Electronics Engineering

To remove the noise

600 volts is five times the 120 volt primary, so the secondary winding must have 750 turns, which is five times as many turns as the 150 turns of the primary winding.

In digital communications, a digit is either a one or a zero, or an "on" or an "off" signal. The wave shape that is composed of "on's and off's" is the square wave.

e1 the basic data rate of communication in telecom( european ,india as well)...

It is made up to 32 channels of 64 kbps PCM Time Slot (ISDN) each ...

Time frame: 448ns

Bit rate: 125us

Bit no in 1 Time frame: 8

so E1=2048 kbps

0 and 16th channel used for synchronizing, remote access and monitoring

E2 is 4xE1 package=8448 kbps

E3 is 4xE2 package=24368 kbps

E4 is 4xE3 package=139264 kbps

An electric generator converts mechanical energy to electric energy while a motor converts electric energy to mechanical. A generator can actually act as a motor if it losses whatever is making it spin (called "motoring"), which is usually a very bad thing. Motors may be used as generators as well, depending on their design.

Let t=0 denote the start of transmission.At t=1 msec,the first frame has been fully
transmitted.At t=271 msec,the first frame has fully arrived.At t=272 msec,the frame
acknowledging the first one has been fully sent.At t=542 msec,the acknowledgement-bearing
frame has fully arrived.Thus, the cycle is 542 msec.A total of k frames are sent in 542 msec,for an
efficiency of k/542.Hence
(a)k=1,efficiency=1/542=0.18%
(b)k=7,efficiency=7/542=1.29%
(c)k=4,efficiency=4/542=0.74%

The power dissipated by a diode is P = Vf x I watts, where Vf is the forward voltage drop on diode (typically 0.5 volts for silicon diode) and I is the current.

well a voltmeter measures AC and DC voltage. resistance is a term used only for DC voltage, A voltmeter uses a term called Impedance for AC, without it, when you measure across two terminals it would create a short circuit and would blow up the meter. it is basically a safety feature to prevent people from creating short circuits when measuring.

Transformer step-up/step-down voltage is turns-ratio, so if a transformer has 20 primary windings and 100 secondary windings (a turns-ratio of 1 to 5) and the secondary voltage is 25, then is the primary voltage is 5.

The bulb converts energy from the power source into light and heat. It is the load in the circuit.

Soldering iron, solder, flux

A choke is an inductor. The impedance of an inductor is dependent on the frequency of the current flowing through it. The greater the frequency, the higher the impedance.

Therefore an inductor when used as a choke blocks the flow of high frequency current (by presenting a high impedance), while allowing low frequency or direct current to flow through it.

Its function is to block ("choke") high frequencies while passing low frequencies.

Electricians will tell you that a differential amplifier is a type of electronic amplifier that amplifies the difference between two voltages but does not change the particular voltages.

O K is absolute zero. At absolute zero, the electrons of the semi conductors are trapped and are immovable from their electron shell as they are in a low energy state. This makes the pure semiconductor an insulator. One must heat the semiconductor to give the electrons enough energy to move to free them from their electron shell, and thus conduct.

An example of a parallel circuit would be the light bulbs in track lighting. Each bulb has the same voltage applied. The current through any one light bulb equals the voltage divided by the resistance of the bulb. The current also equals the wattage of the bulb divided by the voltage. So if all the bulbs had exactly the same resistance the current would be the same. However, your question says "always" so in general the answer is no. In the case of track lighting if you had a 60 watt bulb in parallel with a 120 watt bulb, the 60W bulb would draw 1/2 amp and the 120W bulb would draw 1 amp. The sum of the current flowing in a parallel circuit equals the sum of the current in each leg of the circuit.

when A resistance is connected across the supply voltage, total input vooltage will be drop in the resistance when the resistances are connected across the supply voltage, total input vooltage will be devidedacross the resistances. IF R value will be high ,drop also high. IF R value wll be low ,voltage drop will be less.

1. punchcard reader/punch
2. 7 channel 1/2 inch magnetic tape drive
3. 9 channel 1/2 inch magnetic tape drive
4. fixed head magnetic disk drive
5. movable head magnetic disk drive
6. modem
7. teletype
8. CRT terminal
9. etc.

internal resistance is always infinite in ideal current source .the internal resistance is in shunt with current source

It depends how "ideal" your ideal diode is.

The first approximation of an ideal diode is that it's a device that allows current to flow one way, and not the other way, while doing this with no losses.

The second approximation of a diode implies the inherent 0.7V drop across the device, as well as one way current flow, but with no loss resistance.

The third approximation of an ideal diode is a device which allows current to only flow one way through it, with a 0.7V loss across is, as well as a small internal resistance of a few ohms.

These all vary from a real diode because these are all linear, in every sense. You can neither calculate the voltage nor the current across/through a diode. Instead, you must use the diode's characteristic curve (given on datasheet), and linearize it. You can get any amount of current to flow through the diode the CORRECT direction, by changing the voltage used. A number of microamps will take perhaps, a third of a Volt.

DC coupling on the input/output will always give higher gain because AC coupling involves inserting a capacitor, which adds impedance and augments the signal. AC coupling is sometimes necessary though, for such purposes as eliminating DC offset on the base of a transistor, or eliminating a DC offset in a single supply opamp circuit.

Take two electrolytic capacitors of the same voltage and capacity, connect the positive leads together and connect the negative leads to the circuit. Just keep in mind that this will reduce the cap. value by half (2, 1000uF caps = 500uF) Also the voltage of the circuit should not exceed the voltage of one of the caps.