Question

Find minimum \(R_s\) so that LED light does not get damaged (power rating of LED is \(2\,mW\)).

• A. \(1\Omega\)
• B. \(2\Omega\)
• C. \(3\Omega\)
• D. \(4\Omega\)

Answer 1

The Correct Option is B

Concept: Power relation: \[ P = VI \] Series resistor limits current through the LED. The remaining voltage appears across the resistor. Step 1: Find voltage across LED.} Given LED power rating \[ P = 2\times10^{-3} \,W \] Current through LED \[ i = 0.5\,mA = 0.5\times10^{-3} A \] \[ V_{LED} = \frac{P}{i} \] \[ V_{LED} = \frac{2\times10^{-3}}{0.5\times10^{-3}} \] \[ V_{LED} = 4V \]
Step 2: Find voltage across series resistor.} Supply voltage \[ V = 5V \] \[ V = V_{Rs} + V_{LED} \] \[ 5 = V_{Rs} + 4 \] \[ V_{Rs} = 1V \]
Step 3: Find required resistance.} \[ V_{Rs} = iR_s \] \[ 1 = 0.5\times10^{-3} R_s \] \[ R_s = 2\Omega \] Final Result \[ R_s = 2\Omega \]