A voltage divider (two resistors in series) will achieve the required result. The value of the resistors will depend on what current is required to be supplied. It also matters what variation from 3V is allowed since taking current will make the voltage drop.
An electronics parts shop will sell components to make a more efficient system.
Or you could get 3 voltage regulators .. 9V then 5V then 3V in series .. that should keep the voltage stable as long as the 9V supply is
use a dropping resistor to a 2 volt zener diode. The resistor value and wattage of it and the zener are dependant on the 2 volt current drain needed
You need a transformer to do that.
9v
A: WELL you cannot not with 1,2v most LED start emitting at 1.8 v some as high as 5 volts . two need at least two 1.2 volts better is 9 volts then there is a problem voltage wise but the problem is now that you need a limiting resistor to ensure at least 10ma of current
V = I*RIf you solved this for resistance, this means you would have:V/I = RYou set V = .9 volts, I = 1 amp, and solve to get .9 Ohms.
Four 9v batteries connected in a parallel will still emit 9 volts because you are not increasing the voltage, you are increasing the life. To increase the voltage of four 9v batteries, you must connect them in a series; that series will emit 9v X 4(batteries), which equals 36 volts.
Depends on the led forward bias threashold, if its a typical led it will be .7 volts so, .7x6=4.2V, so pick a resistor that will drop around 7 volts. What is the current? Then just to V=IR, 7=IR.
Kilo means 1,000. So there are 9,000 volts in 9 kilovolts
How many volts for a 9 mag
You can only do that with a transformer. Or you can wire three 9 volt batteries in series to get 27 volts.
You don't. Voltages are a nominal rating. 110 volts is within the nominal range at this potential. The voltage centre line is 115 volts. Utility companies are mandated to keep voltages locked within 5% of a centre line voltage. This means that it could be as high as 115 + 5% = 120.75 volts and as low as 115 - 5% = 109.25 volts
Use a full wave bridge rectifier.
it would be 3v
You need to change the blower motor resistor, it cuts the power down from 12 volts to lesser volts, so you get 1,2,3,4,5 levels of heat. When the resistor goes out, it only puts out 12 volts, so only high works. It's about a 30 dollar part, and you usually have to remove the glove box, pretty easy really.
Downstream O2 sensor circuit volts high.Downstream O2 sensor circuit volts high.
Ohm's Law: Voltage = Amperes times Resistance 9 volts = amps * 10 ohms amps = .9
220 volts, 110 volts, 440 volts, 400 volts, AC or DC voltage. High voltage like - 220 KV, 400 KV, etc
No. The voltage of the charger's output is only 7.5 volts. This is not high enough to charge a 9 volt battery device.
yes, 126 is a little high.. 1 or 2 volts ok