Electrical Engineering

Actually, everything follows Ohm's Law, you just have to consider dynamic resistance, i.e. the resistance at the precise operating point of the measurement.

A bulb has a positive temperature coefficient, or negative, depending on your point of view. My point of view is that, as temperature increases, the resistance increases, meaning a positive temperature coefficient. An alternate point of view, perhaps the common point of view, is that, as temperature increases, power decreases, meaning a negative temperature coefficient. Pick whichever point of view you want to use.

A typical 60 watt 120 volt bulb might have a 16 ohm cold resistance. (I just measured one.) If it kept that resistance, it would have an operating current by Ohm's law of 7.5 amperes, and a power of 900 watts. However, we know that it runs at 60 watts. Back calculating, that is a current of 0.5 amperes, and a resistance of 240 ohms hot. That a big difference from 16 ohms cold.

Now, that explains temperature coefficient. Lets look at compliance with Ohm's law, and the "theory" that some things do not comply...

Ohm's Law states that voltage is current time resistance. Period. It applies in every case. No exception. The problem is that you have to consider all of the parameters when you take that measurement, and the mistake people make is in assuming the resistance is constant. It is not. Not even for a simple resistor.

If you place two 60 watt bulbs in series across that same 120 volt supply, you will discover that they will run somewhat cooler, and that they have slightly less that 240 ohms resistance, but you can still calculate that resistance and you can still apply ohm's law. In this case, you will find that the power would be slightly more than 15 watts each, not the exactly 15 watts that you would expect based on constant resistance.

On a side note, reinforcing this concept, lets look at the forward bias characteristics of an ordinary silicon diode. This is a point of much contention, but I stand my ground on this...

As you increase the current through the diode, the voltage increases slightly. If it were a perfect diode, the voltage would not increase. If it were a resistor, it would track very close to Ohm's Law. Since it is a diode, it is somewhere in between. Does it obey Ohm's Law? Yes, it does. You simply pick a point, measure the conditions, and you can find the equivalent resistance of the diode at the point of current, and you will find that Ohm's Law will tell you what the voltage should be.

The same thing applies for AC circuits with inductive and capacitive reactance. Now we start talking about impedance and phase angle but, guess what, Ohm's law still applies - you just have to take everything into account.

In fact, not everything follows Ohm's Law. An incandescent lamp's filament (tungsten) is an example of a material that does not follow the law. This is because the current flowing through the filament is not proportional to the voltage applied to it. For a material to obey Ohm's Law, there mustbe a linear relationship between current and voltage.

In fact, Ohm's Law makes no mention of resistance. It merely states that 'the current flowing in a circuit is directly proportional to the applied voltage, provided external conditions, such as temperature, remains constant'. No mention of resistance; full stop.

The so-called 'Ohm's Law equation' only represents Ohm's Law when the resistance is a constant. Under all other conditions, the equation simply indicates what the resistance happens to be for specific values of voltage and current -it does not confirm Ohm's Law is being followed.

A star-delta motor should be supplied star-delta.

Wye is simply another name for star. Star-delta motors use a star or wye configuration to start, and a delta configuration to run. This reduces the voltage on the windings during the high current starting time.

In order to bias a bipolar junction transistor on, you need to forward bias the base-emitter junction at the same time you forward bias the collector-emitter junction, and the ratio of collector current over base current must be somewhat less than hFe, the transistor's gain. This is known as saturated, or non-linear mode, operation. In practice, we drive the base much harder than the calculated required current, so as to minimize dependency on varying hFe's for various transistors.

Turning the transistor off is a simple matter of eliminating the base current.

In the case of the NPN transistor, the base and collector would need to be more positive than the emitter. In the case of the PNP, they would need to be more negative.

At 440 volts, a 40 hp motor pulls about 68 amperes. Well, sort of...

The definition of a horsepower for an electric motor is 746 watts. Multiply 746 by 40, and divide by 440, and you do get 68 amperes.

1hp=746W

hence, 40hp=29840W

given that the motor draws 440volts

but current = power/voltage.

therefore, current=29840/440 = 67.81818182 amps.

The complexity, however, is that a motor is an inductive load and it has reactance. This reactance will change the phase angle between voltage and current, causing current to lag voltage by some amount. Power factor is a measure of the ratio of apparent power to true power, and is essentially the cosine of the phase angle between voltage and current. This power factor will make the true power be higher than the apparent power, but horsepower is based on apparent power.

Bottom line is that the amperes in this case will be dependent on the power factor, perhaps about 75 amperes, maybe.

Also, such a motor is bound to be a three phase motor. The amperes will be spread out among the three phases so, for 68 amperes, it would be about 23 amperes per phase, and for 75 amperes, it would be about 25 amperes per phase.

In terms of power factor, it might be closer to unity than anticipated, if the motor has power factor correction capacitors installed.

Keep in mind that starting current will be much higher than running current. Also note that large motors of this type would typically be started in wye configuration, so as to present lower voltage across the windings, and then run in delta configuration, for full power operation.

I'm going to stick with about 25 amperes per phase, full power run mode, unless someone else has something to add, of course.

An added complication is that '40 hp' refers to the output power of the motor. Unless you know the efficiency, you cannot work out the input power. Incidentally, 68 A single phase doesn't work out at 25 A per line for a 3-phase motor!!!!! It works out a little over 39 Aper line, because P = (root 3) x U x I x power factor!

Yes. A capacitor stores charge from any source, including AC.

The difference between DC and AC, however, is that the capacitor will be constantly changing in charge, in step with the AC. Due to the nature of the capacitor, the current through the capacitor will lead the voltage by some amount, depending on capacitance and resistance. {In the ideal case of a perfect capacitor, conductors, and AC power source, the current will lead the voltage by 90 degrees phase angle.}

This is called capacitive reactance.

Another way for a capacitor to store charge from an AC source, of course, is to place a rectifier diode in front of the capacitor. This, then, becomes an AC to DC converter.