The maximum rated power of the LED is 2 text{ | vedclass

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(A) Given: Maximum power of $LED$ $P_{max} = 2 	ext{ mW}$,Input voltage $V_{in} = 5 	ext{ V}$,Resistance $R = 1 	ext{ k}\Omega$,and current through

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(A) Given: Maximum power of $LED$ $P_{max} = 2 	ext{ mW}$,Input voltage $V_{in} = 5 	ext{ V}$,Resistance $R = 1 	ext{ k}\Omega$,and current through $R$ is $I_R = 0.5 	ext{ mA}$.$1$. The voltage across the $LED$ $(V_{LED})$ is the same as the voltage across the resistor $R$ because they are in parallel. Thus,$V_{LED} = I_R 	imes R = 0.5 	ext{ mA} 	imes 1 	ext{ k}\Omega = 0.5 	ext{ V}$.$2$. The maximum current allowed through the $LED$ is $I_{LED} = \frac{P_{max}}{V_{LED}} = \frac{2 	ext{ mW}}{0.5 	ext{ V}} = 4 	ext{ mA}$.$3$. The total current flowing through the series resistor $R_S$ is $I_S = I_{LED} + I_R = 4 	ext{ mA} + 0.5 	ext{ mA} = 4.5 	ext{ mA}$.$4$. The voltage across $R_S$ is $V_{R_S} = V_{in} - V_{LED} = 5 	ext{ V} - 0.5 	ext{ V} = 4.5 	ext{ V}$.$5$. The minimum resistance $R_S$ is $R_S = \frac{V_{R_S}}{I_S} = \frac{4.5 	ext{ V}}{4.5 	ext{ mA}} = 1 	ext{ k}\Omega$. Since $1 	ext{ k}\Omega$ is not among the options,and the question asks for the minimum value to ensure safety,we re-evaluate the circuit. If the diode is considered ideal with a forward voltage drop of $0.5 	ext{ V}$,the calculation holds. Given the options,there might be a typo in the provided options or the circuit parameters. Based on standard competitive exam patterns for this specific problem,$1 	ext{ k}\Omega$ is the calculated value.

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