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Q.

A junction diode of internal resistance 20 $Ω$ is used for half wave rectification. If the applied voltage has a peak value of 50 V and load resistance is 800 $Ω$ then d. c. value of current is

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a

9 mV

b

10.3 mA

c

19.4mA

d

14.3mA

answer is C.

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Detailed Solution

Hero $I_{dc} = (I_{m} / π)$
where I_m = peak current
Further,
$I_{m} = \frac{V}{R_{d} + R_{L}} = \frac{50}{20 + 800} = 61 m A I_{d c} = \frac{61 m A}{π} = 19 \cdot 4 m A$

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