Without getting too technical, this simply means that the circuit experiences some additional current through the resistors when the LED signal comes on. So there you have it. Install one of these resistors into the circuit near your LED signal and you will increase the amount of current the OEM relay sees when the LED signal lights up Step 4: Ensure the Load Equalizer Wires can Reach. Once you've found your turn signal bulb wires, make sure your load equalizer can reach an internal metal surface of your vehicle and that the wiring can reach the turn signal wires. If not, the load equalizer wires can be extended to reach from the mounting location to the turn signal wiring. A better way is to control the current through the LED or diode to be what you want. A very simple way to do this is to use a resistor. Regarding your part c, connecting three 2V 2 V LEDs in series to a 12V 12 V supply. The purpose of the resistor in that circuit is to limit the current, not the voltage.

Slight correction: 240VAC is RMS voltage, and has a peak voltage of 240 * sqrt (2) = 340Vpk. Your resistor needs to be 18.8k to limit current to 18 mA, and will dissipate ~6.1W peak or 4.3Wrms. You still need (at least) a 5W resistor, however. Jul 20, 2016 at 0:13. Google Capacitive Dropper Led for the typical setup.

All the calculators in step 2 are just doing some simple math that you can do at home: The formula to calculate resistance in a circuit is: R=V/I or, more relevant to what we're doing: (Source Volts - LED Volts) / (Current / 1000) = Resistance * So if we have a 12v battery powering a 3.5V 25mA LED our formula becomes: (12 - 3.5) / (25 / 1000 Pricing and details on the GTR Lighting Resistors: &utm_source=youtube&ut
When a series LED connection is in question, you will just need to replace the "LED forward voltage" with "total forward voltage" in the formula, by multiplying FV of each LED by the total number of LEDs in the series. Suppose there are 3 LEDs in series then this value becomes 3 x 3.3 = 9.9. LED Current or I refers to the current rating of the
However, if our voltage was 12V, we would have to rework our calculations to keep the same amount of current flowing through the LED. Our duty cycle would need to drop to 14.167% (1.7V divided by 12V) and our minimum PWM frequency would decrease to 14.285kHz (the inverse of [10us divided by 14.167%]). HOWEVER!, this is cause for concern. In the The ballast resistor can be calculated using the formula: R = V −V LED I = 12− 2 0.03 = 333Ω R = V − V L E D I = 12 − 2 0.03 = 333 Ω. The resistor must have a resistance of 333 Ω. If the precise value is not available, choose the next higher resistance value to keep the current below the LED limits. It is quicker to wire them in your scenario than any other way. As you add LEDs to your parallel array, the voltage requirements stay the same. As you add LEDs to your parallel array, the current requirements increases the amount that each LED requires (5 LEDs @ 20mA = 100mA requirement/draw). 100 Ohm 1/4 watt resistor.
Uni-T UT204 Auto-Ranging AC DC Ture RMS Auto/Manual Range Digital Handheld Clamp Meter Multimeter Test Tool. Measures both ac & dc current up to 400 amps, ideal for vehicles, rvs, marine boats & etc. Ac/dc voltage, current, resistance, frequency, continuity, duty cycle & diode check. $53.99.
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    • do 12v leds need resistors