The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance RS is 0.5 mA. The minimum value of the resistance of RS, to ensure that the LED is not damaged is k . [figure]
- 6
- 2
- 4
- 5
Answer
(2) 2
Solution
[figure]
The upper wire is at the higher potential (positive terminal of the cell). The diode in series with R has its cathode on this wire ⇒ it is reverse biased, so no current flows in that branch.
⇒ The whole 0.5 mA flows through the LED.
For the LED to work at its rated power 2 mW:
VLED= PI= 2 10-30.5 10-3=4 V
Voltage across RS: 5-4=1 V
∴ RS= 10.5 10-3=2 103 =2 k
(A smaller RS would put more than 2 mW into the LED.)
Answer: (2)