Q.39. The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of

Q: What is the correct answer for JEE Main Physics Q39?

The correct answer is B. Given: Maximum rated power of LED $P_{LED,max} = 2$ mW $= 2 \times 10^{-3}$ W Input voltage $V_{in} = 5$ V Current through resistance $R_s$ is $I = 0.5$ mA $= 0.5 \times 10^{-3}$ A To ensure the LED is not damaged, the current flowing through it must not exceed the current corresponding to its maximum rated power. Assuming the LED is operating at its maximum power, we can find the voltage across

Q: Q.39. 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 is 0.5 mA. The minimum value of…

The correct answer is B. Given: Maximum rated power of LED $P_{LED,max} = 2$ mW $= 2 \times 10^{-3}$ W Input voltage $V_{in} = 5$ V Current through resistance $R_s$ is $I = 0.5$ mA $= 0.5 \times 10^{-3}$ A To ensure the LED is not damaged, the current flowing through it must not exceed the current corresponding to its maximum rated power. Assuming the LED is operating at its maximum power, we can find the voltage across

Question

Q.39. 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  is 0.5 mA. The minimum value of the resistance of , to ensure that the LED is not damaged is ____ K\Omega.

FirstInTest · Accepted Answer

Given:
Maximum rated power of LED $P_{LED,max} = 2$ mW $= 2 	imes 10^{-3}$ W
Input voltage $V_{in} = 5$ V
Current through resistance $R_s$ is $I = 0.5$ mA $= 0.5 	imes 10^{-3}$ A

To ensure the LED is not damaged, the current flowing through it must not exceed the current corresponding to its maximum rated power. Assuming the LED is operating at its maximum power, we can find the voltage across the LED ($V_{LED}$) and the current through it ($I_{LED}$). However, the problem statement gives the current through $R_s$ as $0.5$ mA, which implies this is the operating current for the circuit.

Let's assume the current $I = 0.5$ mA is the maximum safe operating current for the LED in this circuit configuration.

From the power formula for the LED: $P_{LED} = V_{LED} 	imes I_{LED}$.
If $P_{LED,max} = 2 	imes 10^{-3}$ W and $I_{LED} = 0.5 	imes 10^{-3}$ A (current through $R_s$ is the same as through LED in series):
$V_{LED} = \frac{P_{LED,max}}{I_{LED}} = \frac{2 	imes 10^{-3} 	ext{ W}}{0.5 	imes 10^{-3} 	ext{ A}} = 4 	ext{ V}$.

Now, apply Kirchhoff's Voltage Law (KVL) to the series circuit:
$V_{in} = V_{R_s} + V_{LED}$
$V_{R_s} = V_{in} - V_{LED}$
$V_{R_s} = 5 	ext{ V} - 4 	ext{ V} = 1 	ext{ V}$.

Now, calculate the resistance $R_s$ using Ohm's Law:
$R_s = \frac{V_{R_s}}{I} = \frac{1 	ext{ V}}{0.5 	imes 10^{-3} 	ext{ A}}$
$R_s = \frac{1}{0.0005} = 2000 	ext{ \Omega} = 2 	ext{ K\Omega}$.

The solution's first line states: "Diode in reverse bias so no current in it." This is incorrect. The diagram shows the diode in forward bias (anode connected to positive terminal via $R_s$). If it were reverse biased, no current would flow, and the LED wouldn't light up. The rest of the solution proceeds as if it's forward biased, calculating $V_{bulb}$ and $R_s$. I will ignore the incorrect initial statement.

The solution's calculation: $P_{bulb} = 2 	imes 10^{-3} = V_{bulb} 	imes I_{bulb}$. $V_{bulb} = 4$V. $V_{Rs} = 5-4 = 1$V. $R_s = 1 / (0.5 	imes 10^{-3}) = 2$ k$\Omega$. This matches my calculation.

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