Q134 / 180 · AIIMS 2006 · Physics

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

  • A
    $40 \mathrm{k} \Omega$
  • B
    $4 \mathrm{k} \Omega$
  • C
    $200 \Omega$
  • Correct answer
    D
    $400 \Omega$.

Solution

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 $R=4 \mathrm{~V}$. $$ \therefore i R=4 \mathrm{~V} \Rightarrow R=\frac{4 \mathrm{~V}}{10 \times 10^{-3} \mathrm{~A}}=400 \Omega \text {. } $$

Hence, the correct answer is option: 'D'.

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