On three phase 208 volts one leg does not have the potential of 208 volts. It takes two legs to provide the 208 volts. The potential is across AB, BC and CA. Voltage to the grounded neutral A-N, B-N and C-N will produce a potential of 120 volts. In a wye configured secondary three phase four wire you have the potential of 120/208 volts.
The primary factors in the calculation are: the type of occupancy, along with the available floor space and the available means of egress.
When creating or modifying a building, the building use and occupancy load determine the fire resistance requirements. Or the desired type and occupancy load may help determine the necessary area and means of egress.
Buildings are classified by the ICC (International Code Council) by group and division.
Groups - defined
Group A - Assembly
Group B - Business
Group E - Educational
Group F - Factory and Industrial
Group H - Hazardous
Group I - Industrial
Group M - Mercantile
Group R - Residential
Group S - Storage
Group u - Utility
Each group mostly has more than one division, categorized by number; I.E.
Group A Division 1 , which is the most stringent category for fire resistant materials and means of egress code rules, is more stringent than Group A
Like wise , Group U Division 2, which is the lowest requirements is less stringent
than Group U Division 1 requirements.
In summary, the lower the letter and number, the more stringent are the fire
resistive , and means of egress components.
To know the fire resistive construction and means of egress requirements of your building, consult the building code book for your state ; which may supersede
To be absolutely sure ,consult with your local building department with information about what kind of use the building is intended for , how big of an occupancy intended , and whether or not it is connected or near to another business of same or different classification.
The instrument used to measure electrical current is called an ammeter, which is actually a shortened form of 'amp meter'. The current is measured in amperes. In scientific labs, a much more sensitive instrument called a galvanometer is used to measure very small currents.
It's got spring loaded clips. Press to either side and down to remove the bulb.
CR2025 = Diameter: 20 mm. Thickness: 2.5 mm. Capacity: 160 mAh, CR2032 = Diameter: 20 mm. Thickness: 3.2 mm. Capacity: 220 mAh
50 Kv = 50,000 volts
Generally a single-phase transformer will have twowindings. One of the Low voltage side and one on the high voltage side. North-American distribution transformers will have three: one high-voltage winding, and two low-voltage windings connected in series.
...a single-phase transformer can also have several primary and several secondary windings. The primary windings can be connected in series or in parallel with each other, as can the secondary windings. For example, taking the primary winding as an example, it could consist of two 120-V rated windings: if connected in series, it could be supplied with 240 V without exceeding its voltage rating; if connected in parallel, it could be supplied with 120 V without exceeding its voltage rating. Multiwinding single-phase transformers allow for a variety of connections.
That depends on how much voltage is applied, because Volts x Amps = Watts. So find out your Voltage and multiply it by the amperage to get your wattage.
60Hz is standard for AC (hence the term Alternating Current) and used for long distance power transmissions and even household currents. Usually what is coming out of the socket is around 120 Volts (240 for appliances and other countries). So applying this formula to this is simple: 120Vx5A=600 Watts or 240Vx5A=1200 Watts. So on so forth...
Simply multiplying volts times amps gives what's called Volt-Amperes or apparent power. Volt-Amperes equal watts (real power) in largely resistive circuits. Circuits with motors, transformers and inductors have power factors that must be taken into consideration when calculating watts. For example, in a motor circuit with a power factor .6, the watts would be Volts X Amps X .6
Bathroom ceiling vent fans carry no warnings about continuous use. A fan in good condition, used in a good environment, will not heat excessively. However, a defective, older, or worn fan, or a fan that is binding or prevented from turning, or operated where airflow is blocked, can overheat, possibly dangerously. If a fan is making any kind of grinding or squealing noise, it should not be operated. A fan that does not turn freely should not be turned on. Such defective fans should be replaced, and it is good practice to replace any older ceiling fan (over six years) as a matter of course. They operate in harsh conditions, often ventilating moist or dirty, smoky air. A good quality, new fan will likely be quieter and more efficient, and it won't have the wear problems that can cause overheating. Potentially it is a fire hazard. Many vent fans use an open frame motor that depends upon air flow past it for cooling, and some vent fans are enclosed in plastic housings.
Over time, dust and lint can collect on this open frame motor, insulating it from the air flow and causing it to heat up. Eventually, if it gets hot enough, it can fail electrically and blow a spark. This spark can ignite the dust and lint, and if the housing is plastic the housing can ignite as well. This becomes a serious problem.
More here, about an actual case where this happened:
No, but close to it.
In regards to just the heating of the water, Tank-type, and tank-less, electric water heaters are about 99% efficient* because the heating elements are immersed in water, and little energy is wasted between heat source and water.
However, all tank-type water heaters loose heat from their reservoirs (tanks). This heat loss lowers the overall efficiency.
Most electric heaters fall between 90-95% total. This is the number that is usually reported by the manufacturers.
* Efficiency is not the same as cost savings. For example electric heaters have higher efficiency, but cost more to operate than same-size gas water heater. Also tankless gas heaters, gas water heaters with blowers, and gas heaters with high recovery may claim higher efficiency, but they consume more fuel per hour of operation.
Efficiency of heating elements does not consider inefficiencies in production, transmission, refining, and delivery of electricity. Estimates say coal-fired power plants waste half the energy of coal during electric generation. However hydroelectric power plants are very efficient.
Tank-type gas water heaters are 69% efficient, since hot combustion gas escapes out the flue after hydrogen content of fuel is burned. This number does not consider inefficiencies during production, transmission, refining, and delivery of gas.
Tank-less gas water heaters are 79% efficient, but tank-less burn more gas per gallon of hot water than tank-type heaters. Hot combustion gasses are released out the flue and are not actively recycled without slowing combustion which turns off unit. Newer type tank-less with blower recycles heat into a tank operate at 96% efficiency, and are made for oversize consumption of both energy and hot water.
High efficiency tank-type gas heaters operate at 96% efficiency by using blowers to circulate hot combustion gas through coiled tube located inside tank. Same amount of combustion gas is released since only heat is recycled. The use of electric blower affects overall cost.
Environmental efficiency: Electric water heater does not release CO2 at each home, instead CO2 is localized at power plant. Gas water heaters release CO2 at each home which factor in environmental efficiency.
Efficiency of all gas water heaters falls and cannot be fully restored if untreated hard water deposits sediment over heat transfer surface. Generally, gas heaters exposed to hard water require softener which affects overall efficiency.
Efficiency of tank-type electric water heater remains at 99% until sediment reaches element, causing element to burn out. Full 99% efficiency is restored by cleaning out tank and replacing element. Generally, tank-type electric heater does not require water softener.
Downsizing water heater, reducing consumption, selecting water heater with fewer parts, reading the manual, and draining tank each 6 months to avoid sediment are key to saving energy.
1 kilowatt-hour is 1000 watt-hours and 60 watt bulb consume during 1 hour 60 watt-hours of electricity, so then it costs 0.6 cent =>60/1000=0,06*price of 1 kilowatt-hour = 0.6 cent
There are several type of circuit breakers now a day we are using these are as follows:
1. M.C.B. (Miniature circuit Breaker)
Rating : 1, 2, 4, 6, 10, 16, 20, 25, 32, 63 Amperes
2. M.C.C.B. (Miniature current circuit Breaker)
Rating : 10, 16, 20, 25, 32, 63, 100, 200, 250, 400 Amperes.
3. A.C.B. (Air Circuit Breaker)
Rating : 400, 800, 1000, 1200, 1500, 1800, 2000 Amperes.
4. A. B. Switch (Air Breaker)
used in High tension line.
5. SF6 Breaker (Contact break in the Sf6 medium)
used in High tension line.
There is no voltage in three phase wire. The ability of wire to carry voltage is dependant upon the insulation that surrounds the wire. The thicker the insulation the higher the voltage potential can become. Three standard insulation voltages are 300, 600 and 1000 volts.
First of all, there is no such thing as a voltage 'on' a wire. 'Voltage' is another word for 'potential difference', so a voltage can only exist between two wires. Voltages in three-phase systems are generally specified in terms of their line voltages-i.e. the voltage between any two line conductors. These depend on the electrical standards used in the country in which you live. In the UK, for example, three-phase transmission lines will have line voltages of 400 kV, 275 kV, or 132 kV, while distribution lines will have line voltages of 33 kV and 11 kV, and low-voltage distribution line voltage is 400 V.
The most important thing for safety is to wire the pump to the correct voltage marked on the pump.
All else being equal (i.e. the cost of the pump), use 220 -- the higher the voltage the less electricity lost to resistance and heat.
ANS 2 Most modern pool or spa pumps are convertible (at the motor connection) Connecting to 220 is definitely more efficient.
Ans 3: 110 v is safer, with a lower risk of shock. For a pump of 1 HP or less the power lost in the wiring is very small if the right wire is used.
You cannot. It must be taken out of the circuit and then tested on its own.
That's not 100% true because, if it has wires at its ends, you can cut through one wire with an appropriate tool and then test the capacitor "out of circuit". If the capacitor is ok you can then re-join the two cut wire ends by applying a blob of solder carefully. (But, to avoid damaging the capacitor, use a suitable heat sink to shield the body of the capacitor from the heat of the soldering iron.)
With direct current a capacitor also works like a special type of resistance. Whilst being charged up, it will show low resistance. As it slowly (or quickly) charges, the resistance will grow larger and larger. Whenever I repair circuitry and I have doubts about a capacitor (in the uF area) I simply use my multimeter on its Ohms setting. If a capacitor has shorted, then the result will be 0 Ohm. If the capacitor is working, or partially working, the resistance will gradually increase until it is out of range of the multimeter.
Use an ohm-meter first to test the on-board capacitor and then use it to test a similar capacitor off-board, to see if the results sort of match up.
Most often they will not match completely as on-board you also measure the effect of all other components connected into circuit with the capacitor. It might point you in the right direction though.
On a separate thought, if you really cannot remove it, or disconnect one of its connections, then why test it at all? If it really can't be removed to replace it, then it makes no sense to test it!
A capacitor can be tested using multimeter without removing it from circuit. but in order to check it, its polarities should be noted and then keep the positive terminal of multimeter on positive of capacitor and negative terminal on negative. It is vital to note that the readings will be affected by the remainder of the circuit. To test for capacitor function in circuit demands a good understanding of the circuit operation.
Of course there are ways to test capacitors, both in circuit and out. While a truly accurate test involved taking the cap out of circuit, a basic test can certainly be done in circuit.
Out of circuit, one can either connect to a VM, or better yet, an oscilloscope, and measure the time for voltage to decay to zero across the capacitor. This time should equal the time given by the equation for the time constant, and is dependant on the values associated with that particular capacitor.
For RC circuits, this equation equals:
Ï„ = R Ã— C. It is the time required to charge the capacitor, through the resistor, to 63.2 (â‰ˆ 63) percent of full charge; or to discharge it to 36.8 (â‰ˆ 37) percent of its initial voltage. These values are derived from 1 âˆ’ e âˆ’ 1 and e âˆ’ 1 respectively.
It is important to keep in mind that one must apply a voltage across the capacitor at its rated value. Thus, if it is a 400V capacitor driving a tube amp, for instance, it must be driven at around 400V. Driving it at 12V will lead to useless results.
The only proper way to check for a capacitor value and or leakage is with a proper test bridge: set it to the capacitor's DC rating with it removed from the circuit completely. Any other way is just waste of time.
Additionally, a common in-circuit test for a electrolytic capacitor is to measure its Equivalent Series Resistance (ESR) which can be done with an ESR meter. This is a quick and easy way to locate failing electrolytic capacitors, especially in power supply circuits.
An effective method of testing any component in-circuit is with an in-circuit curve tracer. If you have an oscilloscope with X-Y input mode you can easily build one of these on your own. They do take some getting used to before you can use it effectively and are most useful for good board vs. bad boardcomparison.
It depends. If it's an inductive ammeter (the kind that clamps around a wire), it won't work at all. If it is the type of ammeter that is actually placed in the circuit, it will work but it won't be accurate.
Actually, modern 'clamp on' ammeters WILL measure d.c. currents. It uses the Hall Effect to measure the current.
max of $30
You cannot convert Volts (or kv, 1000 volts) to watts (or mega watts, 1,000,000 watts) because volts are measure of electric potential difference between two points and watts are a measure of energy/time. However, WATTS = VOLTS x AMPS so... if you have 1000 AMPS flowing over a resistance/load with a difference in potential of 1000 volts (1 KV), you have 1,000,000 WATTS (1 MEGAWATTS) of energy consumed/time. So if a motor has 1KV potential accross its terminals and it is consuming 1,000 AMPS, it is a 1 MegaWatt motor (a large one indeed). To get energy, you have to multiply this 1MegaWatt x the time the motor runs and x a conversion factor to get to the appropriate unit of energy. Yes, I've been called a nerd before.
You have to be careful here. A heater will be advertised as "X" watts, but that is only true if you connect it to the voltage source it is supposed to be connected to. If you plug it into a higher or lower voltage source than intended, it will produce a different number of watts.
Electric heaters are just resistors. When you run electricity through them, they get hot. If you run more electricity through that resistor, it will produce more heat. If you run less electricity through it, it will produce less heat.
As an example, you can find "1500W/120V" water heater elements at the hardware store. This means that if you plug it into a 120V source, it will produce 1500W of heat, and it will pull 1500W/120V = 12.5A of current.
You can calculate the resistance of the heater by taking voltage times voltage divided by watts, so this "1500W/120V" heater is really just a resistor of this many Ohms:
120V * 120V / 1500W = 9.6 Ohm
That Ohm value is physical property of the device. It will not change. If you were to take this heater now and plug it into a 240V supply, you can calculate the amps with voltage divided by resistance:
240V / 9.6 Ohm = 25 Amps
And, for watts, you can take voltage times voltage divided by ohms:
240V * 240V / 9.6 Ohm = 6000W
Sorry for the long text, but it's crucial that you understand this.
If your heater is 1500W and is INTENDED to be running on 240V, you have a 38.4 Ohm resistor. Running that resistor at the lower 208V will produce only 1126W of heat and will pull just 5.4 Amps of current.
However, if your heater is 1500W and is indented to be running on 120V, then you have a 9.6 Ohm resistor. You will almost certainly start a fire if you plug it into a 208V supply, because you will be pulling close to 22 Amps and producing 4500W of heat.
A search on the Internet is probably your best bet.
A 1000 watt microwave at 120 volts draws 8.3 amps. Make the circuit a dedicated circuit, use a 15 amp breaker and a #14 copper conductor which is rated at 15 amps to complete the circuit path.
Use either DC to Dc converter or voltage regulators for the required voltages.AnswerA common method is to use a voltage dividercircuit. This comprises a number of resistors, connected in series, across the power supply. This creates a series of voltage drops across each resistor and, by choosing resistors of appropriate value, the desired load voltage can be achieved.
For example, if three identical resistors are connected in series across 9 V, then the voltage across each resistor will be 3 V, and the load can be connected across any one of these resistors.
In practise, selecting the appropriate values of resistance is more complicated than this simple example, because (1) the resistors themselves mustn't overload the power supply, and (2) the load itself, being connected in parallel with one of the voltage divider resistors, affects the overall resistance of the voltage divider circuit and must, therefore, be taken into account when designing the circuit. This is called the 'loading effect' and, to put it simply, its effect is minimised providing the resistance of the load is VERY much larger than that of the voltage divider resistor it is connected across.
This sounds like you're trying to run a 3-volt device off a 9-volt battery. I would do it in one of three ways.
First way is if there's nothing in the circuit that needs 9 volts. Your best bet here is to replace the 9-volt battery clip with either a holder for two AAA cells or two A-76 button cells. Trying to reduce the output of a 9-volt battery to 3 volts will produce lots of heat. You'll get a fairly short battery life, too. AAA cells are about the same size as a 9-volt; button cells are way smaller. There is a 3.3-volt regulator called the 7803SR - Murata makes it - but they cost $10 each.
Second way is if you need both 3v and 9v...in that case, I would install a pair of AAA cells alongside the 9v cell and wire the 3v supply to the circuit that needs it.
And third is if this is a flashlight. In that case it's even easier: replace the 3v bulb with a 9v bulb (yes, there are 9v flashlight bulbs) and be happy.
Typical pricing of public utility will be always less than gas generator.
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