AC-DC Inverter DC24V to AC110-220V

Ref Price:

Loading Port:: Shanghai

Payment Terms:: TT OR LC

Min Order Qty:: -

Supply Capability:: 10000pcs pc/month

Inquire Now

Add to My Favorites

OKorder Service Pledge

Quality Product, Order Online Tracking, Timely Delivery

OKorder Financial Service

Credit Rating, Credit Services, Credit Purchasing

You Might Also Like

SGB 10H grade insulating dry-type power transformer

Oil-immersed power transformer of 35kV

Power Inverter DC11-15V AC100-120V, AC220V

Oil immersed Power Transformer 66kV

Oil immersed Power Transformer 110kV

Oil immersed Distribution Transformers 6-10kV

Power Transformer 100kV

S9-M-50~2500/33KV Three Phase Oil-immersed No-excitation Voltage-regulating Power Transformer

1. Application

AC/DC inverter is designed for switching DC24V to AC110-220V. They can be usedoutdoors or on home appliance as emergency power. Usually, cord with AC powersocket and the DC power cord with DC battery. When the load connects to the DCpower cord with DC power, if the commercial power falls, the inverter will convert thebatttery voltage to AC voltage and go on supplying. When the commercial power restall,the inverter will switch to charge the battery by itself. When the battery is fully charged,it will stop automatically.

2. Feature
AC voltage available for AC 100V, 110V, 120V, 220V, 230V or 240V at request.Each unit contains two AC sockets available for two pin plugs.Automatic function change for inverter or batttery charge.Selective 4 AC voltage for each unit, such as when AC 110V, it will be available for AC110V, 120V, 130V and 140V.Different led indicator for function of inverter and charger.Each unit contain a AC cord for AC input voltage besides.Model ZUP-300 and ZUP-300A unit contains an extra power cord for DC battery.Products applicable for:
A. Personal computer.
B. Various video/audio equipment. (TV, casset, tape record, etc.)
C. Small motor equipment.
D. Various lighting equipment. Protection:
a. Protection for short circuit and polarity reverse of battery.
b. Overload fuse protection for charging current, input voltage and output voltage.

3. Specification

Type	DC Voltage	Capacity	Charging Current
ZUP-300VA	DC12V, 24V	max.300W	max.25A
ZUP-500VA	DC12V, 24V	max.500W	max.35A
ZUP-1000VA	DC12V, 24V	max.1000W	max.35A
ZUP-1500VA	DC12V, 24V	max.1500W	max.45A
ZUP-2000VA	DC12V, 24V	max.2000W	max.60A
ZUP-3000VA	DC24V, 48V	max.3000W	max.90A
ZUP-5000VA	DC24V, 48V	max.5000W	max.150A

Two outlets are incorporated for the simultaneous use of both the 110V and 220V outlets.The sockets are of a type that is capable of accepting two different types of plugsprongsand blades of two different forms.Built-in overload protection circuit and overdischarge protection circuit.

Q:An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 8.00 V. The RMS current delivered to the answering machine is 570 mA. If the primary (input) coil of the transformer has 600 turns, then how many turns are there on the secondary (output) coil?What is the power drawn from the electric outlet, if the transformer is assumed to be ideal?What is the power drawn by the transformer, if 13.5 percent of the input power is dissipated as heat in the coils and in the iron core of the transformer?: Hello Samantha, in case of ideal transformer the power at the out put would be the same as that in the input. To get the power we need voltage and current. Voltage at the out put is 8 V and current is 570 mA. Hence power consumed by the machine is 4.560 W. So the power drawn in case of ideal transformer will be the same as 4.560 W If the efficiency of the transformer is 86.5% (100-13.5) then the input power will be 5.27 W (approx) To get the number of turns in the secondary let us use Ns/Np Vs/Vp Hence Ns 600*8/120 40 turns.

Q:What is the rated current of the 80KVA transformer?: Rated current formula: I = P / 1.732 × U = 80000 / 1.732 × 380 ≈ 122 (A) (Current contains active and reactive current)

Q:The specific steps and procedures for installing the transformer: Hello, you said it should be the frequency transformer installation steps it. Steps are as follows: first equipped with skeleton Oh, and then use the winding machine around the enameled wire, and then add the fuse (if the customer specified to), then dip tin, dip tin is the dressing, the package is finished shake the film, shake the film is Plus shell, such a transformer to do a good job. I hope I can help you.

Q:80 kilowatts of electrical appliances need much transformer?: 80 kilowatts of electrical appliances need 100kVA transformer. Transformer (Transformer) is the use of electromagnetic induction principle to change the AC voltage of the device, the main components are primary coil, secondary coil and core (core). The main functions are: voltage conversion, current conversion, impedance conversion, isolation, voltage regulator (magnetic saturation transformer) and so on. According to the purpose can be divided into: power transformers and special transformers (electric furnace change, rectifier, frequency test transformer, voltage regulator, mine transformer, audio transformers, IF transformers, high-frequency transformers, impact transformers, instrument transformers, electronic transformers , Reactors, transformers, etc.). Circuit symbols commonly used as the beginning of the number. Example: T01, T201 and so on.

Q:Transformer is not the main and vice two coils? Resistance into which ah?: The significance of no - load power factor of power transformer The power transformer is likely to run without load, while the low power factor will increase the grid supply voltage. When the power transformer is unloaded, the no-load current is mainly used to establish the transformer core excitation and provide the excitation loss. The excitation current is part of the current current is reactive current, the general design, the power transformer power factor is low. When the power transformer winding side of the application of sinusoidal AC, the magnetic field is also sinusoidal changes in the magnetic field. However, the core of the transformer is a non-linear magnetically permeable material, so the field current that produces the magnetic field must be non-sinusoidal. Generally appear as "steeple". Figure 1 for a single-phase transformer no-load current waveform and its harmonic spectrum. Transformer no-load operation, the output power is zero, no-load power is equal to no-load loss. In the case of loss of fixed, the smaller the no-load current, the higher the no-load power factor. Therefore, the power transformer no-load power factor is higher, the better the no-load current is better. Increasing the no-load power factor is one of the main technical research directions of power transformer manufacturers, and reducing the no-load current is an important technical means to improve the no-load power factor.

Q:!!! I took a cardboard cylinder from a used firework, dug out all the crap, wrapped 160 turns of tinned copper wire on the outside tight and evenly so it is a beautiful solenoid, held it in place with 2 rubber bands. That was the secondary. Then I wrapped an insulated steel wire around a screwdriver shaft 10 times, stretched it out a little inserted a chopstick into this primary and inserted the chopstick into the firework. I then held it all in place with tissue paper and tape and wired the primary up to a pulsed DC circuit which was made of 3.0 volts wired to a steel file which I rubbed the electrode across. The secondary was wired to a current detector but I got no signal! Why is there zero inductance in my transformer and why is there no current at all? It can't even light up a small lightbulb!!: There is a little more to transformer design than this. Your secondary should have used insulated wire (varnished or enamel would do) as even one turn shorting will give problems. Also, you need to consider the magnetic circuit formed by the steel core. It is usual to use soft iron for this as otherwise the hysteresis will be too large. The entire magnetic flux needs to be concentrated into the secondary so you will need to consider the magnetic path. The transformer is essentially just two (or more) inductors, sharing a common magnetic path. Any two inductors placed reasonably close to each other will work as a transformer, and the more closely they are coupled magnetically, the more efficient they become.

Q:Branch circuit cables are rated for 75?C and feeder cables are rated for90°C. This will be a 3-phase, 575-volt system with four induction motors, specified asfollows:Motor 1: 60 hp 0.90 p.f. squirrel cage motorMotor 2: 60 hp 0.90 p.f. squirrel cage motorMotor 3: 40 hp 0.85 p.f. wound rotor motorMotor 4: 7-1/2 hp 0.80 p.f. wound rotor motorA) Assuming no line losses, find the capacity of the transformer required to supplypower to this system.B) Assume that it is desired to improve the overall power factor for this system to 0.95lagging. Determine, in kVAR, the required capacitance for this power factorcorrection.C) Assume these motors have their windings connected in a delta configuration. Whatwould be the line voltage if they were connected in wye?: I would look in the NEC for the full load amperes of these motor sizes. These are: 7.5 HP9 amps 40 HP.41 amps 60 HP.62 amps The 7.5 HP motor KVA will be KVA 575 * 9 * 1.732 8.9631 The 7.5 HP motor KW will be KW 575 * 9 * 1.732 * 0.8 7.17048 The KVAR of this motor is Sqrt(8.9631^2 - 7.1705^2) 5.3778 The 40 HP motor KVA will be KVA 575 * 41 * 1.732 40.8319 The 40 HP motor KW will be KW 575 *41 * 1.732 * 0.8 5 34.707115 The KVAR for this motor is Sqrt(40.8319^2 - 34.7071^2) 21.5095 The 60 HP motor KVA will be KVA 575 * 62 * 1.732 61.7458 The 60 HP motor KW will be KW 575 *62 * 1.732 * 0.8 5 55.5712 The KVAR for this motor is Sqrt(61.7458^2 - 55.5712^2) 26.9141 The total KVA requirement for all motors running at once is 8.9631 + 40.8319 + 61.7458 + 61.7458, which is 173.2866 KVA (note this is the requirement from a transformer, not the size of the transformer The total KW of all the motors is 7.1701 + 34.7071 + 55.5712 + 55.5712 153.0196 KW The total KVAR of all the motors is 5.3778 + 21,5095 + 26,9141 + 26.9141 80.7145 KVAR The power factor for all the motor is KW / KVA 153.0196 / 173.2866 0.883 I took a short cut at this point and used an application i wrote to calculate the required capacitor KVAR. The results follows: At 0.95 power factor, the motors' KVA will be 161.065 The motors' KW will be the same. The motors' KVAR will be 50.2926 The required capacitor's KVAR will be 53.988 The reactance of the capacitor will be 10.65 ohms The capacitance will be 3.49059248 E -4 Farads You can calculate these values as I did above, if you want to. EDIT Forgot the last answer. The line voltage is the same for both delta and wye. The leg voltage for the wye motor will be 575 / 1.732 331.98 volts TexMav

Q:I have been reading wikipedia and have gotten confused.Va.Number of Coils on side a---- ._________________Vb.Number of Coils on side bwikipedia mentioned how power companys use transformers for converting to high voltage power li nes then use transformers to bring the voltage back down.Focusing on the transformer that jacks the voltage upwirea-[side a-Transformerxyz- side b]-wirebso assum wireb is the high voltage power line.--Which side of transformerxyz would have more coils ? My guess is that side b has more coils then side a--Is it safe to say that the wattage of side a is equal to side b?--they said something about magnetic flux, does this just mean that the material connecting side a to side b has to be magnetic?--they said that the insulation wears down in transformers over time due to heat. what is causing this heat?--they said something about leakage and spacing of coils, what does this mean?tyVVVVVVVm: The number of turns is proportional to the voltage on that side. So if you connect a voltage to the side with 100 turns, and the other side has 1000 turns, you will get out 10 times the voltage you put in. However, the current available is only 1/10 as much, so the output power is the same as the input power (wattsvolts*amps). To transfer power efficiently, especially at low frequencies, both coils of the transformer are wound around the same piece of magnetic material. All wire has some electrical resistance. When current flows through it, voltage is produced (voltsamps*ohms), and the combination of voltage and current produces heat. The wires making up the coils are covered with insulation, and there is insulation between the coil and the core, and between the two windings, but the voltages may be 100,000 volts or more, and this is enough to cause sparks between windings or from the windings to the frame if there is a spot with no insulation.

Q:Can a european power outlet handle a 5000 watt power transformer?: No. Most are on 16A circuits, which can practically handle 3KW or so. UK recepticals are limited to 13A, so will be less. You might be able to use an electric stove outlet, if you know how to use one safely.

Q:I wanted to watch the transformers cartoons but because there is so many I don't know which to watch first, seconds etc.Can you please tell me in order which to watch?I'm watching Transformers Season 1 episode 1 atm.THANKS: If you want to watch the show that started the whole Transformers, Go on google and look for Transformers Generation 1. That will give you the original transformers cartoon to come out. You can google for a list of them or wiki it. But the List i Got Goes like this. #1. Transformers Generation 1 #2. Transformers: The Headmasters #3. Transformers: Super-God Masterforce #4. Transformers: Victory #5. Transformers: Zone After that starts the beast wars series of transformers then the New Age transformers such as armada,energon, and cybertron, and finally the current airing on cartoon network Transformers: The Animated Series. I personally enjoyed generations one the best, as i watched when i was younger but i hope that helps you out in your search or on what ever you wanted to know.

1. Manufacturer Overview
Location
Year Established
Annual Output Value
Main Markets
Company Certifications

2. Manufacturer Certificates
a) Certification Name
Range
Reference
Validity Period

3. Manufacturer Capability
a)Trade Capacity
Nearest Port
Export Percentage
No.of Employees in Trade Department
Language Spoken:
b)Factory Information
Factory Size:
No. of Production Lines
Contract Manufacturing
Product Price Range