Electrical Engineering

Yes.

The output voltage will not always be 240 volts - it will fluctuate. Plus a typical wind turbine throws out some harmonics due to the AC/DC/AC design necessary to convert it from speed of blade rotating to a constant 60 Hz.

If you are building this all yourself, for you own home use, and especially if you wish to disconnect from the grid, It might be best to build the turbine to AC/DC and charge a battery bank. A reverse charging protection schemed needs to be set up for when the DC voltage dips below the battery voltage (when the wind isn't blowing). A large home-sized DC-AC converter would be used to power the home off the battery bank, then. This scheme would protect the home from the harmonic output from the wind turbine (which could damage sensitive electronics like computers, LCDs, etc.).

It makes and breaks four separate contacts. Think of a switch for example that connects one wire to one wire. A 4 pole contactor is like 4 separate switches going on together.

A knockout is a pre stamped hole in a distribution panel. The metal slug in the pre stamped hole requires knocking or prying to remove it from the panel. Once the slug is removed from the hole, a cable connector is fitted through and secured with a lock nut on the inside of the panel. A wire cable is then threaded through the cable connector and held securely in place by the connectors wire clamp.

ic voltage regulator is good for line or load regulation

switching on the tourch for example is when as you switch it on it will move the switch connecting it to a circuit that lights up the bulb and when you switch the switch off the switch will move breaking the circuit

the quick answer is you cant. squirrel cage induction motors need AC power to generate the magnetic field in the rotor...if however you have a motor that has brushes in it (drills, mixers,etc) then this motor can be run off ac or dc..the commutator brushes do the work of moving the current to the proper coils at the proper time.

It produces a conscious and recognizable disruption of a patient's speech production and language output

A: Overload is a term used to describe a condition whereby a circuit is overstressed by additional loads imposed to it. Overload can be as small quantity or a very big quantity it depends on the circuit as it is stressed.

A local breaker backup relay is used to check the operation of distribution circuit breakers and to trip the feeder circuit breaker if the distribution circuit breaker fails to trip on an overload.

A capacitor can charge to its' maximum OR the voltage applied to it, whichever is LESS.

one insulator unit can bare 11kV. such that 66kV tx line needs 6 insulator units.

Cold climate starters are capable of delivering higher current during cold start. Cold start system is always a combination of the starter motor with a suitable battery.

Generally inventory turnover period is calculated as: Sales/Inventory Also by, Cost of Goods Sold/ Average Inventory

To charge a battery, the charger voltage must be higher than the battery voltage. If a AA battery (or any other) has a normal voltage of 1.2V the charger voltage must be at least 1.2V. The type of voltage supply and its current capability is immaterial. No, the charging voltage have to be larger than the battery voltage, to charge 2 AA batteries, that is 3 volt if connected in series, so a voltage of at least 4 volt is needed

Carbon, due to it's high melting point.

First, we assume you are asking for the resistance of the bulb. Second, there can be only one possible resistance. The bulb has one and only one resistance. The equation for solving this problem may be a quadratic or even a cubic equation, which may have multiple roots, but only ONE can be a real-world answer. Third, let's list everything we know about the circuit. Let's label the power source Vcc. We'll label the known resistor R, and the resistance of the bulb Rb. We'll label the power dissipated by the bulb as simply P. We'll also label the current flowing through the bulb as I. What do we know about power? The power dissipated by a resistor is given by three well-known formulas: P = IV, P = I2R, or P = V2/R. (The second and third are derived from the first.) The second one will come in handy for us right now. We can write: [Equ. 1] P = I2Rb What else do we know? Since it's a series circuit, the current flowing through the bulb, I, is the same as the current flowing through the resistor. And we know from Ohm's Law that I = V/RT, where RT is the sum of the two resistances in the circuit: R + Rb. So, we can write: [Equ. 2] I = Vcc/(R + Rb) So, substituting for I in Equ. 1 with the I we found in Equ. 2 we have: P = [Vcc/(R + Rb)]2Rb = [Vcc2/(R + Rb)2]Rb = Vcc2Rb/(R2 + 2RRb + R2b) Phew! The algebra's getting a bit hairy. We need to do a few more manipulations. We can swap the P on the left side of the equal sign with the denominator of the fraction on the right side of the equal sign, so we write: R2 + 2RRb + R2b = Vcc2Rb/P We now multiply both sides of the equation by P and we write: PR2 + 2PRRb + PR2b = Vcc2Rb Stay with me now; we're almost there. If we subtract Vcc2Rb from both sides and gather similar terms, we can write: PR2b+ (2PR - Vcc2)Rb + PR2 = 0 We can now substitute all the known values into the equation above. We know Vcc = 131, R = 133, and P = 22.9. So, we can write: 22.9R2b - 11070Rb + 405078 = 0 So, we have quadratic equation, which will have two roots for Rb. Using the quadratic formula, we determine that there are two real roots: Rb = 443.5 and 39.9. The second result is thrown out since it is not consistent with Equ. 1 and Equ. 2.

resistors - and they have to be physically large enought to dissipate the heat of the voltage x current you will drop out.

The EMF of a cell is the voltage across the terminals at zero current. This is the quoted cell voltage but as soon as a current is dawn from the cell, the voltage will drop. It's due to the internal resistance of the cell.

In a circuit diagram, a cell is often shown as a voltage source (a perfect source) and a resistor in series to represent the internal resistance. Using Ohms Law, it can be seen that as soon as a current flows, a voltage will be developed across the internal resistance, so reducing the voltage that is seen at the terminals of the cell. The higher the current draw, the higher the voltage drop inside the cell.

Normally, the voltage drop is minimal but in most cells, as it loses charge, the internal resistance rises. Eventually it will reach the point where most of the voltage is dropped across the internal resistance, leaving little to drive the intended load. Often, if a battery is removed from a device and measured, the voltage will be measured as equal to or very close to the quoted cell voltage. It is easy to make a judgment that a battery is good when it is almost dead. The only way to confirm the state of the battery is to measure the voltage at the terminals while the load is attached. The results can be very different to the off load voltage.

Alkaline cells have a low internal resistance compared to other dry cells. This makes them well suited for high current drain applications. The internal resistance also rises more slowly than most other cells, so they remain useful far longer than zinc-carbon types.

When is a parallel circuit beneficial

Kirchoff's Current Law: The signed sum of the currents entering a node is zero.

Assume the top of the battery is a node. The current entering it (from the battery) is equal to the sum of the currents leaving it (to the branches). This adds up to zero.

Regulation has to do with keeping the voltage at a desired value. This is often put in a per unit value, which you can think of as a percent of nominal. So if the desired voltage is 34.5kV, and the actual voltage is 33.8kV, the voltage is .98 per unit. There are regional and national rules on how low or high voltage can be. If the voltage strayes from this, it must be regulated back to an acceptable value.

14 micro farads @ 1.4 seconds.. idk what it is at 2 seconds ;p

The Chinese city of Shanghai is reported to have a Maglev train in commercial use. The Japanese and Germans are working on Maglevs so are the USA.