Circuits

Resistors do that when there's a current running through them.

The majority of countries in the world run the residential AC mains in the vicinity

of 220, 230, or 240 volts.

The highest nominal supply is 240 volts AC, in all of the following countries:

Afghanistan

Brunei

Cook Islands

Cyprus

Falkland Islands

Fiji

Gibraltar

Guyana

Isle of Man

Kenya

Kiribati

Kuwait

Liberia

Malaysia

Nauru

Nigeria

Oman

Papua New Guinea

Qatar

St. Lucia

Seychelles

Tonga

Uganda

The UK

DC

For lead-acid chemistry, as in a car battery, think 13.6 volts DC.

Signal Loss and Bandwidth

Technically, yes. But it would be like putting a Ford engine

and transmission into a Chevy body. Definitely harder and

more expensive than going out and buying the appliance

that you actually need.

An xor gate with 1 i/p being the original clk signal.The other i/p is the clk delayed by cycle_time/4.The delay can be achieved by buffer.The o/p is now double the clk freq.

Conduct a Hi-Pot test on the full length of cable (ensure safety and qualified operator) for a predetermined time at a predetermined voltage. Verify ground straps are properly earthed before test. Test results for a good stress cone installation will be about 35-50 Micro-amps.

You need to drop 6 volts across the resistor.

-- The resistor you need is 6/(the current in amps that your 6vdc device uses to operate) ohms.

Example:

If the device uses 1/2 Amp when it's running, then you need a 6/0.5 = 12-ohm resistor.

-- And the power-dissipation rating of the resistor has to be at least 36/resistance watts.

Example:

For the 12-ohm resistor in the last example, it needs to be a (3-watt or more) resistor.

For a sinusoidal waveorm, RMS (effective, heating) value = 2/pi x (peak voltage).

It's not 2/pi for waveforms with other shapes.

2/pi = roughly 63.7%

Power = Voltage*Current. Multiply the current and the voltage. Keep your units in mind. If your voltage is Volts, and your current is in Amps, your power will be in Watts. If you are using milliamps, your power will be in milliwatts. You can also use P=I2*R. The current squared, mulitplied by the resistance of the circuit. Or P=V2/R, the voltage squared divided by the resistance of the circuit. The last two of these can be derived from the basic equation V=I*R and P=V*I. Here's a little helper for you too. "Twinkle twinkle little star, power equals I squared R".

You want a filter to remove energy at only certain frequencies, and not to

have any affect on all the others. Resistors just dissipate energy, regardless

of what its frequency may be.

what is iron core inductors

Environment Circuit

A capacitor nominally has no inductance, which is lucky

because there is no such device to mesure it with.

You have a 2-to-1 step-up transformer. The voltage across the secondary

winding is 200 volts.

The power in the secondary winding is the power required by whatever

200-volt device you connect across that winding.

The power drawn by the primary winding from the 100-volt AC supply is

somewhat more than the power delivered to the device by the secondary,

since some power is lost in the transformer wire and core. That's why the

transformer hums and gets warm.

In a series circuit, the current has only one path to take.

a power drill

The most contributing factor to core losses in transformers is the material on which they are wound. Transformers wound on iron cores are roughly 75% efficient but they can transfer large amounts of power at low frequency. Transformers wound on ferrite are typically better then 90% efficient but can't be used to transfer the same amount of power unless the frequency is increased. Switch mode power supplies do this.

After that, the voltage, impedance and current are all related. If you increase the current (by increasing the voltage) there are more losses because the magnetic field strength is not directly proportional to the current. The impedance remains the same for the same number of turns. Transformers are wound to the best compromise between efficiency and transfer capability. If you try to increase the voltage too much, the core will saturate and behave like a straight piece of wire.

(a'b+b'a)'

Current in the single core cable would induce a magnetic current in the steel cable, though a transformer effect.

This would heat the steel armored strands, and the circuit would increase more electrical power from the load supply point.

The earthing of the cable glands would complete the circuit and the return current would flow in the earth bonding cable between the two points.

This is called Eddy currents generated in the cable by the twist of the steel armored around the central core, current flowing in one direction

non-linear circuit

The step recovery diode is used as what is termed a charge controlled switch. When the step recovery diode is forward biased and charge enters it, the diode appears as a normal diode and it behaves in much the same way. When diodes switch from forward conduction to reverse cut-off, a reverse current flows briefly as stored charge is removed. When all the charge is removed it suddenly turns off or snaps off. It is the abruptness with which the reverse current ceases that enables the step recovery diode to be used for the generation of microwave pulses and also for waveform shaping.

To explain this in more detail, under normal forward bias conditions the diode will conduct normally. Then if it is quickly reverse biased it will initially appear as a low impedance, typically less than an ohm. Once the charge that is stored in the device is depleted, the impedance will very abruptly increase to its normal reverse impedance which will be very high. This transition occurs very quickly, typically well under a nanosecond.

This property allows the step recovery diode to be used in pulse shaping (sharpening) and in pulse generator circuits. The high harmonic content of the signal produced by any repetitive waveforms from step recovery diode circuits enables them to be used as comb generators where a comb of harmonically related frequencies are generated.