + i2 + i3 + i4 = 0

Law is given by

∑ in = 0n

For reference, this law is sometimes called Kirchhoff's first law, Kirchhoff's point rule, Kirchhoff's junction rule, and Kirchhoff's first rule.

We observe four currents "entering" the junction depicted as the bold black dot in Figure 6.1. Of course, two currents are actually exiting the junction , but for the purposes of circuit analysis it is generally less restrictive to consider what are in actuality positive currents flowing out of a junction to be negative currents flowing into that junction (mathematically the same thing). Doing so allows us to write Kirchhoff's law for this example as:

∑ in = i1 + i2 + i3 + i4 = 0n

Part 3It may not be clear at this point why we insist on thinking of negative currents flowing into a junction instead of positive currents flowing out. But note that Figure 6.1 provides us with more information that we generally can expect to get when analysing circuits, namely the helpful arrows indicating the direction of current flow. If we don't have such assistance, we generally should not pass judgment on the direction of current flow (i.e., placing a negative sign before our current variable) until we calculate it, lest we confuse ourselves and make mistakes.

Nevertheless in this case we have the extra information of directional arrows in Figure 6.1, so we should take advantage of it. We know that currents i2 and i3 flow into the junction and the currents i1 and i4 flow out. Thus we can write

Part 4 : Example

Figure 6.2: Example 1

Consider Figure 6.2 with the following Parameters:

V1 = 15V V2 = 7V R1 = 20Ω R2 = 5Ω R3 = 10Ω

Find current through R3 using Kirchhoff's Current Law.

i1 + i4 = i2 + i3

Kirchhoff's Current Law as written is only applicable to steady-state current flow (i.e., no alternating current, no signal transmission). It can be extended to include time-dependent current flow, but that is beyond the scope of this section.

Kirchhoff's Current Law is used in a method of circuit analysis referred to as nodal analysis to be discussed in lecture 8. A node is a section of a circuit where there is no change in voltage (where there are no components, wire is often assumed to be perfectly conductive).

Each node is used to form an equation, and the equations are then solved simultaneously, giving the voltages at each node.

Solution:

Figure 6.3: Voltages at nodes

This is the same example we solved in Exercise 5.

Figure 6.3 shows Voltages at Nodes a, b, c and d.

We use node a as common node ( ground if you like ). thus Va = 0V.

Part 5 : Example (Continued)From Node b we get:

Vb = − V1 = − 15V

From Node d we get:

Vd = V2 = 7V

It is clear that we must solve V_c, in order to complete Voltage definitions at all nodes. V_c will be found by applying KCL at Node c

Part 6 : Example (Continued)Substitute values into previous equations you get:

Vc(0.35) = 0.65 thus Vc = 1.857V

Thus now we can calculate Current through R3 as follows:

and solving resulting equations Follows:

i1 = i2 + i3

and

We can group like terms to get the following equation:

.

Just as we expected! Note that current here is simplified because of following Voltage definitions and current paths in Figure 6.3.

This method becomes tedious as the complexity of circuits is increased.

Part 7:Further Reading Links:

Kirchhoff's circuit laws

Refferences:

Nilsson, James W. and Riedel, Susan A. Electric Circuits (5th ed.). Addison-Wesley. (1996).ISBN 020155707X

Exercise 6

Figure 6.4: Exercise 6

Consider Figure 6.4 with the following Parameters:

V1 = 15V V2 = 7V R1 = 20Ω R2 = 5Ω R3 = 10Ω

Find current through R3 using Kirchhoff's Current Law.

Part 1

Resistors in Series

Series of resistors means resistors connected end to end in a line.

This means that the resistance for the circuit is different than any one resistor. Take two resistors in series in a circuit with a voltage supply.

To find the overall resistance of the circuit, add the resistances of the resistors.

Part 2Equation 3.1

So what if there were 10 resistors in series? Just add up all of the resistances and you have the equivalent over all resistance. In general this can be expressed:

Equation 3.2

Where R equivalent is the sum of all N of the resistors in series. So it really doesn't matter how many

resistors there are. If they are in series they can be added up into an equivalent resistance.

Part 3

Voltage Divider

There comes a time when the boss or the project demands that you know what the voltage is between these millions of resistors in series. No need to panic though because it isn't too much harder.

Lets take the two resistor problem first. There is a voltage source with two resistors in series. We know that the overall voltage drop ac The Circuit

When we apply an ac voltage to a series RL circuit as shown to the right, the circuit behaves in some ways the same as the series RC circuit, and in some ways as a sort of mirror image. For example, current is still the same everywhere in this series circuit. VR is still in phase with I, and VL is still 90° out of phase with I. However, this time VL leads I — it is at +90° instead of -

Part 4Equation 3.4

This is the drop over the second resistor. But if it is dropping to zero, ground, or the negative side of the source then adding it to zero would give us the same answer as above.

For more than two resistors in series it is just a matter of keeping track of which resistor is on which side and summing appropriately.

Equation 3.5

Where VN is the voltage drop over N resistors out of a total of M resistors. Remember that the resistors where the voltage drop is being calculated should be continuous. If they aren't all that can be said about the answer derived from the equation is that it is part of the whole voltage drop and somewhat worthless otherwise.

90°.

For this circuit, we will assign experimental values as follows: R = 560Ω, L = 100 mH, and VAC = 10 Vrms. We build the circuit, and measure 7.464 V across L, and 6.656 V across R. As we might have expected, this exceeds the source voltage by a substantial amount, and the phase shift is the reason for it.

The Vectors in an LC Series Circuit

The vectors for this example circuit are shown to the right. This time the composite phase angle is positive instead of negative, because VL leads IL. But to determine just what that phase angle is, we must start by determining XL and then calculating the rest of the circuit parameters.

XL =  2πfL      =  6.28 × 1000 × 0.1      =  628    

Z  =  560 + j628 Ω      =  (560² + 628²)½

If the resistors are in the middle of the series then it will be necessary to calculate the voltage drop on one of the sides to be able to calculate the voltage.

     =  (313600 + 394384)½

     =  (707984)½      =  841.4 Ω    

I  =  E/Z      =  10/841.4      =  0.0119 A      =  11.9 mA     VR =  I × R      =  0.0119 × 560      =  6.656 V     VL =  I × XL      =  0.0119 × 628      =  7.464 V    

θ  =  arctan(XL/R)      =  arctan(628/560)      =  arctan(1.121)      =  48.28°

This really completes the description of the series RL circuit with a fixed AC signal applied to it. Starting with the

component values and the frequency of the applied AC voltage, we have described every aspect of this circuit's behavior at that frequency

ross the two resistors is the same as the voltage the source is supplying in our example world. So the voltage drop across one resistor would be a portion of the overall drop. What proportion would we use to figure out the answer? One resistor over the two added together times the over all voltage drop:

Equation 3.3

Remember, this is the voltage drop across the first resistor. If you want the actual voltage there you still need to do some adding or subtracting to get it. Say that you have a 12V source and a drop over the first resistor of 3V. Then you actually need to subtract 3V from 12V to get the actual voltage between the resistors.

At this point it seems that everything isn't quite as simple as it started. With our example and equation for two resistors in series something else can happen. What if the second resistor was set in the first resistor's place in the equation? Well, simply we would get the other side of the proportion:

Part 5 Part 6

It becomes clear then that, two equal resistors will divide the source voltage into two equal voltages (half of the source's voltage is dropped across each resistor). If the ratio of the resistance values is 3 to 1, there will be 3/4 of the source voltage dropped across the

higher resistance, and of the source voltage dropped across the lower resistance.

Three equal resistances in a series circuit with a single voltage source would drop 1/3 of the source voltage across each resistor. If the three had 1-2-3 proportionality (100,200,300 ohms

for instance) they would drop , and

of the source voltage each. That is:

× VTotal, ×

VTotal,and × VTotal.

Current

Where does current come into any of this? Current, in this case, plays a similar role to that of the current in the Simple Resistive Circuits. Once the equivalent resistance of all the resistors in a series is found, effectively making a simple circuit again, then the current can be found with:

Equation 3.6

Just as a reminder. But the interesting thing is that the current through all resistors in series is the same. If the resistor is 30Ω it has the same current flow as the resistor with 500Ω, so long as it is in series. Thinking about everything above we are adding up all of the resistors to make a single equivalent resistor. So current isn't different from 30Ω to 500Ω because together the resistance is 530Ω. That resistance is used then to calculate the current.

Part 7

More Examples

Figure 3.1: Series resistors

Figure 3.1 shows a Series resistive circuit with the following parameters. Vs=100Volts ; R1=15; R2=30; Find V1 and V2.

Solution: from Equation 2.3 we see that.

.

Similarily:

.

Thus it can be said that The Supply Voltage has been divided between R1 and R2 by

and respectively.

Resistors in Parallel

Introduction

The best way to understand Parallel circuits is to start with the definition. A circuit is parallel to another circuit or several circuits if and only if it share common terminals. That is if both of the branches touch each other endpoints they are in parallel. Here is an example:

Part 2

Voltage Rule

If two or more branches are parallel then the voltage across them is equal. So based on this we can conclude that VR1=VR2=5volts. However unlike series resistors, the current across the branches is not necessarily equal.

Equivalent resistance

For series resistors to find the total resistance we simply add them together. For parallel resistors its a little more complicated. Instead we use the

Figure 4.1: A Parallel circuit

R1, R2, and the voltage source are all in parallel. To prove this fact consider the top and bottom parts of the circuit.

Figure 4.2: Components in parallel share a common

nodes

The areas in yellow all are connected together, as well as the areas in blue. So all the branches have the same terminals, which means that R1, R2, and the source are all in parallel.

If we take this discussion of the water flow analogy. Electric current can be seen as water and the conductors as water pipes. Something interesting happens as the current reaches the common node of Resistors that are connected in parallel, The total current is divided into the parallel branches.

following equation:

However for the case of only two resistors and only two resistors we can use this simplified form

Equation 4.2: Total Parallel Resistance

It is well to note at this point that The total Resistance of parallel connected Resistors will always be Less than the smallest of the individual Resistors.

Current Rule

In Series Connection we deduced that Voltage is divide amongst resistors. For Parallel connected Resistors, Current is divided. So here is a mathematical formula as we did with voltage division principle.

Equation 4.3: Current Divider Formula

Using this formula you can workout the currents flowing through individual Resistors.

Part 3

Application

We have spent three lectures hacking on about What & Why Resistors & resistive circuits in two connection schemes are used, ( i.e Series and parallel connections ). The question now is, where & how in Real life do these connections happen?

One simple application of these connection schemes is the Shunt application. In Electric Measurement industry, most often enough, we wish to measure Currents and Voltages of Very High Magnitudes ( e.g some ranges of 500kV and upwards or 1000kA and upwards ). The problem is that metering devices have delicate electronic components and usually have small Voltage and Ampere operating ratings.

Solution to the above is to have a metering device connected in parallel to a

Part 4

Figure 4.3: Application of Parallel Resistive circuits. Shunt

connection

If we know what the ampere rating of a device and what the total current is then we can work out the shunt current and thus the Shunt Resistor.

resistor, this resistor is thus called a "shunt" resistor since it is there to protect the metering device as shown in the next figure in part 4.

Part 5: Examples

Figure 4.4: Example 3.1

Figure 3.4 shows a Parallel resistive circuit with the following parameters.Vs = 10Volts ; R1 = 3Ω;R2 = 7Ω; Find Req; I1andI2.

Solution: from Equation 4.2 we see that.

Here are the solutions to the above

Part 6: Examples

Figure 4.5: Example 3.1

Figure 4.5 shows a Parallel resistive circuit with the following parameters.Is = 5Amps ; R1 = 2Ω; R2 = 3Ω; Find: I1;I2 and Vs.

Solution: from Equation 3.2 we see that.

Here are the solutions to the above problem:

problem:

.

. .

Thus it can be said that The Supply Current has been divided between R1 and R2 .

We know that when solving these problems, we look at the Data given and thus we can see how we need to manipulate our equations in order to achieve our objective.The Following Example Highlights this point, see that you can follow the Method used and the reasoning behind.

First Find: Req:

.

Then;

.

..

Part 7

Do you Remember?

Let's take some time to Reflect on Material covered thus far. We have learned a great deal about simple resistive circuits and the possible connections they afford us. Here I think you'll want to remember:

Voltage: (V or v - Volts)The electrical potential between two points in a circuit.

Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit.

Power: (W - Watts)Simply P = IV. It is the current times the voltage. Source: A voltage or current source is the supplier for the circuit. Resistor: (R measured in Ω - Ohms)A circuit element that "constricts"

current flow.

Total Series Resistance: ( )

Voltage Divider :

Current through Resistors connected in Series is the same for all resistors.

Two resistors connected in Parallel:

Current Divider Principle:

Part 8: Exercise 4

Here are some questions to test yourself with.

1. Given 2 Resistors: R1 = R2 = 5Ω in parallel find The Req. 2. Given 3 Resistors: R1 = 2Ω; R2 = 3Ω and R3 = 7Ω in parallel and Supply

Current is 15Amps. Find: Req ; I1; I2 & I3 and Supply Voltage across these Resistors.

3. Given 4 Resistors: R1 = 2Ω is connected in series to a parallel branch consisting of R2 = 3Ω ; R3 = 7Ω and R4 = 4Ω Find: Total Resistance as seen by the Voltage source.

4. Is it possible to connect Voltage sources in Parallel, If so what conditions must be met?

UNIT II : TRANSIENT RESONANCE IN RLC CIRCUIT

SERIES RC CIRCUIT:

When we applied a dc voltage to a resistor and capacitor in series, the capacitor charged to the applied voltage along an exponential curve, and then just sat there. This is not the case when an ac voltage is applied to this combination as shown in the schematic diagram to the right. Here, the input voltage is constantly changing, so the capacitor will constantly charge and discharge as it continually tries to oppose the changes.

Essentially, R and C in this circuit now form a voltage divider for ac. We can expect that part of the applied voltage will appear across R, and part will appear across C. But how much voltage will appear across each component?

As a practical example of such a circuit, assume VAC = 10 vrms at a frequency of 1 kHz. C = 0.01 µf and R = 15K. If we were to build this circuit with an accurate audio frequency generator, we would measure a voltage of 6.855 vrms across the resistor, and 7.275 vrms across the capacitor. This of course adds up to considerably more than the 10 vrms supplied by the generator. How is this possible?

The question is compounded by the fact that C will cause a phase shift between voltage and current. Since this is a series circuit, the current must necessarily be the same everywhere in the circuit. Therefore the voltage across the capacitor will lag that current by 90°, while at the same time the voltage across the resistor will be in phase with the current. How can we deal with such a mess?

Mapping Voltages with Vectors:The real problem here is that 90° phase shift between vC and vR. Since they are not in phase and some voltage is dropped across each component, the phase relationship between generator voltage and generator current must be somewhere between the two extremes. We need to be able to determine that relationship, as well as to determine the combined effect of R and XC in this circuit.

One solution is to map the component voltages graphically, as shown to the right. Using X-Y coordinate axes, the positive X axis is defined as the zero-degree reference, and counter-clockwise rotation is defined as the direction of increasing positive angles.

Since the circuit current is necessarily the same throughout the circuit, it is used as the reference phase angle. Resistive voltage, vR, is in phase with the current as shown in red. Capacitive voltage, vC, is at -90° as shown in blue. In this way, we can represent the voltages across R and C as vectors, having direction as well as magnitude. Now it is clear that the composite voltage comprising both vR and vC must be the vector sum of the two, as shown in violet.

To find the composite voltage, we must apply the formula for the diagonal of a rectangle, and find the square root of the sum of the squares:

vG = (vR² + vC²)½     = (6.855² + 7.273²)½

    = (47 + 53)½     = 100½     = 10

Since our generator produces an output of 10 vrms, this confirms our measurements.

Impedance:

Another point to consider here is that this circuit contains both resistance (R) and reactance (XC). This is neither a pure resistance or a pure reactance. We need a name for this combined characteristic, and a means of calculating it.

The name is easy enough; the combined characteristic is named impedance, and is represented by the letter Z. To calculate Z, we must first note that, in accordance with Ohm's Law, R = vR/I and XC = vC/I. But we already know that the voltages are 90° out of phase. Since the circuit current, i, is the same everywhere, then R and XC must be 90° out of phase as well. Thus, we must apply the same method for finding Z as we did for finding the composite voltage. The full set of calculations for this circuit, then, using the circuit values given above, is:

XC = 1/2πfC     = 1/6.28 × 1000 × 0.00000001     = 1/6.28 × 10-5     = 15.92K   

Z = (R² + XC²)½

    = (15² + 15.92²)½     = (225 + 253.45)½     = 478.45½     = 21.87K   

I = VG/Z     = 10V/21.87K     = 0.457 mA   VR = I × R     = 0.457 mA × 15K     = 6.855 V   VC = I × XC     = 0.457 mA × 15.92K     = 7.275 V

BASIC RLC CIRCUIT:

When we add a resistance to a series LC circuit, as shown in the schematic diagram to the right, the behavior of the circuit is similar to the behavior of the LC circuit with no resistance, but there are some variations. To see how the added resistance affects the operation of the circuit, we'll use the same parameters as with the Series LC Circuit, plus the resistor:

f = 1 MHz e = 10 vrms L = 150 µh C = 220 pf R = 100 Ω

This time, our measured voltages come out as follows:

vL = 39.1v vC = 30.0v vR = 4.15v

Does this make sense? We now know we must deal with the difference between vL and vC, which is just over 9 volts. But then we have a vR of over 4 volts. Did we somehow convert that 10 vrms input voltage to a 13 volt drop? Or have we overlooked something else?

The Vectors:

As before, we must take into account the different phase angles between voltage and current for each of the three components in the circuit. The vector diagram to the right, while not to scale, illustrates this concept.

Since this is a series circuit, the current is the same through all components and is therefore our reference at a phase angle of 0°. This is shown in red in the diagram. The resistor's voltage, vR, is in phase with the current and is shown in green. The blue vector shows vL at +90°, while the gold vector represents vC, at -90°. Since they oppose each other diametrically, the total reactive voltage is vL - vC. It is this difference vector that is combined with vR to find vT (shown in cyan in the diagram). We already know that vT = 10 vrms. Now we can see that vT is also the vector sum of (vL - vC) and vR. In addition, because of the presence of R, the phase angle between vT and i will be arctan((vL-vC)/vR), and can vary from -90° to +90°.

Basic RL Circuit

When we apply an ac voltage to a series RL circuit as shown to the right, the circuit behaves in some ways the same as the series RC circuit, and in some ways as a sort of mirror image. For example, current is still the same everywhere in this series circuit. VR is still in phase with I, and VL is still 90° out of phase with I. However, this time VL leads I — it is at +90° instead of -90°.

For this circuit, we will assign experimental values as follows: R = 560Ω, L = 100 mH, and VAC = 10 Vrms. We build the circuit, and measure 7.464 V across L, and 6.656 V across R. As we might have expected, this exceeds the source voltage by a substantial amount, and the phase shift is the reason for it.

The Vectors in an LC Series Circuit

The vectors for this example circuit are shown to the right. This time the composite phase angle is positive instead of negative, because VL leads IL. But to determine just what that phase angle is, we must start by determining XL and then calculating the rest of the circuit parameters.

XL =  2πfL      =  6.28 × 1000 × 0.1      =  628    

Z  =  560 + j628 Ω      =  (560² + 628²)½  

   =  (313600 + 394384)½

     =  (707984)½      =  841.4 Ω    

I  =  E/Z      =  10/841.4      =  0.0119 A      =  11.9 mA     VR =  I × R      =  0.0119 × 560      =  6.656 V     VL =  I × XL      =  0.0119 × 628      =  7.464 V    

θ  =  arctan(XL/R)      =  arctan(628/560)      =  arctan(1.121)      =  48.28°

This really completes the description of the series RL circuit with a fixed AC signal applied to it. Starting with the component values and the frequency of the applied AC voltage, we have described every aspect of this circuit's behavior at that frequency.

Resonance effect:The resonance effect occurs when inductive and capacitive reactances are equal. [Notice that the LC circuit does not, by itself, resonate. The word resonance refers to a class of

phenomena in which a small driving perturbation gives rise to a large effect in the system. The LC circuit must be driven, for example by an AC power supply, for resonance to occur (below).] The frequency at which this equality holds for the particular circuit is called the resonant frequency.

Series resonance

Here R, L, and C are in series in an ac circuit. Inductive reactance () increases as frequency increases while capacitive reactance () decreases with increase in frequency. At a particular frequency these two reactances are equal in magnitude but opposite in phase. The frequency at which this happens is the resonant frequency () for the given circuit.

Hence, at  :

Converting angular frequency into hertz we get

Here f is the resonant frequency. Then rearranging,

In a series ac circuit, XC leads by 90 degrees while XL lags by 90. Therefore, they cancel each other out. The only opposition to a current is coil resistance. Hence in series resonance the current is maximum at resonant frequency.

At , current is maximum. Circuit impedance is minimum. In this state a circuit is called an acceptor circuit.

Below , . Hence cct is capacitive. Above , . Hence cct is inductive.

Parallel resonance

Here a coil (L) and capacitor (C) are connected in parallel with an ac power supply. Let R be the internal resistance of the coil. When XL equals XC, the reactive branch currents are equal and opposite. Hence they cancel out each other to give minimum current in the main line. Since total current is minimum, in this state the total impedance is maximum.

Resonant frequency given by: .

Note that any reactive branch current is not minimum at resonance, but each is given separately by dividing source voltage (V) by reactance (Z). Hence I=V/Z, as per Ohm's law.

At fr,line current is minimum. Total impedance is maximum. In this state cct is called rejector circuit.

Below fr, cct is inductive. Above fr,cct is capacitive.

Applications of resonance effect1. Most common application is tuning. For example, when we tune a radio to a

particular station, the LC circuits are set at resonance for that particular carrier frequency.

2. A series resonant circuit provides voltage magnification. 3. A parallel resonant circuit provides current magnification. 4. A parallel resonant circuit can be used as load impedance in output circuits of RF

amplifiers. Due to high impedance, the gain of amplifier is maximum at resonant frequency.

5. A parallel resonant circuit can be used in induction heating.

SERIES RESONANCE:

A similar effect happens in series inductive/capacitive circuits. (Figure below) When a state of resonance is reached (capacitive and inductive reactances equal), the two impedances cancel each other out and the total impedance drops to zero!

Simple series resonant circuit.

With the total series impedance equal to 0 Ω at the resonant frequency of 159.155 Hz, the result is a short circuit across the AC power source at resonance. In the circuit drawn above, this would not be good. I'll add a small resistor (Figure below) in series along with

the capacitor and the inductor to keep the maximum circuit current somewhat limited, and perform another SPICE analysis over the same range of frequencies: (Figure below)

Series resonant circuit suitable for SPICE.

series lc circuitv1 1 0 ac 1 sin r1 1 2 1c1 2 3 10u l1 3 0 100m .ac lin 20 100 200 .plot ac i(v1) .end

Series resonant circuit plot of current I(v1).

As before, circuit current amplitude increases from bottom to top, while frequency increases from left to right. (Figure above) The peak is still seen to be at the plotted frequency point of 157.9 Hz, the closest analyzed point to our predicted resonance point of 159.155 Hz. This would suggest that our resonant frequency formula holds as true for simple series LC circuits as it does for simple parallel LC circuits, which is the case:

A word of caution is in order with series LC resonant circuits: because of the high currents which may be present in a series LC circuit at resonance, it is possible to produce dangerously high voltage drops across the capacitor and the inductor, as each component possesses significant impedance. We can edit the SPICE netlist in the above example to include a plot of voltage across the capacitor and inductor to demonstrate what happens: (Figure below)

series lc circuit v1 1 0 ac 1 sin r1 1 2 1c1 2 3 10u l1 3 0 100m .ac lin 20 100 200 .plot ac i(v1) v(2,3) v(3) .end

Plot of Vc=V(2,3) 70 V peak, VL=v(3) 70 V peak, I=I(V1#branch) 0.532 A peak

According to SPICE, voltage across the capacitor and inductor reach a peak somewhere around 70 volts! This is quite impressive for a power supply that only generates 1 volt. Needless to say, caution is in order when experimenting with circuits such as this. This SPICE voltage is lower than the expected value due to the small (20) number of steps in the AC analysis statement (.ac lin 20 100 200). What is the expected value?

  Given: fr = 159.155 Hz, L = 100mH, R = 1  XL = 2πfL = 2π(159.155)(100mH)=j100Ω  XC = 1/(2πfC) = 1/(2π(159.155)(10µF)) = -j100Ω  Z = 1 +j100 -j100 = 1 Ω  I = V/Z = (1 V)/(1 Ω) = 1 A  VL = IZ = (1 A)(j100) = j100 V  VC = IZ = (1 A)(-j100) = -j100 V  VR = IR = (1 A)(1)= 1 V  Vtotal = VL + VC + VR   Vtotal = j100 -j100 +1 = 1 V

The expected values for capacitor and inductor voltage are 100 V. This voltage will stress these components to that level and they must be rated accordingly. However, these voltages are out of phase and cancel yielding a total voltage across all three components of only 1 V, the applied voltage. The ratio of the capacitor (or inductor) voltage to the applied voltage is the “Q” factor.

  Q = VL/VR = VC/VR

Parallel Resonance:

The resonance of a parallel RLC circuit is a bit more involved than the series resonance. The resonant frequency can be defined in three different ways, which converge on the same expression as the series resonant frequency if the resistance of the circuit is small.

Q factor

In physics and engineering the quality factor or Q factor is a dimensionless parameter that compares the time constant for decay of an oscillating physical system's amplitude to its oscillation period. Equivalently, it compares the frequency at which a system oscillates to the rate at which it dissipates its energy. A higher Q indicates a lower rate of energy dissipation relative to the oscillation frequency, so the oscillations die out more slowly. For example, a pendulum suspended from a high-quality bearing, oscillating in air, would have a high Q, while a pendulum immersed in oil would have a low one. The concept originated in electronic engineering, as a measure of the 'quality' desired in a good tuned circuit or other resonator.

Usefulness of QThe Q factor is particularly useful in determining the qualitative behavior of a system. For example, a system with Q less than 1/2 cannot be described as oscillating at all, instead the system is said to be overdamped (Q < 1/2), gradually drifting towards its steady-state position. However, if Q > 1/2, the system's amplitude oscillates, while simultaneously decaying exponentially. This regime is referred to as underdamped.

Special values of Q critically damped , : the boundary between exponential and oscillatory response.

The simplest equal-C, equal-R Sallen–Key filter is critically damped.[citation needed] The second-order filter with the flattest passband frequency response

(Butterworth filter) has .[citation needed]

The second-order filter with the flattest group delay (Bessel filter) has .[citation needed]

Physical interpretation of Q

The bandwidth, Δf, of a damped oscillator is shown on a graph of energy versus frequency. The Q factor of the damped oscillator, or filter, is f0 / Δf

Physically speaking, Q is 2π times the ratio of the total energy stored divided by the energy lost in a single cycle or equivalently the ratio of the stored energy to the energy dissipated per one radian of the oscillation.

Equivalently (for large values of Q), the Q factor is approximately the number of oscillations required for a freely oscillating system's energy to fall off to 1 / e2π, or about 1/535, of its original energy.

When the system is driven by a sinusoidal drive, its resonant behavior depends strongly on Q. Resonant systems respond to frequencies close to their natural frequency much more strongly than they respond to other frequencies. A system with a high Q resonates with a greater amplitude (at the resonant frequency) than one with a low Q factor, and its response falls off more rapidly as the frequency moves away from resonance. Thus, a high Q tuned circuit in a radio receiver would be more difficult to tune with the necessary precision, but would have more selectivity; it would do a better job of filtering out signals from other stations that lay nearby on the spectrum.

TUNED CIRCUITS:

Scope

This article aims to give the reader an overall idea of the use of tuned LC circuits, and the theory behind them. The article is especially written for people who collect and restore vintage radios.

Self-induction

To understand tuned circuits, we first have to understand the phenomenon of self-induction. And to understand this, we need to know about induction.The first discovery about the interaction between electric current and magnetism was the realization that an electric current created a magnetic field around the conductor. It was then discovered that this effect could be enhanced greatly by winding the conductor into a coil. The effect

proved to be two-way: If a conductor, maybe in the form of a coil was placed in a changing magnetic field, a current could be made to flow in it; this is called induction.

So imagine a coil, and imagine that we apply a voltage to it. As current starts to flow, a magnetic field is created. But this means that our coil is in a changing magnetic field, and this induces a current in the coil. The induced current runs contrary to the applied current, effectively diminishing it. We have discovered self-induction. What happens is that the self-induction delays the build-up of current in the coil, but eventually the current will reach its maximum and stabilize at a value only determined by the ohmic resistance in the coil and the voltage applied. We now have a steady current and a steady magnetic field. During the buildup of the field, energy was supplied to the coil, where did that energy go? It went into the magnetic field, and as long as the magnetic field exists, it will be stored there.

Now imagine that we remove the current source. Without a steady current to uphold it, the magnetic field starts to disappear, but this means our coil is again in a variable field which induces a current into it. This time the current is in the direction of the applied current, delaying the decay of the current and the magnetic field till the stored energy is spent. This can give a funny effect: Since the coil must get rid of the stored energy, the voltage over it rises indefinitely until a current can run somewhere! This means you can get a surprising amount of sparks and arching when coils are involved. If the coil is large enough, you can actually get an electric shock from a low-voltage source like an ohmmeter.

Self-inductance is measured in henry (H or Hy). A henry is almost as enourmous value as a Farad, and coils are often measured in milli, micro and even nanohenry.

Now to tuned circuits

So, a coil is a component that effectivily stores current. A capacitor is a component that stores voltage. If we connect those two together, we get interesting results. Imagine a coil (L) and a capacitor (C) connected in parallel:

Let us assume we charged the capacitor with a voltage before connecting it to the coil, so now we apply that voltage to the coil. This means a current starts to flow, and a magnetic field is created. As the magnetic field builds up and reaches maximum, an increasing amount of current flows in the circuit. Eventually, the capacitor becomes discharged; the voltage across it is zero, but now a magnetic field has been built in the coil and a current

is flowing in the circuit, the charge has been transferred and transformed: From a voltage charge in the capacitor to a current charge in the coil.

As the voltage is now gone, the current starts to decay, but the self-induction keeps it running till the magnetic field has been spent. This current charges the capacitor, now with the opposite polarity as before; the charge is transferred back to the capacitor:

When the current stops, the voltage starts a new current in the opposite direction, etc. etc. The charge swings like a pendulum back and forth between the coil and the capacitor, changing polarity twice in each cycle. If there were no losses, this would go on forever. In real life, some energy is lost in each cycle, and unless the lost energy is somehow replaced, the tuned circuit will ring out, much like a bell:

Like a pendulum, the tuned circuit swings at a certain frequency, here determined by the values of the components. The formula is simple:

Types of tuned circuits

Many variations and complex couplings exist, but there are just two basic couplings, series and parallel:

The parallel circuit has its highest impedance at the resonance frequency. Placed across a signal, it allows only the resonance frequency to pass.The series circuit has its lowest impedance at the resonance frequency. Placed across a signal, it shorts out the resonance frequency, allowing all other frequencies to pass.

Bandwidth

Of course nothing is as simple as it seems. The tuned circuit has a bandwidth, that is, frequencies near the actual resonance frequency will also more or less achieve resonance. The bandwidth curve of a single resonant circuit looks like this:

The width of this curve depends on the Q parameter of the circuit. The Q is basically a figure for the effectivity of the tuned circuit. A circuit with a high loss will have a low Q and a high bandwidth. A circuit with low losses will have a high Q and a narrow bandwidth. Losses come mostly from two things:

The components themselves. Most notoriously, ohmic resistance in coils, but especially at high frequencies, there are many materials used in components and wiring that can introduce losses.

Loading. A tuned circuit is not very interesting unless it is incorporated in some kind of circuitry. Signal is fed to it and signal is tapped from it. This leads to loading of the resonant circuit.

Coupled resonance

In complex circuits like radios, the simple resonant curve of one tuned circuit is not always enough. To transmit the frequency range needed for audio, a bandwidth of at least twice the audio range is needed, but outside that a steep rolloff is needed. For this purpose several sets of tuned circuits coupled together in a so-called critical coupling is used, like in IF transformers. Two critically coupled tuned circuits have a resonance curve that looks like this:

Various couplings can be encountered, forming low-pass, high-pass, band-pass or band-suppression filters, but these specialized circuits are beyond the scope of this article.

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