BLOG 13 - ELECTRICAL CIRCUIT 2

THIS BLOG CONTAIN ABOUT THE SINUSOIDAL AND PHASE.

Sinusoidal Functions and Circuit Analysis

The sinusoidal functions (sine and cosine) appear everywhere, and they play an important role in circuit analysis. The sinusoidal functions provide a good approximation for describing a circuit’s input and output behavior not only in electrical engineering but in many branches of science and engineering.

The sinusoidal function is periodic, meaning its graph contains a basic shape that repeats over and over indefinitely. The function goes on forever, oscillating through endless peaks and valleys in both negative and positive directions of time. Here are some key parts of the function:

• The amplitude V_A defines the maximum and minimum peaks of the oscillations.
• Frequency f₀ describes the number of oscillations in 1 second.
• The period T₀ defines the time required to complete 1 cycle.

The period and frequency are reciprocals of each other, governed by the following mathematical relationship:

Here is a cosine function you can use as the reference signal:

You can move sinusoidal functions left or right with a time shift as well as increase or decrease the amplitude. You can also describe a sinusoidal function with a phase shift in terms of a linear combination of sine and cosine functions. Here is a cosine function and a shifted cosine function with a phase shift of π/2.

Phase shifts in a sinusoidal function

A signal that’s out of phase has been shifted left or right when compared to a reference signal:

Right shift: When a function moves right, then the function is said to be delayed. The delayed cosine has its peak occur after the origin. A delayed signal is also said to be a lag signalbecause the signal arrives later than expected.

Left shift: When the cosine function is shifted left, the shifted function is said to be advanced. The peak of the advanced signal occurs just before the origin. An advanced signal is also called a lead signal because the lead signal arrives earlier than expected.

Here are examples of unshifted, lagged, and lead cosine functions.

To see what a phase shift looks like mathematically, first take a look at the reference signal:

At t = 0, the positive peak V_A serves as a reference point. To move the reference point by time shift T_S, replace the t with (t – T_S):

where

The factor ϕ is the phase shift (or angle). The phase shift is the angle between t = 0 and the nearest positive peak. You can view the preceding equation as the polar representation of the sinusoid. When the phase shift is π/2, then the shifted cosine is a sine function.

Express the phase angle in radians to make sure it’s in the same units as the argument of the cosine (2πt/T₀ – ϕ). Angles can be expressed in either radians or degrees; make sure you use the right setting on your calculator.

When you have a phase shift ϕ at the output when compared to the input, it’s usually caused by the circuit itself.

Expand a sinusoidal function and find Fourier coefficients

The general sinusoid v(t) involves the cosine of a difference of angles. In many applications, you can expand the general sinusoid using the following trigonometric identity:

Expanding the general sinusoid v(t) leads to

The terms c and d are just special constants called Fourier coefficients. You can express the waveform as a combination of sines and cosines as follows:

The function v(t) describes a sinusoidal signal in rectangular form.

If you know your complex numbers going between polar and rectangular forms, then you can go between the two forms of the sinusoids. The Fourier coefficients c and d are related by the amplitudeV_A and phase ϕ:

If you go back to find V_A and ϕ from the Fourier coefficients c and d, you wind up with these expressions:

The inverse tangent function on a calculator has a positive or negative 180° (or π) phase ambiguity. You can figure out the phase by looking at the signs of the Fourier coefficientsc and d. Draw the points c and d on the rectangular system, where c is the x-component (or abscissa) and d is the y-component (or ordinate).

The ratio of d/c can be negative in Quadrants II and IV. Using the rectangular system helps you determine the angles when taking the arctangent, whose range is from –π/2 to π/2.

Connect sinusoidal functions to exponentials with Euler’s formula

Euler’s formula connects trig functions with complex exponential functions. The formula states that for any real number θ, you have the following complex exponential expressions:

The exponent jθ is an imaginary number, where j = √-1.

The imaginary number j is the same as the number i from your math classes, but all the cool people use j for imaginary numbers because i stands for current.

You can add and subtract the two preceding equations to get the following relationships:

These equations say that the cosine and sine functions are built as a combination of complex exponentials. The complex exponentials play an important role when you’re analyzing complex circuits that have storage devices such as capacitors and inductors.

The Phasor 

In the last tutorial, we saw that sinusoidal waveforms of the same frequency can have aPhase Difference between themselves which represents the angular difference of the two sinusoidal waveforms. Also the terms “lead” and “lag” as well as “in-phase” and “out-of-phase” were used to indicate the relationship of one waveform to the other with the generalized sinusoidal expression given as: A(t) = Am sin(ωt ± Φ) representing the sinusoid in the time-domain form.

But when presented mathematically in this way it is sometimes difficult to visualise this angular or phase difference between two or more sinusoidal waveforms so sinusoids can also be represented graphically in the spacial or phasor-domain form by a Phasor Diagram, and this is achieved by using the rotating vector method.

Basically a rotating vector, simply called a “Phasor” is a scaled line whose length represents an AC quantity that has both magnitude (“peak amplitude”) and direction (“phase”) which is “frozen” at some point in time.

A phasor is a vector that has an arrow head at one end which signifies partly the maximum value of the vector quantity ( V or I ) and partly the end of the vector that rotates.

Generally, vectors are assumed to pivot at one end around a fixed zero point known as the “point of origin” while the arrowed end representing the quantity, freely rotates in an anti-clockwise direction at an angular velocity, ( ω ) of one full revolution for every cycle. This anti-clockwise rotation of the vector is considered to be a positive rotation. Likewise, a clockwise rotation is considered to be a negative rotation.

Although the both the terms vectors and phasors are used to describe a rotating line that itself has both magnitude and direction, the main difference between the two is that a vectors magnitude is the “peak value” of the sinusoid while a phasors magnitude is the “rms value” of the sinusoid. In both cases the phase angle and direction remains the same.

The phase of an alternating quantity at any instant in time can be represented by a phasor diagram, so phasor diagrams can be thought of as “functions of time”. A complete sine wave can be constructed by a single vector rotating at an angular velocity of ω = 2πƒ, where ƒ is the frequency of the waveform. Then a Phasor is a quantity that has both “Magnitude” and “Direction”. Generally, when constructing a phasor diagram, angular velocity of a sine wave is always assumed to be: ω in rad/s. Consider the phasor diagram below.

Phasor Diagram of a Sinusoidal Waveform

As the single vector rotates in an anti-clockwise direction, its tip at point A will rotate one complete revolution of 360o or 2π representing one complete cycle. If the length of its moving tip is transferred at different angular intervals in time to a graph as shown above, a sinusoidal waveform would be drawn starting at the left with zero time. Each position along the horizontal axis indicates the time that has elapsed since zero time, t = 0. When the vector is horizontal the tip of the vector represents the angles at 0o, 180o and at 360o.

Likewise, when the tip of the vector is vertical it represents the positive peak value, ( +Am ) at 90o orπ/2 and the negative peak value, ( -Am ) at 270o or 3π/2. Then the time axis of the waveform represents the angle either in degrees or radians through which the phasor has moved. So we can say that a phasor represent a scaled voltage or current value of a rotating vector which is “frozen” at some point in time, ( t ) and in our example above, this is at an angle of 30o.

Sometimes when we are analysing alternating waveforms we may need to know the position of the phasor, representing the Alternating Quantity at some particular instant in time especially when we want to compare two different waveforms on the same axis. For example, voltage and current. We have assumed in the waveform above that the waveform starts at time t = 0 with a corresponding phase angle in either degrees or radians.

But if a second waveform starts to the left or to the right of this zero point or we want to represent in phasor notation the relationship between the two waveforms then we will need to take into account this phase difference, Φ of the waveform. Consider the diagram below from the previous Phase Difference tutorial.

Phase Difference of a Sinusoidal Waveform

The generalised mathematical expression to define these two sinusoidal quantities will be written as:

The current, i is lagging the voltage, v by angle Φ and in our example above this is 30o. So the difference between the two phasors representing the two sinusoidal quantities is angle Φ and the resulting phasor diagram will be.

Phasor Diagram of a Sinusoidal Waveform

The phasor diagram is drawn corresponding to time zero ( t = 0 ) on the horizontal axis. The lengths of the phasors are proportional to the values of the voltage, ( V ) and the current, ( I ) at the instant in time that the phasor diagram is drawn. The current phasor lags the voltage phasor by the angle, Φ, as the two phasors rotate in an anticlockwise direction as stated earlier, therefore the angle, Φ is also measured in the same anticlockwise direction.

If however, the waveforms are frozen at time t = 30o, the corresponding phasor diagram would look like the one shown on the right. Once again the current phasor lags behind the voltage phasor as the two waveforms are of the same frequency.

However, as the current waveform is now crossing the horizontal zero axis line at this instant in time we can use the current phasor as our new reference and correctly say that the voltage phasor is “leading” the current phasor by angle, Φ. Either way, one phasor is designated as the reference phasor and all the other phasors will be either leading or lagging with respect to this reference.

Phasor Addition

Sometimes it is necessary when studying sinusoids to add together two alternating waveforms, for example in an AC series circuit, that are not in-phase with each other. If they are in-phase that is, there is no phase shift then they can be added together in the same way as DC values to find the algebraic sum of the two vectors. For example, if two voltages of say 50 volts and 25 volts respectively are together “in-phase”, they will add or sum together to form one voltage of 75 volts.

If however, they are not in-phase that is, they do not have identical directions or starting point then the phase angle between them needs to be taken into account so they are added together using phasor diagrams to determine their Resultant Phasor or Vector Sum by using the parallelogram law.

Consider two AC voltages, V1 having a peak voltage of 20 volts, and V2 having a peak voltage of 30 volts where V1 leads V2 by 60o. The total voltage, VT of the two voltages can be found by firstly drawing a phasor diagram representing the two vectors and then constructing a parallelogram in which two of the sides are the voltages, V1 and V2 as shown below.

Phasor Addition of two Phasors

By drawing out the two phasors to scale onto graph paper, their phasor sum V1 + V2 can be easily found by measuring the length of the diagonal line, known as the “resultant r-vector”, from the zero point to the intersection of the construction lines 0-A. The downside of this graphical method is that it is time consuming when drawing the phasors to scale. Also, while this graphical method gives an answer which is accurate enough for most purposes, it may produce an error if not drawn accurately or correctly to scale. Then one way to ensure that the correct answer is always obtained is by an analytical method.

Mathematically we can add the two voltages together by firstly finding their “vertical” and “horizontal” directions, and from this we can then calculate both the “vertical” and “horizontal” components for the resultant “r vector”, VT. This analytical method which uses the cosine and sine rule to find this resultant value is commonly called the Rectangular Form.

In the rectangular form, the phasor is divided up into a real part, x and an imaginary part, y forming the generalised expression  Z = x ± jy. ( we will discuss this in more detail in the next tutorial ). This then gives us a mathematical expression that represents both the magnitude and the phase of the sinusoidal voltage as:

Definition of a Complex Sinusoid

So the addition of two vectors, A and B using the previous generalised expression is as follows:

Phasor Addition using Rectangular Form

Voltage, V2 of 30 volts points in the reference direction along the horizontal zero axis, then it has a horizontal component but no vertical component as follows.

• • Horizontal Component = 30 cos 0o = 30 volts
• • Vertical Component = 30 sin 0o = 0 volts
• This then gives us the rectangular expression for voltage V2 of:  30 + j0

Voltage, V1 of 20 volts leads voltage, V2 by 60o, then it has both horizontal and vertical components as follows.

• • Horizontal Component = 20 cos 60o = 20 x 0.5 = 10 volts
• • Vertical Component = 20 sin 60o = 20 x 0.866 = 17.32 volts
• This then gives us the rectangular expression for voltage V1 of:  10 + j17.32

The resultant voltage, VT is found by adding together the horizontal and vertical components as follows.

• VHorizontal = sum of real parts of V1 and V2 = 30 + 10 = 40 volts
• VVertical = sum of imaginary parts of V1 and V2 = 0 + 17.32 = 17.32 volts

Now that both the real and imaginary values have been found the magnitude of voltage, VT is determined by simply using Pythagoras’s Theorem for a 90o triangle as follows.

Then the resulting phasor diagram will be:

Resultant Value of VT

Phasor Subtraction

Phasor subtraction is very similar to the above rectangular method of addition, except this time the vector difference is the other diagonal of the parallelogram between the two voltages of V1 and V2 as shown.

Vector Subtraction of two Phasors

This time instead of “adding” together both the horizontal and vertical components we take them away, subtraction.

The 3-Phase Phasor Diagram

Previously we have only looked at single-phase AC waveforms where a single multi-turn coil rotates within a magnetic field. But if three identical coils each with the same number of coil turns are placed at an electrical angle of 120o to each other on the same rotor shaft, a three-phase voltage supply would be generated. A balanced three-phase voltage supply consists of three individual sinusoidal voltages that are all equal in magnitude and frequency but are out-of-phase with each other by exactly 120o electrical degrees.

Standard practice is to colour code the three phases as Red, Yellow and Blue to identify each individual phase with the red phase as the reference phase. The normal sequence of rotation for a three phase supply is Red followed by Yellow followed by Blue, ( R, Y, B ).

As with the single-phase phasors above, the phasors representing a three-phase system also rotate in an anti-clockwise direction around a central point as indicated by the arrow marked ω in rad/s. The phasors for a three-phase balanced star or delta connected system are shown below.

Three-phase Phasor Diagram

The phase voltages are all equal in magnitude but only differ in their phase angle. The three windings of the coils are connected together at points, a1, b1 and c1 to produce a common neutral connection for the three individual phases. Then if the red phase is taken as the reference phase each individual phase voltage can be defined with respect to the common neutral as.

Three-phase Voltage Equations

If the red phase voltage, VRN is taken as the reference voltage as stated earlier then the phase sequence will be R – Y – B so the voltage in the yellow phase lags VRN by 120o, and the voltage in the blue phase lags VYN also by 120o. But we can also say the blue phase voltage, VBN leads the red phase voltage, VRN by 120o.

One final point about a three-phase system. As the three individual sinusoidal voltages have a fixed relationship between each other of 120o they are said to be “balanced” therefore, in a set of balanced three phase voltages their phasor sum will always be zero as:  Va + Vb + Vc = 0

LEARNING

In this topic I learn about the phasor are Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC waveform. Phasor diagrams can be drawn to represent more than two sinusoids. They can be either voltage, current or some other alternating quantity but the frequency of all of them must be the same while in the sinusoidal Directly finding the steady-state response without solving the differential equation. According to the characteristics of steady-state response, the task is reduced to finding two real numbers, i.e. amplitude and phase angle, of the response. The waveform and frequency of the response are already known.

BLOG 12 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE Natural Response of First Order RC and RL Circuits

Natural Response of an RL Circuit

If we consider the circuit:

It is assumed that the switch has been closed long enough so that the inductor is fully charged. This means that all voltages and currents have reached constant values. Thus only constant (or d.c.) currents can appear just prior to the switch opening and the inductor appears as a short circuit.

As the inductor appears as a short circuit there can be no current in either R₀ or R. Hence all of the source current, I₀, appears in the inductive branch and the voltage across this branch is zero.

To find the natural response we need to see what happens when the source is disconnected. Hence we say when t = 0 the switch is opened. This then reduces the above circuit to:

To find i(t) we use Kirchoff's voltage law to obtain an expression involving i, R, and L. Summing the voltages around the closed loop gives:

This is known as a first order differential equation and can be solved by rearranging and then 'separating the variables'. This gives us:

Then by integrating both the right hand side and left hand side and including a constant of integration i(0) gives:

hence taking inverse logs:

if we use 0^- to represent the time just prior to switching and 0⁺ just after switching. Due to the characteristics of an inductor an instantaneous change of current in an inductor is not possible, therefore the current just after switching is equal to the current just prior to switching, then:

this then gives us the final value for the current of:

This leads us to define the time constant for a RL circuit:

We can then derive the voltage across the resistor from a direct application of Ohm's Law:

Natural Response of an RC Circuit

By following the above steps we can calculate the current and voltage in the circuit show below:

The switch remains to the left until the capacitor is fully charged then at time, t = 0 the switch is changed to the right position, so the capacitor is effectively connected to only the resistor. This gives us the equations:

and:

where time constant, 

LEARNING

Our topic is all about the RL and RC Circuit I learn in this topic are Resistive Circuit => RC Ci t => RC Circuit algebraic equations => differential equations

Same Solution Methods (a) N s (a) Nodal Analysis (b) Mesh Analysis. The natural response is due to the initial condition of the storage component ( C or L). The forced response is resulted from external input ( or force). In this chapter, a constant input (DC input) will be considered and the forced response is called step response. When a dc voltage (current) source is suddenly applied to a circuit , it can be modeled as a step function , and the resulting response is called response .

BLOG 11 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE Maximum Power Transfer, Capacitor and Inductors.

Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum.

This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.

Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):

With this value of load resistance, the dissipated power will be 39.2 watts:

If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:

Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:

If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverterloading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).

The Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.

Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, "output impdance" : "load impedance" is known as damping factor, typically in the range of 100 to 1000. [rar] [dfd]

Maximum power transfer does not coincide with the goal of lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.

Capacitor Circuits

Next let us consider a single capacitor of capacitance C, here the relationship between the current flow and applied voltage is given by

unlike the resistor, the current flow is proportional to the voltage gradient (with respect to time) and consequently this introduces a phase shift between the two. The impedance response for a circuit containing a single capacitor is shown in time, phasor and bode notations below.

Circuit Component	Frequency Response

clearly the current is 90° out of phase with the voltage, the Bode plot shows that this relationship holds for all frequencies although the magnitude of the signal drops as the frequency increases. To explain this behaviour we need to understand how the capacitor resists the passage of current. A measure of this resistance to current flow is given by the capacitative reactance Xc which has units of Ohms. This quantity Xc has both magnitude and phase and calculated using

it may be predicted by invoking Ohms law which tells us that the circuit resistance (in this Xc) is equal to

This shows us that the quantity ‹Xc always has a -90° angle attached to its magnitude and it's usually written as above or in the complex form -j Xc. 

Inductor Circuits

Finally we consider the response of an inductor with an inductance (L). Here the relationship between the current flow and applied voltage is given by

like the capacitor, the current can be seen to be out of phase with the voltage. The impedance response of a circuit containing a single inductor is shown below in time phasor and bode forms.

Circuit Component	Frequency Response

In an analogous manner to the capacitor the 'resistance' to current flow is given by the inductive reactance Xl which has both a magnitude and phase:

it may be predicted by using Ohms law which tells us that the circuit resistance (in this case Xl) is equal to

This shows us that the inductive reactance always has a 90° angle attached to its magnitude and is usually written in complex form as jXl or in polar form

We now have the basic information required to analyse circuits containing combinations of the above components in series or parallel. As the majority of circuits of interest in electrochemical analysis are combinations of resistors and capacitors we will only consider these in the later sections, although the extension to examine inductive circuits requires no further developments.

Learning

I learn in this topic is that the Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power. The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency. capacitive reactance in AC circuits.phase relationships in capacitors in AC circuits. true power and reactive power in a capacitor .inductive reactance in AC circuits . phase relationships in inductors in AC circuits .true power and reactive power in an inductor. trigonometric functions. inverse trigonometric functions.

BLOG 10 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE THEVENIN'S THEOREM and NORTON THEOREM

Norton's Theorem

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

. . . after Norton conversion . . .

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

As with Thevenin's Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (I_Norton) and Norton resistance (R_Norton).

As before, the first step is to identify the load resistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R₁ is strictly a function of B₁'s voltage and R₁'s resistance: 7 amps (I=E/R). Likewise, the current through R₃ is now strictly a function of B₂'s voltage and R₃'s resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (I_Norton) in our equivalent circuit:

Remember, the arrow notation for a current source points in the direction opposite that of electron flow. Again, apologies for the confusion. For better or for worse, this is standard electronic symbol notation. Blame Mr. Franklin again!

To calculate the Norton resistance (R_Norton), we do the exact same thing as we did for calculating Thevenin resistance (R_Thevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R₂; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin's Theorem apply to Norton's as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit again and again, applying nothing more complex than simple parallel circuit analysis to determine what's happening with each trial load.

Thevenin Theorem

This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in Thevenin theorem. Suppose we have to calculate the electric current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The current through the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch is connected and the internal resistance of the source is nothing but the equivalent resistance of the circuit looking back into the terminals where, the branch is connected. So the Thevenin theorem can be stated as follows,

1. When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is Thevenin equivalent voltage and,
2. The equivalent resistance of the circuit network looking back into the terminals, is Thevenin equivalent resistance.
3. If we replace the rest of the circuit network by a single voltage source , then the voltage of the source would be Thevenin equivalent voltage and internal resistance of the voltage source would be Thevenin equivalent resistance which would be connected in series with the source as shown in the figure below.

To make Thevenin theorem easy to understand, we have shown the circuit below,

Here two resistors R₁ and R₂ are connected in series and this series combination is connected across one voltage source of emf E with internal resistance R_i as shown. One resistive branch of R_L is connected across the resistance R₂ as shown. Now we have to calculate the current through R_L.

First, we have to remove the resistor R_L from the terminals A and B.

Second, we have to calculate the open circuit voltage or Thevenin equivalent voltage V_T across the terminals A and B.

The electric current through resistance R₂,

Hence voltage appears across the terminals A and B i.e.

Third, for applying Thevenin theorem, we have to determine the Thevenin equivalent electrical resistance of the circuit, and for that; first we have to replace the voltage source from the circuit, leaving behind only its internal resistance R_i. Now view the circuit inwards from the open terminals A and B. It is found the circuits now consist of two parallel paths - one consisting of resistance R₂ only and the other consisting of resistance R₁ and R_i in series.

Thus the Thevenin equivalent resistance R_T is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance R_L is connected across terminals A and B, the network behaves as a source of voltage V_T and internal resistance R_T and this is called Thevenin equivalent circuit. The electric current through R_L is given as,

Thevenin Equivalent Circuit

Learning

Our topic is all about the Thevenin Theorem and I learn from this topic is the Thevenin's Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load. To get easily the answer of the thevenin theorem is we need to follow the steps on how to solve the problem there are the steps of thevenin theorem (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be. (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points. (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit. (4) Analyze voltage and current for the load resistor following the rules for series circuits. I learn in Norton Theorem is to Norton's Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load. The norton theorem have step to get the problem easily there are the step to determine the norton theorem (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be. (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points. (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit. (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.