Question

A 12 vdc power supply powers a blue high intensity led with a forward voltage of 4.5 volts. The led is rated to withstand a maximum current of 20 mA. What size resistor should be connected in series

Answer 1

Answer:

Resister of 0.15 W is to be added.

Step-by-step explanation:

In a given circuit we have the attachments

A power supply of 12 V

High intensity LED with forward voltage = 4.5V

LED is rated to withstand a maximum current of 20 mA or $20.10^{-30} A$

We have to calculate the size of a resister to be connected in series.

Here we will apply the Ohm's law for the calculations

Total voltage of the circuit = voltage of LED + voltage across resistance.

$12=4.5+20.10^{-3}.r$

r = (12 - 4.5)/20.10^{-3}

r = 7.5×10^{3}/20 = 375 ohm

Therefore power of the resistance = i².r

$P=i^{2}\times r$

         $=(20\times 10^{-3})^{2}\times 375$

        = 0.15 W

Therefore a resister of 0.15 W is to be added.