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  • A light emitting diode has a voltage drop of <img alt="\mathrm{2\ V}" src="https://entran

A light emitting diode \mathrm{(LED)} has a voltage drop of \mathrm{2\ V} across it and passes a current of \mathrm{10\ mA}. When it operates with \mathrm{6\ V} battery through a limiting resistor \mathrm{R}, the value of \mathrm{R} is

Option: 1

\mathrm{40\ k\ \Omega}


Option: 2

\mathrm{4\ k\ \Omega}


Option: 3

200 \ \Omega


Option: 4

400 \ \Omega


Answers (1)

best_answer

Current in the circuit \mathrm{=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}}
and voltage in the circuit \mathrm{=6-2=4\ V}
from Ohm’s law,
\mathrm{\begin{aligned} &\mathrm{ V=I R} \\ &\mathrm{ \therefore=\frac{V}{I}=\frac{4}{10 \times 10^{-3}}=400\ \Omega} \end{aligned}}

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Rishabh

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