Discussion
In the following circuit, the diode current flows through the resistance 10 \(\Omega\) :
If the voltage drop across the diode is 0.7 Volt, then the diode used should have minimum power rating nearly :
Answer: C
The diode needs a minimum rating of about 280 mW.
Replace the 33.4 V source with its two 60 \(\Omega\) resistors by a Thevenin equivalent seen by the diode branch:
\[ \begin{aligned} V_{TH} &= 33.4 \times \frac{60}{120} = 16.7\ \text{V} \\ R_{TH} &= 60 \parallel 60 = 30\ \Omega \\ I &= \frac{16.7 - 0.7}{30 + 10} = 0.4\ \text{A} \\ P_{D} &= 0.7 \times 0.4 = 0.28\ \text{W} \end{aligned} \]
The diode dissipates its forward drop times its current, 280 mW, so it must be rated for at least that. 240 mW and 180 mW would underrate it, and 350 mW, while safe, is not the minimum.
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