LED CIRCUIT

195

05:02:38

Simple stuff

published 1 year ago

Robert_Kidd

1 year ago

The LED is rated at 2V and 20mA. To achieve this you need to drop 3V across the resistor. Using R=V/I we have 3/0.02=150 Ohms. Change resistor to 150 Ohms.

Redstone_guy

1 year ago

@Robert. You are right. An LED rated for 2V and 20mA, using R=V÷I, is 2÷0.02=100Ω. If the resistor remained at 1kΩ, the total load resistance would be 1.1k. Using Kirchhoff’s Current Law, since this is a series circuit, if the LED drops 20mA, so does the 1k resistor. That means that in order for the LED to get its required current, but the resistor is rated at 1kΩ, then, using V=I×R, the voltage would have to be 0.02A×1100Ω=22V. If billyrow3 doesn’t adjust the resistor to 150Ω, then they should change the voltage to 22V.

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