The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R connected in series with the diode so that the diode carries the maximum current allowed by its power rating?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10\,\Omega \)
The power in the diode is its voltage drop times the current, \( P=V_{\mathrm{D}}I \). The drop stays at 0.5 V. So the power rating sets the largest current the diode can carry. The resistor must take the rest of the battery voltage.
The maximum current allowed is:
\( \begin{aligned} I_{\max} &= \frac{P}{V_{\mathrm{D}}} \\ &= \frac{100\times10^{-3}\,\mathrm{W}}{0.5\,\mathrm{V}} \\ &= 0.2\,\mathrm{A} \end{aligned} \)
From the figure, the battery is 2.5 V, so the potential difference across \( R \) is \( 2.5\,\mathrm{V}-0.5\,\mathrm{V} \). The resistance is:
\( \begin{aligned} R &= \frac{V-V_{\mathrm{D}}}{I_{\max}} \\ &= \frac{2.5\,\mathrm{V}-0.5\,\mathrm{V}}{0.2\,\mathrm{A}} \\ &= 10\,\Omega \end{aligned} \)