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Light Emitting Diodes (LED) solid state devices are current dependant. LED have ratings for the max current allowed to the LED device. A typical current allowed to a LED is 20ma. Ohms Law is used to determine the resistance required to limit the current to your desired current. The formula is used in this manner: E=12VOLTS the desired current is I = 20ma. R = E/I = 600 ohms. Where E=2 volts dc and I=20ma R= E/I = 100 Ohms.
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How much voltage is required to run 0.054 a of current through a 15 resistor?
0.81 V - just answered the other answer to this question (2.8 x 10^2V) and got it wrong
When a diode is properly forward biased what is the voltage drop across it equal to?
The forward biased voltage drop of a diode depends on the type of diode and the current through the diode. A typical silicon diode will exhibit a voltage drop between 0.6v and 1.4v depending on current. An LED might range from 2v to 3v. A germanium diode might go a low as 0.2v. Bottom line; it varies.
What value of resistor to light a LED on 230 volts?
To figure this out, you need to know the expected forward voltage and current of the LED. Lets assume 2ma and 2V. (Actually, 2ma is small, but I intend to make a point.) By Kirchoff's Voltage Law, you know that the signed sum of the voltage drops going around a series circuit must add up to zero. This means that the voltage across the resistor must be 228 volts. (-230 + 228 + 2 = 0) By Kirchoff's Current Law, you know that the signed sum of the currents entering a node is zero. As a consequence, you also know that the current at every point in a series circuit is the same. Therefore, the current through the resistor is also 2ma. By Ohm's Law, you know that resistance is voltage divided by current, so you know that the resistor is 228V divided by 2ma, which is 114K. The nearest standard value in the E12 scale is 100K. Recalculate the current for 100K, and you get about 2.25ma. (You could also use 120K, and I'll let you run the calculations yourself.) Don't stop here. There are some issues... By the power law, you know that power is voltage times current, so you know that the power dissipated by the resistor is 228V times 2.25ma, which is 513mW. I would put a one watt resistor in there. However, consider this. 2ma is a low current LED. Some of them pull 25ma. The power in the resistor in that case is about 6.5W, which is getting pretty high. Secondly, you need to consider the reverse breakdown voltage on the LED. I assume that when you said 230V, you meant AC, not DC, which means that there is going to be 230V (actually, a peak value of 325) across that LED for one half the line cycle. You need to check the datasheet and make sure the LED can handle that. If not, you need to put an ordinary signal diode, such as a 1N4148, in parallel with the LED, in the reverse direction, so that it clamps the reverse voltage at about 0.7 volts. (Don't worry about the reverse breakdown on the 1N4148, because the LED will protect it, on opposite half line cycles.) Last, but not least, you need to consider the safety of the operator. 230V is a high voltage, and LED's are not the most rugged thing around. If the LED breaks, you need to consider if its internal wiring could come into contact with the operator. I would certainly demand a UL listed device in this application.
How do I design a voltage divider circuit that has 2 VDC 5 VDC and 8 VDC available with a 10 VDC source?
Without knowing permissible current draw by the divider or its maximum power dissipation the actual resistor values cannot be determined, but the ratios of the resistor values can be determined from the required voltage drops.The divider will be composed of 4 resistors starting at the 10VDC rail:2VDC drop, ratio = 2V/10V = 0.23VDC drop, ratio = 3V/10V = 0.33VDC drop, ratio = 3V/10V = 0.32VDC drop, ratio = 2V/10V = 0.2Therefore you will need 2 resistors (R1 & R4) that are 0.2 * the total resistance of the voltage divider and 2 resistors (R2 & R3) that are 0.3 * the total resistance of the voltage divider.But as stated at the beginning you can get no further without additional requirements being specified.
What is the disadvantage of using the centre tapped full wave rectifier?
it can be driven by a single source voltage, such as the untapped secondary of a transformer or directly from the power line. The peak reverse voltage that can be tolerated is 2x the reverse breakdown of the diodes.
What voltage does an led bulb need?
Minimum 2V. Some can run as high please check sheerled.co.uk/
What are the five voltages that can be produced by ATX or BTX?
12V, 8V, 5V, 3.3V, and 2V.
Need a 12v 12ah battery to power a device that normally uses a 7 point 2v 1400mah li-on battery can it be done and would it be practical?
Yes. You will need a voltage regulator circuit to produce the 7.2v from the 12v source.
How do the milliamp ratings of LEDs 20ma and a 12 volt 1000ma power supply figure into the resistor equations or do they not?
A: Since you know the current flow you need the voltage drop for this particular led at 20ma. All LED require a current and voltage to operate properly assuming a voltage drop of 2v then for a 12v source it becomes 12-2=10v 10v/.02=500 ohms in series to limit the current is required
What happen when you give 2v or 3v to ic 7805 as input?
the output of 7805 is 2v or 3v if we give the input as 2v or 3v.
Is there a way to wire six 2V solar panels so that they put out 12V?
Since the output of a solar cell is DC they can be wired up in series so that they total 12 VDC. The - terminal of one connects to the + terminal of the next cell and so on for all six cells. This is the same as flashlight batteries being put end to end to create a higher voltage.
Can someone explain how to differentiate 1400 v plus 2v 7?
Do you mean 1400v + 2v^7 (is 2v to the power 7)? When you differentiate, e.g. 2v^7, you bring the 7 to the front and reduce the power by 1. If so the answer is 1400 + 7*2v^6 (where * means multiply) or 1400 + 14v^6
What is the minimum voltage for LED tail lights?
A standard LED will need about 2V and would probably draw about 10mA to give a normal level of brightness. Obviously these factors can be varied to suit the particular application. Care must be taken to stay within the tolerances of the LED itself or burn-out will result.
How do you factor 100V square minus 25W square?
100 V2 - 25 W2 = (10V + 5W) (10V - 5W)Another way to show it:100 V2 - 25 W2 = 25 (4 V2 - W2) = 25 (2V - W) (2V + W)
How the kinetic energy of an object changes when the speed of object doubles?
Kinetic energy at velocity "V" is (0.5) x mass x V x V Kinetic energy at velocity "2V" is (0.5) x mass x 2V x 2V Ratio of KE at velocity 2V and KE at velocity V is [(0.5) x mass x 2V x 2V] / [(0.5) x mass x V x V] = 4 So if the velocity double, KE quadruples
Where should a resistor be placed when using an LED?
In series with the LED. The value would be whatever it takes to obtain the required voltage and current across the LED. As an example, if you had an LED that required 25ma at 2v, and you wanted to use a 9v battery, you would need a resistance of 280 ohms. (This is (9-2) / 0.025, a simple application of Ohm's law.)
What is the rainbow trouts population?
2v
