Electrical Engineering

You can measure the electrical current with an amp meter. Amperage measures the current flow.

An autorecloser is a circuit breaker equipped with a mechanism that can automatically close the breaker after it has been opened due to a fault.

They are used in coordinated protection schemes for overhead line power distribution circuits. These circuits are prone to transitory faults such as nearby lightning strikes, wind-borne debris, squirrels climbing insulators, and the like. With a conventional circuit breaker or fuse, a transient fault would open the breaker or blow the fuse, disabling the line until a technician could manually reclose the circuit breaker or replace the blown fuse. But an autorecloser will make several pre-programmed attempts to re-energize the line. If the transient fault has cleared, the autorecloser's circuit breaker will remain closed and normal operation of the power line will resume. If the fault is some sort of a permanent fault (downed wires, tree branches lying on the wires, etc.) the autorecloser will exhaust its pre-programmed attempts to re-energize the line and remain tripped off until manually commanded to try again.

Autoreclosers are made in single-phase and three-phase versions, and use either oil or vacuum interrupters. Controls for the reclosers range from the original electromechanical systems to digital electronics with metering and SCADA functions. The ratings of reclosers run from 40 amperes at 2400 volts up to 1200 amperes at 35000 volts.

Not legally if one wanted to do the modifications themselves. All sorts of liability issues involved including would your property insurance cover damage caused by a non-UL listed device.

Perhaps a manufacturer has made a UL listed 240 volt dimmer. Check the major electrical device manufacturers websites.

In a series circuit each light completes its part of the circuit and connects to the next light. So, if one light fails, the circuit is broken and the flow of current to all lights must stop.

A farad is a unit of measure of capacitance. A capacitor of C farads charged up to V volts has charge of Q = CV (measured in coulombs). But this is still not power. If the capacitor is then completely discharged in say two seconds half of the power is provided than if it is discharged twice as fast (in one second). This is current. That is I = C dV/dT (amperes). dV or delta V = V if it is completely discharged. dV = 0.7*V if it is discharged to 0.3*V for example. Power is the current multiplied by the voltage (watts). So if the capacitor is completely discharged in T amount of time the power delivered is P = C * V * V / T. Or if the capacitor is charged up from zero charge to V volts in time T then the formula above is the P consumed to charge up the capacitor of C farads. You can think of capacitance as volume as in liters, cubic centimeters, or gallons. Using this analogy current would be flow rate such as gallons per minute, cc's per minute, or liters per second. I am not positive on this but I think voltage in this analogy could be thought of as pressure such as psi or bars. So power would be something like gpm * psi. In this analogy you are asking how do you convert gallons to a usable unit of power.

An AC supply is not used to magnetize a bar of steel because the alternating current continuously changes direction, causing the magnetic field to fluctuate. This results in the magnetization being ineffective, as the steel would not retain a stable magnetic orientation. For permanent magnetization, a direct current (DC) supply is preferred, as it creates a steady magnetic field that can align the domains in the steel and maintain the magnetization.

Water heater. anything frost free, a bad thermstat if you have electric heat one possible thing to check is the service wire connection coming into the house at the meter base, corrosion on the contact lugs can cause excessive resistance/power usage.

It's the 40W tube! because it's nonlinear: indeed it generates harmonics which increase the apparent power and thus the apparent energy.

It depends on the electrical standards in your country. For example, in Europe, the nominal voltages for low-voltage three-phase, four-wire, distribution systems is 400 V (line-to-line) and 230 V (line-to-neutral).

No, there is not. Three phase ,or generally speaking multiphase, circuits are available only for AC (alternating current) systems, because there is no phase shift in DC (direct current) systems. This is because DC power is not sinusoidal (current flows in one direction).

Use this formula to find amperage when kW is shown. Amps = kW x 1000/1.73 x Volts x pf. Use a power factor of .85 for the formula. Once the current is calculated re-ask the question for cable size for the calculated current.

block diagram

A switch

That depends how they are connected. For example, if they are connected in series, just add the individual resistances.

It is virtually impossible to measure this, because it is constricted by time, season, locational load growth, extreme weather (including specific locations), and many other factors.

Here is some information to give you an idea of how this must be calculated, and why it is so difficult:

Transmission losses are also known as I^2 * R losses, because the losses are due to the inherent resistivity of power lines. Power "lost" in the transmission system = I^2 * R for each particular line. Power line resistivity is effected by how balanced the loading is (this is assumed to be perfectely balanced in many locations), the configuration of the three phases to each other, the type of conductor, transposition vs. non transposition, and the list continues, so each individual line in the power system must be taken into account. The other factor is the actual amount of current that is flowing, which is dependent on the amount of load at a particular time (which is not constant throughout the day).

Now, say one part of the country experiences severe weather, damaging several major power lines. When this occurs, local utilties often rely on neighboring utilities to provide power. This power may have to travel through more lines, incurring more transmission losses, than it typically would.

What if several lines are out of service due to work (reconductoring, replacing insulators, retentioning the line,...). This will cause different transmission losses for the power that would "typically" flow through these lines.

I live in a farming area, where lots of power is used for irrigation. If the season is unusually rainy, farmers will irrigate less, which allows the local utilities to sell power to other utilities outside our typical service area, which results in different transmission losses than normal.

If a power plant is added or removed, this impacts where power is flowing from, thus reduces / increases transmission losses to specific loads.

You can start to see the complexity involved in your question. I have never seen a number quoted for this, since it would be at best an order of magnitude estimate, and at worst (say, calculated during an "abnormal" year) completely worthless, even though the amount of data necessary to calculate this, and the inherent added costs (adding high precision meters literally everywhere, collecting the data, and calculating losses), would be extreme.

Assuming that the voltage between two nodes are the same, current changes with branches. If you add branches, this means the current will be divided, depending on the resistance strength. Greater the resistance, the lower the current. If there is no resistance, it will end up in a short circuit, getting all the current. If there is great resistance, there will be very little current passing.

Mathematically, V=I*R is the formula. If you know the voltage between two nodes, you can calculate the currency (I) in each branch and for the whole nodes by putting in the particular resistance value for the branch or the whole nodes.

Use the symbols "I" for current, "E" for voltage, "R" for resistance.

One of the forms in which Ohm's Law can be written is:

I = E / R.

Applying this equation to the question:

I = 5 / 8 = 0.625 Ampere = 625 milliamperes

The last color code on the resistor is the tolerance marking it can be from zero up to 10% and higher , then there is other resistances that will influence the measurement, like meter probe tips and even the resistance in the probe wire especially on lower value resistors one can test it by connecting the two probes it is almost impossible to get a zero reading

The two main reasons are (1) the ease and efficiency by which a.c. voltages can be changed, using transformers, and (2) the ease and efficiency by which a.c. can be changed into d.c., using rectifiers.

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*Correction - Each maglev train is unique and may include any number of cars in its consist, but typically a relatively small number simply due to platform lengths and power limitations. So this number may range from 2 to as high as 20, but expect 2 to 8 cars depending on the system.

A transformer is a device to convert high voltage AC to low voltage AC & vice-versa. It works on the principle of induction. Since induction occurs only in an AC supply, a transformer cannot work on DC.. So, it is not possible to convert DC supply to AC using transformer. You would need an inverter to convert DC to AC.