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A light emitting diode (LED) has a voltage drop of 2 volt across it and passes a current of $10 \mathrm{~mA}$ when it operates with a 6 volt battery through a limiting resistor $R$. The value of $R$ is
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The correct answer is:
$400 \Omega$.
As LED is connected to a battery through a resistance in series.
The current flowing, $10 \mathrm{~mA}$ is the same.
The voltage drop across LED $=2 \mathrm{~V}$
$\therefore$ As the battery has $6 \mathrm{~V}$, the potential difference across $R=4 \mathrm{~V}$.
$$
\therefore i R=4 \mathrm{~V} \Rightarrow R=\frac{4 \mathrm{~V}}{10 \times 10^{-3} \mathrm{~A}}=400 \Omega \text {. }
$$
The current flowing, $10 \mathrm{~mA}$ is the same.
The voltage drop across LED $=2 \mathrm{~V}$
$\therefore$ As the battery has $6 \mathrm{~V}$, the potential difference across $R=4 \mathrm{~V}$.
$$
\therefore i R=4 \mathrm{~V} \Rightarrow R=\frac{4 \mathrm{~V}}{10 \times 10^{-3} \mathrm{~A}}=400 \Omega \text {. }
$$
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