Electricity,a power supply has a positive and negative charge,this charge is used by metal,and the metal,or any electric conducter charges from the,negative charge,and the conponent goes out the,positive charge
I assume you wish to connect this power supply to the mains which in my case means 220V 50Hz and that this is supposed to be a DC supply.
Assuming this the first problem that you need to solve is stepping this voltage down to 35V DC region. There are two methods in which you can do this.
The first involves using a 60watt 50/60Hz trans to step down your voltage to 35V peak voltage after which you can use a full bridge rectifier followed by a large capacitor to convert it to DC. The biggest problem with this method is that a 60W 50Hz transformer can be rather bulky and quite expensive.
The second option which I prefer is to first rectify the 220V ac to 311V DC using a full bridge rectifier. The next step is to then design a DC-DC converter to convert the 311V to 35V. My best recommendation is a flyback converter. Using a high frequency in the flyback design (ie 50kHz) reduces the size of the coupled inductor (transformer) of the flyback converter significantly. The biggest problem you may encounter in this is generating the 50kHz pwm signal to drive the flyback if you are not familiar with PWM chips or microcontrollers.
Once you have a nice stiff 35 DC buss you can use an L200 chip which can be used to do both voltage and current regulation. If you download the L200 data sheet there is a nice example circuit on how to use it on page 8 figure 23.
This is not exactly a project for the feint hearted
It depends. The 2A current, did you measure that while the soldering iron was on? Or is it rated at 2A current consumption on the device itself (on the powercable or the stem of the soldering iron). Generaly speaking you can calculted the real power consumption by using P=V*I (thus 2A*24V = 48W). But do remeber that this power consumption is in the steady state, that is, after its switch on and all transient effects have died down. To be safe allow for 3A-4A switching currents that occurs at power on.
The question is moot, here's why: First of all, the replacement must supply the correct voltage. This cannot change. You always replace a 12V supply with a 12V supply. If the replacement supply has the correct amperage, it will also have the correct volt-amperes, since (by our definition) the voltage is the same. Say the old supply was 12V, 2A. This is a 24VA supply (12 X 2 = 24). The new supply also must be 12V, so which is more important, amperes, or VA? As you can see, if we make sure the new supply is 2A, it will also mean it supplies 24VA (12 X 2 = 24). If we make sure the new supply is 24 VA it will obviously supply 2A (24/12 = 2). So, assuming the voltage stays the same, matching either VA or amperes automatically means the other value is correct.
Inductance = Magnetic Flux/Current = [ML2T-2A-1]/[A] = [ML2T-2A-2] So, Dimensional Formula of Inductance = [ML2T-2A-2]
If it were just 12V to 5V we would be talking about a simple regulator. Since we are also talking about 1A to 2A, we are talking about some kind of inverter, perhaps a pulse width modulated power converter.
No, neither will work and they may even be damaged.
No No No. If your supply can give .2A, and you need 2 Amps, your supply's not going to cut it.
Unfortunately no, if the device calls for 2000ma you will need a 2A (amp) power supply to adequately power it.
No B/c ur unit Circuitry design for 1A u can put in 2A MAX 15A may cause burn ur unit
It depends. The 2A current, did you measure that while the soldering iron was on? Or is it rated at 2A current consumption on the device itself (on the powercable or the stem of the soldering iron). Generaly speaking you can calculted the real power consumption by using P=V*I (thus 2A*24V = 48W). But do remeber that this power consumption is in the steady state, that is, after its switch on and all transient effects have died down. To be safe allow for 3A-4A switching currents that occurs at power on.
2a to the second power. If you combine the like terms, (a to the second power + a to the second power), it would be the same as 2a to the second power.
The question is moot, here's why: First of all, the replacement must supply the correct voltage. This cannot change. You always replace a 12V supply with a 12V supply. If the replacement supply has the correct amperage, it will also have the correct volt-amperes, since (by our definition) the voltage is the same. Say the old supply was 12V, 2A. This is a 24VA supply (12 X 2 = 24). The new supply also must be 12V, so which is more important, amperes, or VA? As you can see, if we make sure the new supply is 2A, it will also mean it supplies 24VA (12 X 2 = 24). If we make sure the new supply is 24 VA it will obviously supply 2A (24/12 = 2). So, assuming the voltage stays the same, matching either VA or amperes automatically means the other value is correct.
(2a + 1)(4a2 - 2a + 1)
6a to the second power minus 8ab + 2a
Probably not. The other way around would be fine: a device that only needs 750mA will work fine on a 2A power supply, but one that needs 2A will NOT work on only 750mA, which is less than half the current required.
The verbal expression of 3b to the second power plus 2a to the 3rd power is "three b squared plus two a cubed."
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
Heaters have an efficiency around 99%.