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 [tex]20.10^{-30} A[/tex]
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.
[tex]12=4.5+20.10^{-3}.r[/tex]
r = (12 - 4.5)/20.10^{-3}
r = 7.5×10^{3}/20 = 375 ohm
Therefore power of the resistance = i².r
[tex]P=i^{2}\times r[/tex]
[tex]=(20\times 10^{-3})^{2}\times 375[/tex]
= 0.15 W
Therefore a resister of 0.15 W is to be added.
