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A crystal diode having internal resistance $r_{f} = 20 Ω$ is used for full wave rectification. If the applied voltage is V=50sinwt and load resistance is $R_{L} = 800 Ω$ , then find $i_{d c} a n d i_{r m s}$ .

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38.85A,43.14A

28.4A, 32.5A

12.5A,13.5A

68.85A,92.15A

answer is A.

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Detailed Solution

$\begin{array}{l} i_{max} = \frac{v_{max}}{R_{f} + R_{L}} = \frac{50}{20 + 800} = 61 m A \\ i_{d c} = \frac{2 i_{max}}{π} = \frac{2 \times 61}{π} = 38.85 m A \\ i_{r m s} = \frac{I_{0}}{\sqrt{2}} = \frac{61}{\sqrt{2}} = 43.14 m A \end{array}$

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