Electronics Engineering

The most common type is resistive division. Say we have a source of V volts and we wish to apply a portion of the V volts (aV where a is between 0 and 1) to a load of resistance R. We can place a resistor R2 in series with load R to drop off the unwanted part of the voltage V. The math is a = R / (R + R2). This only works if the resistors are constant. For a changing load other methods are necessary.

Say the load needs 5 volts and it draws 10 milliamperes at 5 volts, but the supply is 10 volts. The effective load resistance is 500 ohms. A 500 ohm resistor in series with the load would drop off the unwanted 5 volt excess.

Suppose the load changes by 10%, but we wish to hold load voltage to 5%. We can place a 500 ohm resistor in parallel with the load and use a 250 ohm resistor to drop the voltage to 5 volts. The voltage change is now only 5% for a 10% load change.

To measure Voltage which is the force pushing current :)

Computers do not understand decimal notation. All information (both instructions and data) must be converted to a binary representation before the machine can understand it. We use the symbols 0 and 1 (binary notation) but the machine has a variety of physical representations it can use to encode binary data, including transistors, flux transitions, on/off switches and so on.

Watts = Volts X Amps. Thus watts and volts are related but neither is higher in energy. Energy is measured as power times times, in other words watts times seconds (called Joules) or kilowatt-hours (called Units).

No, the voltage is determined by the technology of the cell:

Ni-Cad 1.2 v

Alkaline 1.5 v

Zinc chloride 1.6v

Lithium 3 v

Extrinsic

With full double sideband AM the bandwidth of the modulated signal is twice that of the baseband information signal.

With suppressed carrier single sideband AM the bandwidth of the modulated signal is identical to that of the baseband information signal.

With vestigial sideband AM the bandwidth of the modulated signal is somewhere between the above two cases, depending on how much of the vestigial sideband is included.

the maximum permissible speed above which the generator is likely to damage or may cause circuit to fail.

yes.

If you use wire of certain resistance first the get the result and if u use other wire which is of more or less resistance then the voltage drop in the wire changes so the output also changes.

Yes it does, but the difference may be inconsequential. The specific circuit in question and type of wire should be included for a specific answer.

Ohms law is E=IxR. Kirkoffs law has to do with the distribution of voltage and amperes. The amperes in a series circuit is always the same and the voltage changes. In a parallel circuit the voltage is always the same but the amperes change. You can find all the values in a circuit with limited information using these two laws.

The difference between the p-type and the n-type semiconductor is that the p-type semiconductor has more holes than electrons while the n-type semiconductor has more electrons than holes.

The capacitors allow the signal to pass through, while 'blocking' the DC voltage level that the signal is 'riding' on. Are you asking to remove the capacitor and connect it straight through? If you had a multistage amplifier, then the DC riding voltage would try to get amplified as well, and the next stage amplifier would probably 'max out' and you'd wind up with just solid DC output, or components further along in the circuit could be damaged.

If you're asking if the capacitor was taken out (like if it blew) then no signal would get through.

in full wave bridge rectifier, the input and out put voltages are same but in case of two diode rectifier the input and output voltages can be different as per requirement a there is a transformer in the circuit.

The former is lighter and the later is heavier.

• avoid crosstalk
• safety, a short between them could cause hazardous voltages where none are expected
• protect low voltage equipment from damage
• etc.

The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.

Signal amplification is when receptor proteins interact with molecules known as signal molecules. This makes the signals stronger so that things get done more quickly.

the power will be reduced to the haflf because P=V2/R so when the resistance doubles the power reduced to the half of itsoriginal value

Amps times volts = watts Watts measures the rate of power usage.

watts times hours = watt hours Watt hours is a measure of the amount of power used.

Impossible to give a specific answer without more information. Is it a BJT or a FET?

Most general answer is you first have to understand the circuit using the transistor and what it's supposed to do. Then,

1. Check that the power supply voltages are correct.

2. Check that the DC biases are correct.

3. Check that the input signal is correct.

4. Check the output.

But you have to know in advance what to expect.

the electrons don't actually go anywhere, they move in a circular mode... the electrons bounce from one atom to another so that there is always balance... as soon as a valence electron moves to another atom another electron replaces the one just lost...

They will dim because you have more resistance in the circuit. More resistance means less current which means less light. The power is P=V*I = I/R*I = I²/R

Think of blowing through one straw. Now attach a second straw end on with no leaks between joints and then attach a third straw in a long line. Which one is the easiest to blow?

no the strength of the magnetic field does not decrease because of the number of coils increases.

You can make a voltage divider set of resistors and tap off the desired voltage. Some voltage regulators can be fed much higher voltages. The regulator will provide a constant settable voltage with short circuit and overload protection. There are DC to DC converters. They first turn the DC into AC then rectify it and produce DC.