Electronics Engineering

Because real inductors, as well as the wires used to connect them to the

capacitors and to the excitation, have some resistance, and where there is

resistance, there is dissipation of energy.

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A: As temperature increases its resistance increases. Like all silicon diodes it will reach a point where the temperature coefficient is zero but it is at such elevated temperature to make it invaluable

A multimeter - is a piece of testing equipment used in electronics and electrical diagnostics.

we will get rectified voltage at the output.in case we replaced the DC source with AC source

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There will be 20 amps flowing through the 5 ohm resistor. We could do some math and all to figure voltage drops and some other things, but let's cut to the chase and see what's happening. Based on what we know about series circuits and about parallel circuits, we can shred this in nothing flat. So let's. First, we're told 30 amps flows in the circuit. That's the total current, and it will be the current through the first 10 ohm resistor. It has to be. The the 30 amps "splits" to flow through the parallel network of the 10 ohm and 5 ohm resistors. That's 30 amps that has to "split" and some will go through the 10 ohm resistor and some will go through the 5 ohm resistor. With me? Sure. Now for the "trick" here. Since the 5 ohm resistor has only half the resistance of the 10 ohm resistor, twice as much current will flow through it as the 10 ohm resistor. Make sense? Yup. Let's finish this. Since there is twice as much current flowing in the 5 ohm resistor 'cause it's got half the resistance of the 10 ohm resistor, if we have "x" amount of current flowing in the 10 ohm resistor, then we'll have "2x" amps of current flowing in the 5 ohm resistor. That's "3x" amps total, and the "3x" amps equals 30 amps. See through it now? There will be 10 amps flowing through the 10 ohm resistor, and 20 amps flowing through the 5 ohm resistor. Piece of cake.

A: Ico is directly related to temperature that is leakage that will offset the loading point .

You may find it helpful to use Ohm's law and the definition of electrical power.

Any variation of the charge within a p-n diode with an applied voltage variation yields a capacitance wich must be added to the circuit model of a p-n diode. The capacitance associated with the charge variation in the depletion layer is called the junction capacitance, while the capacitance associated with the excess carriers in the quasi-neutral region is called the diffusion capacitance. Both types of capacitances are non-linear so that we will derive the small-signal capacitance in each case. We will find that the junction capacitance dominates for reverse-biased diodes, while the diffusion capacitance dominates in strongly forward-biased diodes. The total capacitance is the sum of both.

No one individual was responsible for improving both (or even either) technology. Radar and sonar were worked on by many investigators over decades to bring them along to the point where they are the effective, reliable tools we look to as we attempt to look at things in the sky, below the water and in the ground. What is arguably the most effective technology in bringing radar and sonar forward is the advancements in electronics, and in computer technology.

A diode blocks current flowing in one direction, and passes a current going into the other direction (with a little potential energy loss).

DC is a current flowing into one direction all the time. therefore if you put a diode in a dc circuit it will either always block or always let it pass (provided a certain amount of supply voltage).

if you have an ac circuit the current alternates (ac=alternating current) most of the cases the current flows half of the time in one direction and half of the time into the other direction. so you will have both cases 50% blocking 50% passing through.

so if you put an LED into an ac circuit, no matter what way you put it in you will always see light.

The side

A: Extremely important since at the level there is infinite current available at zero volts, In reality it can never be ac hived but it can be assumed for consideration

Each piece of hookup wire can be thought of as a zero-ohm resistor. In fact even the hookup wire has resistance but it's so small it it can be neglected.

The simplest form of an electrical circuit is when an energy source (such as a battery) is connected to a load (such as a light bulb or a resistor). The connections must be made in such a way as to allow the energy to flow from the source, through the load and back into the source to form a loop.

Forward saturation in a BJT occurs when the ratio of collecter-emitter current and base-emitter current reaches hFe or dc beta. A that point, the BJT is no longer operating in linear mode.

Electrical voltage can be changed by a transformer.

More voltage than required will light something such as a bulb.

However, you have to consider such things as bulb longevity. Anything over 12 volts will illuminate a 12 volt bulb. If it's connected to an engine alternator and not a pure clean 12 volt power supply, ideally 13.5-14 volts is the voltage you're looking for to power this bulb.

16 volts will work, but it may prematurely burn the bulb out.

That is, unless it is an LED. Even overpowered they will continue to burn bright, but not burn out, but it won't last as long as if it were connected to a true 12 volt supply voltage. But in the end, even an overpowered LED will last longer than an overpowered 12V incandescent bulb.

It will definitely burn out prematurely if a higher than design intent voltage is supplied, however the hours. minutes, seconds lost may not be as high as people may assume them to be, from 12v to 16v is not such a jump especially in light of most automotive applications having output voltages up to 14.8v a car battery typically has 6 x 2.2v cells making 13.2v

Green - Black - Red, but this is not a standard value. You probably want a 5.1K resistor, and that is Green, Brown, Red

The time constant is precisely the product R times C. The result of this product will be a time; it represents the time for the capacitor to discharge to about 37% (1/e, to be precise) of its initial voltage.