Question

A diode whose internal resistance is 40 $\Omega$ is used to supply power to a 1 K$\Omega$ load from a 110 V (RMS) source supply. Then the dc load current will be

• A. 149.58 mA
• B. 18.26 mA
• C. 42 mA
• D. 47.61 mA

Answer 1

The correct option is D 47.61 mA


The diode will work as a half wave rectifier
thus, $I_{dc}=\frac{I_m}{\pi }$

R = 1000+40 = 1040 $\Omega$

Now, $I_m=\frac{V_m}{R}$ = $\frac{110\times \sqrt{2}}{1040}$
= 149.58 mA
$\therefore I_{dc}=\frac{149.58}{\pi }=47.61mA$