Question

A diode whose internal resistance is $20\Omega$ is to supply power to a 1000 $\Omega$ load from a 110 V (rms) source of supply. The total input power to the circuit is _________W.

• A. 5.92

Answer 1

The correct option is A 5.92


Peak load current $I_{0peak}=\frac{V_{rms}\sqrt{2}}{R_f+R_L}=\frac{110\times \sqrt{2}}{1020}$

$I_{max}=0.1525A$

$I_{0rms}=$ rms value of load current

$=\frac{Imax}{2}=\frac{0.1525}{2}=0.0762$
The total input power to the circuit is
$=I_{0rms}^2(r_f+R_L)$
$=(0.0762{)}^2\times (20+1000)$
= 5.92 W