A light emitting diode (LED) has a voltage dr | vedclass

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(A) The $LED$ is connected in series with a resistor $R$ and a $6\,V$ battery.Since the $LED$ and the resistor are in series,the current flowing throu

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$0.4$

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(A) The $LED$ is connected in series with a resistor $R$ and a $6\,V$ battery.Since the $LED$ and the resistor are in series,the current flowing through both is the same,$I = 10\,mA = 10 	imes 10^{-3}\,A$.The voltage drop across the $LED$ is $V_{LED} = 2\,V$.The total voltage of the battery is $V = 6\,V$.According to Kirchhoff's voltage law,the voltage drop across the resistor $R$ is $V_R = V - V_{LED} = 6\,V - 2\,V = 4\,V$.Using Ohm's law,$V_R = I 	imes R$,we get $R = \frac{V_R}{I}$.Substituting the values: $R = \frac{4\,V}{10 	imes 10^{-3}\,A} = 0.4 	imes 10^3\,\Omega = 400\,\Omega$.Since $1\,k\Omega = 1000\,\Omega$,then $400\,\Omega = 0.4\,k\Omega$.

$A$ 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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