A crystal diode having internal resistance $r_{f} = 40 Ω$ is used for full wave rectification. If the applied voltage is $V = 100 \sin w t$ , load resistance is $R_{L} = 1000 Ω$ , find AC power input and DC power output

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$4.7 W, 3.7 W$

$6.7 W, 9.7 W$

$1.7 W, 6.7 W$

$8.7 W, 10.7 W$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n : V_{0} = 100 V, r_{f} = 40, R_{L} = 1000 \\ P e a k C u r r e n t, I_{0} = \frac{V_{0}}{R_{L} + r_{f}} \\ \Rightarrow I_{0} = \frac{100}{1040} = 96 m A \end{array}$

$\begin{array}{l} D C c u r r e n t, i_{d c} = \frac{2 i_{0}}{π} = \frac{2 \times 96 \times 10^{- 3}}{3.14} = 61.14 \times 10^{- 3} A \\ D C P o w e r O u t p u t, P_{d c} = i_{d c}^{2} \times R_{L} = (61.14)^{2} \times 1000 \times 10^{- 6} = 3.7 W \\ R M S C u r r e n t, i_{r m s} = \frac{I_{0}}{\sqrt{2}} = \frac{96 \times 10^{- 3}}{\sqrt{2}} = 67.89 \times 10^{- 3} A \\ A C P o w e r I n p u t, P_{A . C} = {i^{2}}_{m s} [R_{L} + r_{f}] = (67.89)^{2} \times (1040) \times 10^{- 6} = 4.7 W \end{array}$