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(A) Light Emitting Diode $(LED)$ is a forward-biased $P-N$ junction that emits light.The voltage drop across the $LED$ is $V_{LED} = 2\,V$.The battery

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$0.4$

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(A) Light Emitting Diode $(LED)$ is a forward-biased $P-N$ junction that emits light.The voltage drop across the $LED$ is $V_{LED} = 2\,V$.The battery voltage is $V_{battery} = 6\,V$.The voltage across the limiting resistor $R$ is $V_R = V_{battery} - V_{LED} = 6\,V - 2\,V = 4\,V$.The current flowing through the circuit is $I = 10\,mA = 10 	imes 10^{-3}\,A = 0.01\,A$.Using Ohm's law,$V_R = I 	imes R$,we get $R = V_R / I$.$R = 4\,V / 0.01\,A = 400\,\Omega$.Since $1\,k\Omega = 1000\,\Omega$,we have $R = 400 / 1000 = 0.4\,k\Omega$.

$A$ Light Emitting Diode $(LED)$ has a voltage drop of $2\,V$ across it and passes a current of $10\,mA$ when it operates with a $6\,V$ battery through a limiting resistor $R$. The value of $R$ is.........$k\,\Omega$.

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