Concept:When a diode's cathode (N-side) is connected to the positive terminal of a battery and its anode (P-side) to the negative terminal, the diode is reverse biased and blocks current conduction.
Formula:$$I_{\text{reverse}} \approx 0$$
Solution:- In the provided schematic, the cathode bar points toward the positive terminal of the \( 12\text{ V} \) DC source.
- This places the diode in reverse bias.
- An ideal diode in reverse bias behaves as an open switch, allowing zero current (\( 0.00\text{ A} \)) to flow through the series resistor.
Why other options are incorrect:- Option A: \( 12\text{ V}/100\,\Omega = 0.12\text{ A} \) would be the current if the diode were ideal and forward biased.
- Option B: Incorrect by a factor of 10.
- Option D: \( (12 - 0.7)/100 = 0.113\text{ A} \) would be the forward-biased current taking the silicon barrier potential into account.
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