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  • A full wave p – n diode rectifier uses a load resistor of  No filte

A full wave p – n diode rectifier uses a load resistor of \mathrm{1500\Omega } No filter is used. The forward bias resistance of the diode is \mathrm{10\Omega } The efficiency of the rectifier is:

Option: 1

81.2 %


Option: 2

40.6%


Option: 3

80.6%


Option: 4

40.2%


Answers (1)

best_answer

\mathrm{\text { Here, } r_i=10 \Omega, R_L=1500 \Omega}

Efficiency of the full wave rectifier is

\mathrm{\eta=\frac{P_{d c}}{P_{a c}}=\frac{\left(2 \mathrm{I}_m / \pi\right)^2 R_L}{\left(I_m / \sqrt{2}\right)^2\left(\mathrm{r}_f+R_L\right)}=\frac{0.812 R_L}{r_f+R_L}=\frac{0.812 \times 1500}{10+1500}=0.806=80.6 \%}

Posted by

Devendra Khairwa

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