103.MET November 2020 Q.8(b)

A diode whose internal resistance is 20 Ω is to supply power to 1000 Ω load from 110 V rms source. Calculate i) peak load current ii) DC load current iii) AC load current

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Solution

i) We know that Vmax = Vrms x √2

                                     = 110 x √2

                                     = 155.56 V

We know that I = V / R

Here total resistance R = 20 + 1000 = 1020 Ω

So Peak load current  I = 155.56 / 1020 

                                     = 0.152 A

 

ii) DC load current = Peak load current / π

                                = 0.152 / 3.14

                                = 0.048 A

iii)  AC load current = Peak load current / 2

                                 = 0.152 / 2

                                 = 0.076 A