Concept:For an ideal diode during forward conduction, internal diode drop is zero, and peak current is determined by peak input voltage and load resistance.
Formula:$$I_{\text{peak}} = \frac{V_{\text{peak}}}{R_L}$$
Solution:- Given peak input voltage: \( V_{\text{peak}} = 10\text{ V} \).
- Given load resistance: \( R_L = 100\text{ }\Omega \).
- Using Ohm's law for the ideal conducting diode:
- $$I_{\text{peak}} = \frac{10\text{ V}}{100\text{ }\Omega} = 0.10\text{ A}$$
Why other options are incorrect:- Option B: \( 0.07\text{ A} \) (\( \approx \frac{0.1}{\sqrt{2}} \)) corresponds to the RMS current, not the peak forward current.
- Option C: \( 1.00\text{ A} \) is off by a factor of 10 from incorrect division.
- Option D: \( 0.20\text{ A} \) would correspond to a \( 50\text{ }\Omega \) load rather than a \( 100\text{ }\Omega \) load.
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