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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

User Nizarazu
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1 Answer

5 votes

Answer:

Resister of 0.15 W is to be added.

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)* r


=(20* 10^(-3))^(2)* 375

= 0.15 W

Therefore a resister of 0.15 W is to be added.

User Anjosc
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