Physics - Semiconductors Question with Solution | TestHub

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