**Resistor** = (Battery Voltage \u2013 LED voltage) / desired LED current. So assuming a 12-volt power source and a white LED with the desired current of 10 mA; The

**formula** becomes

**Resistor** = (12-3.4)/. 010 which is 860 ohms. Since this is not a standard value I would use an 820-ohm

**resistor**.

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**LEDs** typically require 10 to 20mA, **the** datasheet for **the LED** will detail this along with **the** forward voltage drop. For example an ultra bright blue **LED** with a 9V battery has a forward voltage of 3.2V and typical current of 20mA. So **the resistor** needs to be 290 ohms or as close as is available.Jan 29, 2014

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To **reduce voltage** in half, we simply form a **voltage** divider circuit between 2 **resistors** of equal value (for example, 2 10K\u03a9) **resistors**. To divide **voltage** in half, all you must **do** is place any 2 **resistors** of equal value in series and then place a jumper wire in between the **resistors**.