Electrical Engineering

Differentially compounded motors are commonly used in applications requiring precise speed regulation and consistent torque under varying loads, such as in electric traction systems, cranes, and elevators. They are particularly beneficial in situations where the motor needs to handle significant load changes without substantial speed fluctuations. Their design allows them to maintain stable performance, making them suitable for industrial machinery and heavy equipment.

I have found that to fix an electrical problem it is best to break it down into smaller pieces. Example: Do I have power between A & B. Yes, OK Do I have power between B & C. Yes, OK Do I have power between C & D. No, Why? Found bad connection - problem fixed.

(a) what is the total capacitance of this arrangement (B) the charge stored on each capacitor (C) the voltage across the 50 micro farad capacitor and the energy stored in it. 20v and 20+30+50 micro farad

Inverse proportion is when one value increases, as the other decreases. For example the number of builders building a wall and the number of days it takes to build a wall are 2 quantities in inverse proportion because the more builders there are, the less time it takes to build the wall (there are more people working on it, so it will be built quickly). As the number of builders increases, the number of days to build the wall decreases. This is inverse proportion.

Energy as heat

I'm sure I can answer this if given more details, what circuit? current is determined by voltage and impedence(resistance) in a purely resistive circuit it is pretty basic, in alternating voltage circuits the impedence is dependent on the frequency of the voltage/current. inductors and capacitors are designed into a/c circuits to force the load to behave like a resistive circuit which boils down to how much voltage divided by how much resistance you have, 120 volts divided by 10 ohms equals 12 amps. in a parallel circuit it works pretty much the same way voltage placed on the resistor divided by its resistance equals the current flowing through that leg. more details please.

It needs to be in good working order. If there are any cracks or tears in it, then it might not work properly.

Well if you are talking about phases, it would be half of 208V.

No relation is there power factor is a unit less quantity.

The default segment for SP (Stack Pointer) relative memory accesses in the 8086/8088 is SS (Stack Segment).

A: The coil is really a transformer. The 12 enter the coils and charges with current as the point opens up this charge is released as energy about 100 mw. This release is transformed to hi voltage at the secondary by a primary to secondary turn ratio like 1:3333 more or less. Since the primary is open the same ratio will feed the primary that is why we see ringing and the point have capacitor to protect from false firing.

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The coil is a transformer, but we know that DC will not work with a transformer. A supply of 12 volts DC is applied to the coil's primary winding. In series with this coil is a set of breaker points. These points are used to interrupt the 12 volt supply from the coil. When the 12 volt supply is interrupted, the magnetic field around the primary winding collapses. This collapsing field induces a voltage into the secondary coil of the transformer. The secondary coil has many more turns that the primary coil and the voltage induced will be in the 10,000 to 15,000 volt range. This high voltage is needed to make the voltage jump across the air gap in the spark plug.

yes

Yes. Measuring resistance involves imposing a small current through a device and measuring the voltage drop. If there were any other source of voltage, the reading would be in error. You could also damage the ohmmeter if there were an external voltage source. You should also take the device out of circuit before measuring it, so as to not bias the results by the resistance of other devices in the circuit. It all depends on the situation, for which you are responsible for evaluating before you start.

The law of induction says voltage = N d/dt (flux) which for a fixed single frequency AC becomes: voltage (rms) = 2 pi x freq x N x flux (rms), or

volts per turn = 2 pi x freq x flux.

Therefore with 50 Hz and 0.06 Webers of flux the volts per turn is 2 pi x 50 x 0.06 which is 18.85 volts per turn.

The HV winding need 318 turns while the LV winding needs 13 turns.

With a typical peak flux density of 0.7 Wb/m2 rms the cross section area of the iron core would be 0.086 sq. metres or just under 1 ft-square.

The 208 volt configuration is one phase of a three phase source at 240 volts, where the 208 volt circuit is connected between the center tap of one 240 volt phase (usually a grounded neutral, in the style of a standard 120/240 split phase system) and the high delta connection on either of the other two phases.

208 circuit would consist of two phases of a 208 volt wye system or could be all three phases. The voltage between conductors would be 208 volts. The voltage to ground from any phase would be 120 volts. A 240 volt delta system would give you a high leg to ground, somewhere around 190 volts and the other two would be 120 volts to ground.

If you have three 100 ohm resistors, and you want an equivalent resistor of 66.7 ohms, put two resistors in series, and then parallel the third resistor across the first two. Resistors in series: R1 + R2 Resistors in parallel: R1 * R2 / (R1 + R2) This example: Two 100 ohm resistors in series: 100 + 100 = 200

A 100 ohm resistor in parallel with a 200 ohm resistor: 100 * 200 / (100 + 200) = 66.7

It depends on what you mean by '5 mva'. Because you haven't used the correct symbol, no-one will know whether you mean '5 millivolt amperes' or '5 megavolt amperes' -there's a HUGE difference! The correct symbol for 'millivolt ampere' is 'mV.A' and the correct symbol for 'megavolt ampere' is 'MV.A'. I hope you can now see the importance of using the correct symbol.

To answer your question, the load volt amperes should not exceed the transformer's volt amperes. So, in your case, the load shouldn't exceed 5 mV.A or 5 MV.A -whichever you mean.

Transformers are inherently bidirectional.

0. An ammeter is placed in series with the circuit in question; if its' internal resistance is high, it will change the current flow, thus making the measurement meaningless.

For the same reason an ideal voltage meter will have infinite resistance.

increases

Eddy currents are, in general, stray currents of electricity that are created (by induction) in conductors. Any time a magnetic field "sweeps" or "moves across" a conductor, induction occurs, and tiny voltages (with their tiny currents) are set up. These currents, in turn, create small magnetic fields that oppose the magnetic field that induced the eddy currents in the first place. Said another way, any time a voltage is induced in a conductor, eddy currents flow and counter-electromotive force (counter emf or back emf) is induced which opposes the original field. We sometimes have a little narrower view of things in electronics. In many applications, eddy currents are tiny currents in things where we don't want them and can do something to minimize them. In a transformer, the primary windings induce voltage in the secondary winding. The expanding and contracting magnetic field of the primary windings is said to "sweep" the secondary windings, and induction will occur. Voltage will appear in the secondary windings. We often wind our transformers around cores of ferromagnetic material. This material "conducts" magnetic lines of force and makes our transformer much more efficient. But the the magnetic field that sweeps the secondary also sweeps the core, and it induces voltage there, too. (Even though we don't want it to.) The voltage induced in the core causes some small amounts of current flow, and these small currents are called eddy currents. Transformers often use laminated iron for core material. Or they will used what appears to be powdered iron grains that are tightly compressed and stuck together. Both these tactics are designed to provide some insulation between pieces of material in the core, and this will considerably reduce eddy currents. Need a link to the Wikipedia article on eddy currents? You got it.

For voltage drop calculation you must use the current of the load and the distance of the load from the supply source. The circuit breaker is then used to limit the current to the rating of the wire that feeds the load.