Thevenin's Theorem

Named after M.L. Thevenin , a French engineer , Thevenin ‘s is very useful in simplifying the voltages in a network. By Thevenin’s theorem,many sources and components, no meter how they are interconnected, can be represented by an equivalent series circuit with respect to pair of terminals in the network. In Fig.10-2 imagine that the block at the left contains a network connected to terminals A and B Theremin’s theorem states that the entire network connected to A and B can be replaced by a single voltage source Vth in series with a single resistance Rth connected to the same two terminals.
Voltage is the open-circuit voltage across terminals A and B. This means, find the voltage that the network produces across the two terminals with an open circuit between A and B. The polarity of Vth is such that it will produce current from A and B in the same direction as in the original network.
Resistance Rth is the open-circuit resistance across terminals A and B but with all the sources killed. This means , find the resistance looking back into the network from terminals A and B. Although the terminals are open, an ohmmeter across AB would read the value of Rth as the resistance of the remaining paths in the network, without any sources operating.
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The Hall Effect

In 1879, E.H. Hall observed that a small voltage is generated across a conductor carrying current in an external magnetic field. The Hall voltage was very small with typical conductors, and little use was made of this effect. However, with the development of semiconductors, lager values of Hall voltage can be generated. The semiconductor material indium arsenide (In As) is generally used. As illustrated in Fig.13-14, the InAs element inserted in the magnetic field can generate 60 mV with B equal to 10 KG and an I of 100 mA. The applied flux must be perpendicular to the direction of current. With current in the direction of the length of conductor, the generated voltage is developed across the which.
The amount of Hall voltage v/H is directly proportional to the value of flux density B. This means that gauss meter in Fig.13-15 uses an InAs probe in the magnetic field to generate a proportional Hall voltage v/H. This value of v/H is then read by the meter, which is calibrated in gauss. The original calibration is made in terms of a reference magnet with a specified flux density.
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Norton's Theorem

Named after E,L. Norton, a scientist with bell telephone Laboratories Norton’s theorem is used for simplifying a network in terms of current instead of voltages. In many cases, analyzing the division of currents may be easier than voltage analysis. For current analysis, therefore, Norton’s theorem can be used to reduce a network to a simple parallel circuit, with a current source. The idea of a current source is that it supplies a total line current to be divided among parallel branches, corresponding to voltage source applying a total voltage to be divided among series components. This comparison is illustrated in Fig.10-7.


A source of electric energy supplying voltage is often shown with a series resistance which represents the internal resistance of the source, as in Fig.10-7a. This method corresponds to showing an actual voltage source, such as a battery for dc circuits. However, the source may be represented also as a current with a parallel resistance, as in Fig.10-7b. Just as a voltage source is rated at, say, 10V , a current source may be rated at 2 A . For the purpose of analyzing parallel branches, the concept of a current source may be more convenient than a voltage source.
If the current I in Fig. 10-7is a 2-A source, it supplies 2A no matter what is connected across the output terminals A and B. Without anything connected across A and B, all 2 A flows through the shunt R. When a load resistance R/L is connected across A and B, then the 2-A I divides according to the current division rules for parallel branches.
Remember the parallel current divide inversely to branch resistance but directly with conductance. For this reason it may be preferable to consider the current source shunted by the conductance G, as shown in Fig. 10-7c. we can always convert between resistance and conductance, because 1/R in ohms is equal to G in siemens.
The symbol for a current source is a circle with an arrow inside, as shown inFig.10-7b and c, to shown the direction of current . This direction must be the same as the current produced by the polarity of the corresponding voltage source.
Remember that a source produce electron flow out from the negative terminal.
An important difference between voltage and current sources is that a current source is killed by making it open, compared with short-circuiting a voltage source. Opening a current source kills its ability to supply current without affecting any parallel branches. A voltage source is short-circuited to kill its ability to supply voltage without affecting any series components.
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Millman's Theorem

Millman’s theorem provides a shortcut for finding the common voltage across any number of parallel branches with different voltage sources. A typical example is shown in fig. 10-16 For all the branches, the ends at point Y are connected to chassis ground. Furthermore, the opposite ends of all the branches are also connected to the point X. The voltage V/x y, therefore, is the common voltage across all the branches.

Finding the value of V/x y gives the net effect of the all source in determining the voltage at X with respect to chassis ground. To calculate this voltage
 v/x y=V1/R1 + V2/R2 + V3/R3 ……etc
1/R1 + 1/R2 +1/R3
This formula is derived from converting the voltage source to current sources and combining the results. The numerator with V/R terms is the sum of the parallel current sources. The denominator with 1/R terms is the sum of the parallel conductances. The net V/x y then is the form of I/G or 1*R, which is in units of voltage.
For the values in Fig.10-16,
Vxy =32/4 + 0/2 – 8/4
1/4 + 1/2 + 1/4
=8 + 0 -2
Vxy=6 V
Note the branches 3, V3 is considered negative because it world make point X negative. However, all the resistances are positive. The positive answer for Vxy means that point X is positive with respect to Y.
In branch 2, V2 is zero because this branch has no voltage source however, R2 is still used in the denominator.
This method can be used for any number of branches, but they must all be in parallel, without any series resistance between the branches. In a branch with several resistance, they can be combined as one Rt. When a branch has more than one voltage source, they can be combined algebraically for one Vt. Read More!

Magnetic Shielding

The idea of preventing one component from affecting another through their common electric or magnetic field is called shielding. Examples are the braided copper wire shield around the inner conductor of a coaxial cable, metal shield can that encloses an RF coil, or a shield of magnetic material enclosing a cathode-ray tube.
The problem in shielding is to prevent one component from inducing an effect in the shielded component. The shield material are always metals, but there is a difference between using good conductors with low resistance like copper and aluminum and using good magnetic materials like soft iron.

A good conductor is best for two shielding function. One is to prevent induction of static electric charges. The other is to shield against the induction of a varying magnetic field. For static charges, the shield provides opposite induced charges, which prevent induction inside the shield. For a varying magnetic field, the shield has induced to produce induction inside the shield.
The best shield for a steady magnetic field is a good magnetic material of high permeability. A steady field is produced by a permanent magnetic, a coil with steady direct current, or the earth’s magnetic field. A magnetic shield of high permeability concentrates to magnetic flux. Then there is little flux to induce poles in a component inside shield. The shield can be considered as a short circuit for the lines of magnetic flux. Read More!

Magnetic Flux

The entire group of magnetic field lines, which can be considered to flow outward from the north pole of magnet, is called magnetic flux. Its symbol is the Greek letter (phi). A strong magnetic field has more lines of force and more flux than a week magnetic field.
One Maxwell (Mx) unit equals one magnetic field line. In Fig.13-5, as an example, the flux illustrated is 6 Mx because there are 6 field lines flowing in or out for each pole. A 1-1b magnet can provide a magnetic flux of about 5000 Mx. This unit is named for James Clerk Maxwell (1831-1879), an important Scottish mathematical physicist who contributed much to electrical and field theory. Read More!

Magnetic Field Around an Electric Current

In Fig.15-1, the iron filings aligned in concentric rings around the conductor shoe the magnetic field of the wire. The iron filings are dense next to the conductor, showing that the field is strongest at this point. Furthermore, the field strength decreases inversely as the square of the distance from the conductor. It is important to note the following two factors about the magnetic lines of force:

1) The magnetic lines are circular, as the field is symmetrical with respect to the wire in the center.
2) The magnetic field with circular lines of force is in a place perpendicular to the current in the wire.
From points C to D in the wire, its circular magnetic field is in the horizontal plane because the wire is vertical. Also, the vertical conductor between points EF and AB has the associated magnetic field in the horizontal plane. Where the conductor is horizontal, as from B to C and D to E, the magnetic field is in a vertical plane.
These two requirements of a circular magnetic field in a perpendicular plane apply to any charge in motion. Whether electron flow or a motion of positive charges is considered, the associated magnetic field must be at right angles to the direction of current.
In addition the current need not be in a wire conductor. As an example, the beam of moving electrons in the vacuum of a cathode-ray tube has an associated magnetic field. In all cases, the magnetic field has circular lines of force in a plane perpendicular to the direction of motion of the electric charges. Read More!

Alternating Current Application

Figure 16-1 shows the out put from an voltage generator, with the reversals between positive and negative polarities and the variations in amplitude. In Fig. 16-1a, the wave from shown simulates an ac voltage as it would appear on the screen of an oscilloscope, which is an important test instrument for ac voltages. The oscilloscope shows a picture of any as voltage connected to its input terminals, while indicting the amplitude. The details of how to use the oscilloscope for ac voltage measurements are explained in App. D. “Using the oscilloscope”.
In Fig.16-1b the Read More!

Faraday's Law of Induced Voltage

The voltage induced by magnetic flux cutting the turns, of a coil depends upon the number of turns and how fast the flux moves across the conductor. Either the flux or the conductor can move. Specifically, the amount of induced voltage is determined by the following three factors.
1) Amount of flux. The more magnetic lines of force that cut across the conductor, the higher the amount of induced voltage.
2) Number of turns. The more turns in a coil, the higher the induced voltage. The V/ind is the sum of all individual voltages generated in each turn in series.
3) Time rate of cutting. The faster the flux cuts a conductor, the higher the induced voltage. Then more lines of force cut the conductor within a specific period of time.
These factors are fop fundamental importance in many applications. Any conductor with current will have voltage induced in it by a change in current and its associated magnetic flux.
The amount of induced voltage can be calculated by faraday’s law:

 V/ind =N d (Weber’s)
dt (seconds)
Where N is the number of turns and d/dt specifies hoe fast the flux cuts across the conductor. With d/dt in Weber’s per second, the induced voltage is in volts.
As an example, suppose that magnetic flux cuts across 300 turns at the rate of 2 Wb/s.
To calculate the induced voltage,
 V/ind = N d
=300 (2)
V/ind = 600 V
It is assumed that all the flux links all the turns, which is true with an iron core Read More!

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