An LED has a voltage drop of 2 V across it and a current of 10 mA flows when it operates with a 6 V battery through a limiting resistance R. The value of R is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 400\,\Omega \)
The LED and the limiting resistance are in series with the battery, so the same current flows through both and the battery voltage is shared between them.
\( V_R = 6 - 2 = 4\,\mathrm{V} \)
\( I = 10\,\mathrm{mA} = 10^{-2}\,\mathrm{A} \)
\( \displaystyle R = \frac{V_R}{I} = \frac{4}{10^{-2}} = 400\,\Omega \)