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What is the difference between magneto and a.c generator?what is meant by frequency of alterneting current?


A magneto is an electrical generator that uses permanent magnets to produce periodic pulses of alternating current.

Hand-cranked magneto generators were used to provide ringing current in early telephone systems.
Magnetos adapted to produce pulses of high voltage are used in the ignition systems of some gasoline-powered internal combustion engines to provide power to the spark plugs. Use of magnetos for ignition is limited mainly to engines where there is no low-voltage electrical system available, such as lawnmowers and chainsaws. Magnetos have traditionally been used in aviation piston engines even though a low-voltage electrical system is usually available, to keep the ignition system independent of the battery and charging system and to keep the engine running in the event of alternator or battery failure. For redundancy purposes, virtually all piston engine aircraft are fitted with two magneto systems, each supplying power to one of two spark plugs in each cylinder.

Magnetos were rarely used for power generation, although they were for a few specialised uses.

A Generator is a device which converts mechanical energy into electrical energy.

A.C Generator works on the principle of electromagnetic induction (motional emf). In generator an induced emf is produced by rotating a coil in a magnetic field. The flux linking the coil changes continuously hence a continuous fluctuating emf is obtained.

A.C Generator consists of the following parts.
Powerful field magnet with concave poles.
Armature:It is a rectangular coil of large number of turns of wire wound on laminated soft-iron core of high permeability and low hysteresis loss.
Slip rings:The ends of the coil are joined to two separate copper rings fixed on the axle (S1 & S2).
Carbon brushes:Two carbon brushes remain pressed against each of the rings which form the terminals of the external circuit.


The frequency of alternating current depends on the speed of the alternator/generator at the power production station.

In India, its usually 220 Volts at 50 Hz.
In USA, it is 440 Volts at 60 Hz.
So the variation depends on the speed of the alternator at the time of generation and how the phase changes on the wave.

Hence, its possible to change the frequency of AC.

Electric Power Transmission (College Level )

The electrical power generated in a power station situated in a remote place is transmitted to different regions for domestic and industrial uses. Here we discuss the electric power grid, power generation, transmission, and distribution.

  • Power Grid

    What is a power grid?
    The power grid forms a bridge between electrical suppliers and consumers through interconnected networks.
    The electrical power grid consists of three main parts:
    1. Generating plant for electric power.
    2. Transmission of the electric power.
    3. Distribution of the electric power.
  • Electrical Power Generation in Power Stations

    Normally, electric power generating plants are located near the source of power generation like dams, coal mines, etc. They may be in remote locations, and the generated power has to be supplied to the consumer in the city.
    Sources for Power Generation:
    The following types of resources are available for generating electrical energy for distribution:
    Steam Power Plants
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    • Conventional Methods
    1.Thermal:
    Thermal energy or Nuclear Energy used for producing steam for turbines which will drive the alternators (rotating AC generators).
    2. Hydro-Electric:
    Potential of water stored at higher altitudes is utilized as it is passes through water turbines which drive the alternators.
    • Non- Conventional Methods:
    1. Wind Power:
    High velocities of wind are utilized in driving wind turbines coupled to alternators.
  • Sources of Power Generation

    Hydro-electric power stationNuclear Power PlantCoal power plant -Thermal Power StationWind Power
  • Transmission of Power

    Power DistributionHigh Voltage Transmission Lines
  • Transmission of Electric Power

    Once the power is generated from the alternator, it is send to the typical substation in the power plant where they step up the voltage by using the step-up transformers for transmission purposes.
    As the voltage is stepped up, it reduces the transmission losses. It is then sent to the power grid from where it is then transmitted to different cities. All the power generated in different places by different methods is stepped up and sent to a common place called the grid.
    Why do we need to step up the voltage for transmission?
    For long distance transmission, power lines are made of conducting material like aluminum. There is always some power loss associated with these lines.
    If I is the current through the wire and R is the resistance, a considerable amount of electric power (I2R) is dissipated as heat. Hence, the power at the receiving end will be much less than the actual power generated.
    However by transmitting the electrical energy at higher voltage, power loss can be controlled as is evident from the following cases:
    Case 1:
    Power Transmitted At Lower Voltage:
    A Power of 11,000 Watts is transmitted at 220 Volt.
    Formula for Power (P) = V × I
    V-Voltage
    I-Current
    Therefore, Current (I) =P / V
    Current (I) =11,000 / 220.
    Current (I) =50 Ampere.
    If R is the Resistance of the line wire,
    Then the power Loss:
    Formula, Power Loss = I2 × R
    Power Loss = 502 × R watts = (25000 × R) Watts.
    Case 2:
    Power Transmitted At Higher Voltage:
    If 11,000 Watts power is transmitted at 22,000 Volt
    Current (I) =P/V = 11,000/22,000 =0.5 Ampere.
    Power loss = I2 × R = (.5)2× R = (0.25 × R ) watts.
    From the case 1 and 2, we found that when the power is transmitted at higher voltage, the power losses are less. Hence it is evident that if power is transmitted at a higher voltage, the loss of energy in the form of heat can be considerably reduced.
    For transmitting electric power at 11,000W at 220 V the current capacity of line wire has to be 50 A and if transmission is done at 22,000V it is only 0.5 A.
    Thus for carrying larger current (50A), thick wires have to be used. This increases the cost of transmission to support these thick wires, and stronger poles have to be erected, which further adds on to the cost. On the other hand, if transmission is done at high voltages, the wires required are of lower current carrying capacity, so thicker wires can be replaced by thin wires, thus reducing the cost of transmission considerably.
    An example would be if 400MW power is produced at 15,000V in a power station, it could be stepped up to 230,000V before transmission. The power is then transmitted through the transmission lines or network to the power grid.
  • Power Grid

    Power DistributionSub-Station Before the City to Step Down Voltage
  • Distribution of Electric Power

    The electric power grid connects different parts of the country. And the grid distributes the power to the different parts of the country or state through the transmission lines or network connecting different cities.
    Outside the city, the transmitted power from the grid is stepped down in voltage to 110,000 Volt from 230,000 Volt by a step-down auto transformer. Again the power is stepped down to 11,000 Volt from 110,000 Volt by a Step Down power transformer located in the streets and it is distributed to the industrial uses.
    Finally before distribution to the domestic user, the power is stepped down to 230 V or 440 V depending upon the needs of the domestic user.

Wave Nature of Particle - the de Broglie Hypothesis( College Level)

In experiments like photoelectric effect and Compton effect, radiation behaves like particles. de Broglie, a french physicist asked whether in some situations, the reverse could be true, i.e., would objects which are generally regarded as particles (e.g. electrons) behave like waves ? In 1924 de Broglie postulated that we can associate a wave with every material object. In analogy with photons, he proposed that the wavelength associated with such a matter wave is related to the particle momentum through the relationship



where is the Planck's constant

Wavelike behaviour of a macroscopic object is difficult to detect as the wavelength is very small.
However, wave nature of particles may be detected in diffraction experiments where the dimensions of the obstacles are comparable with the wavelength of matter wave incident on the obstacle.



Example :Electron Diffraction from a Crystal

Consider a beam of electron with a speed m/s corresponding to a wavelength nm. Such a wave may be diffracted by gratings with separation of similar order as that of the wavelength. Crystals provide such natural gratings.
Davisson - Germer Experiment :
Experimental confirmation of de Broglie hypothesis was provided in 1926 by Davisson and Germer, who studied diffraction of a beam of electrons from the surface of a nickel crystal.



















A beam of electrons from a heated filament, accelerated through a potential difference is made to strike the surface of a crystal of Ni. Electrons are scattered in all directions and may be detected by an array of detectors located at various angles of scattering. It is found that the intensity of scattered beam is maximum at some particular angles of incidence, in the same manner as the case when a beam of x-rays strikes the crystal
In Davisson - Germer experiment, the electron beam was accelerated through a potential difference of volts. The kinetic energy of the electron is thus 54 eV. The de Broglie wavelength associated with an electron accelerated through a potential difference may be expressed as


Notes:

The dual nature not only is exhibited by radiation but is also associated with matter. In some experiments matter shows wave character.
de Broglie hypothesis poastulates a wavelength of with a particle having a momentum .
 Experimental confirmation of wave nature of matter comes from experiments such as Davisson Germer experiments
on electron diffraction from crystals. It is seen that the intensity of scattered beam is maximum at those points where one would expect Laue spots in x-ray diffraction assuming the electronsare waves with de Broglie wavelength.
 Bohr model can be understood by postulating that stable orbits in atoms are those which are standing waves of
electrons.
One can perform double slit experiment with electrons, similar to the way Young's double slit experiment is
performed with light waves. The intensity pattern obtained on a screen is very similar in both cases.
 According to the principle of complementarity one cannot obtain information on both the wave nature and particle
nature of matter or radiation in the same experiment.
Heisenberg uncertainty principle states that one cannot precisely measure both position and momentum of a
particle in the same experiment.


Uses of DC Load line


The below figure shows the output characteristic curves for the transistor in CE mode. The DC load line is drawn on the output characteristic curves. 

Load line To draw load line, we have to find saturation current and the cutoff voltage. After plotting these values on the vertical and the horizontal axes, a line is drawn joining these two points, which represents DC load line. It represents all possible combinations of the collector current Ic and the collector voltage Vc (or Vce) for the given load resistorRc. 


Saturation point The point at which the load line intersects the characteristic curve near the collector current axis is referred to as the saturation point. At this point of time, the current through the transistor is maximum and the voltage across collector is minimum for a given value of load. Therefore saturation current for the fixed bias circuit, Ic (sat) =Vcc/Rc 


Cutoff point The point where the load line intersects the cutoff region of the collector curves is referred as the cutoff point (i.e. end of load line). At this point, collector current is approximately zero and emitter is grounded for fixed bias circuit.Therefore, Vce (cut) = Vc = Vcc


Operating point The "Q point" for a transistor amplifier circuit is the point along its operating region in a "quiescent ", where no input signal gets amplified A load line is used in graphic analysis of circuits, representing the constraint other parts of the circuit place on a non - linear device, like a diode or transistor. A load line represents the response of a linear circuit to which the nonlinear device in question is connected to. The operating point is where the parameters of the nonlinear device and the parameters of the linear circuit match according to how they are connected while still adhering to their internal systems. In the example on the right, the nonlinear diode is placed in series with a linear circuit consisting of a resistor and a voltage source. The load line represents the relationship between current and voltage in the linear part of the circuit while the exponential represents the relationship between current and voltage in the nonlinear device. Since the current going through three elements in series should be the same, the operating point of the circuit will be at the intersection of the exponential with the load line. In a BJT circuit, the BJT has a different current-voltage(IC-VCE) characteristic depending on the Base current. Placing a series of these curves on the graph shows how the base current will effect the operating point of the circuit. It should be noted that the load line is used for dc analysis, and has no bearing on small-signal analysis once an operating point is identified. Load lines for common configurations Common-Emitter The given load line diagram is for the Common emitter configuration. Common emitter load line. The load line diagram illustrates all possible values of collector current (IC) and the collector voltage (VCE in this case) for a given load resistor (RC). The point on the load line where it intersects the collector current axis is referred to as saturation point. At this point, the transistor current is maximum and voltage across collector is minimum, for a given load. For this circuit, IC-SAT= VCC/RC. The cutoff point is the point where the load line intersects with the collector voltage axis. Here the transistor current is minimum (approximately zero) and emitter is grounded. Hence VCE-CUTOFF=Vcc. The operating point of the circuit in this configuration is generally designed to be in the active region, approximately between middle of the load line and close to saturation point. In this region, the collector current is proportional to the base current, and hence useful for amplifier applications. 

Basic Unit Conversions



Ans:In the field of science, the metric system is used in performing measurements. The metric system is actually easier to use than the English system, as you will see shortly. The metric system uses prefixes to indicate the magnitude of a measured quantity. The prefix itself gives the conversion factor. You should memorize some of the common prefixes, as you will be using them on a regular basis. Common prefixes are shown below:

PrefixSymbolPowerPrefixSymbolPower
mega-
M
106centi-
c
10-2
kilo-
k
103milli-
m
10-3
hecto-
h
102micro-
10-6
deca-
D
101nano-
n
10-9
deci-
d
10-1pico-
p
10-12

Metric - Metric Conversions
Suppose you wanted to convert the mass of a 250 mg aspirin tablet to grams. Start with what you know and let the conversion factor units decide how to set up the problem. If a unit to be converted is in the numerator, that unit must be in the denominator of the conversion factor in order for it to cancel.

Notice how the units cancel to give grams. I've shown the conversion factor numerator as 1 x 10-3 because on most calculators, it must be entered in this fashion, not as just 10-3. If you don't know how to use the scientific notation on your calculator, try to find out as soon as possible. Look in your calculator's manual, or ask someone who knows. Also, notice how the unit, mg is assigned the value of 1, and the prefix, milli-, is applied to the gram unit. In other words, 1 mg literally means 1 x 10-3 g.
Next, let's try a more involved conversion. Suppose you wanted to convert 250 mg to kg. You may or may not know a direct, one-step conversion. In fact, the better method (foolproof) to do the conversion would be to go to the base unit first, and then to the final unit you want. In other words, convert the milligrams to grams and then go to kilograms:

English - Metric Conversions
These conversions are accomplished in the same way as metric - metric conversions. The only difference is the conversion factor used. It would be a good idea to memorize a few conversion factors involving converting mass, volume, length and temperature. Here are a few useful conversion factors:
length: 2.54 cm = 1 inch (exact)

mass: 454 g = 1 lb

volume: 0.946 L = 1 qt

temperature: oC = (oF - 32)/1.8

All of the above conversions are to three significant figures, except length, which is an exact number. As before, let the units help you set up the conversion.
Suppose you wanted to convert mass of my 23 lb cat to kilograms. One can quickly see that this conversion is not achieved in one step. The pound units will be converted to grams, and then from grams to kilograms. Let the units help you set up the problem:

Let's try a conversion which looks "intimidating", but actually uses the same basic concepts we have already examined. Suppose you wish to convert pressure of 14 lb/in2 to g/cm2. When setting up the conversion, worry about one unit at a time, for example, convert the pound units to gram units, first:

Next, convert in2 to cm2. Set up the conversion without the exponent first, using the conversion factor, 1 in = 2.54 cm. Since we need in2 and cm2, raise everything to the second power:


Notice how the units cancel to the units sought. Always check your units because they indicate whether or not the problem has been set up correctly.

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