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

Question

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 $R_{s}$ is 0.5 mA.

The minimum value of the resistance of $R_{s}$ , to ensure that the LED is not damaged is ________ $k Ω$ . [2026]

Smart Achievers · Accepted Answer

(2)

$A diode in reverse bias has no current flowing through it.$

$P_{bulb} = 2 \times 10^{- 3}$

$V_{bulb} = 4 V$

$So 5 V = V_{R s} + V_{bulb}$

$V_{R s} = 5 - 4 = 1 V$

$1 = 0.5 \times 10^{- 3} R_{s} \Rightarrow R_{s} = 2 k Ω$

Solution

Q.
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 $R_{s}$ is 0.5 mA.

The minimum value of the resistance of $R_{s}$ , to ensure that the LED is not damaged is ________ $k Ω$ . [2026]

1 6

2 2

3 4

4 5

Ans.
(2)

$A diode in reverse bias has no current flowing through it.$

$P_{bulb} = 2 \times 10^{- 3}$

$V_{bulb} = 4 V$

$So 5 V = V_{R s} + V_{bulb}$

$V_{R s} = 5 - 4 = 1 V$

$1 = 0.5 \times 10^{- 3} R_{s} \Rightarrow R_{s} = 2 k Ω$

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Solution

Q. 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 ________ kΩ. [2026]

1 6

2 2

3 4

4 5

Ans. (2) A diode in reverse bias has no current flowing through it. Pbulb=2×10-3 Vbulb=4V So 5V=VRs+Vbulb VRs=5-4=1V 1=0.5×10-3Rs⇒Rs=2 kΩ

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Q.
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 $R_{s}$ is 0.5 mA.

The minimum value of the resistance of $R_{s}$ , to ensure that the LED is not damaged is ________ $k Ω$ . [2026]

Ans.
(2)

$A diode in reverse bias has no current flowing through it.$

$P_{bulb} = 2 \times 10^{- 3}$

$V_{bulb} = 4 V$

$So 5 V = V_{R s} + V_{bulb}$

$V_{R s} = 5 - 4 = 1 V$

$1 = 0.5 \times 10^{- 3} R_{s} \Rightarrow R_{s} = 2 k Ω$