Question

Show that the maximum efficiency of a full wave rectifier is 81.2%

Answer 1

The Maximum Efficiency Of A Full Wave Rectifier Is 81.2%

GIVEN

Maximum efficiency of Full wave rectifier circuit

TO PROVE:

The maximum efficiency of a full wave rectifier is 81.2%

Solution:

A full wave rectifier is a rectifier circuit which can transform an ac voltage into dc voltage with the half cycles of the applied ac voltage.

we know that,

$η =\frac{Pdc}{dc} × 100$          

Power of dc can be defined as -

$Pdc = \frac{Vdc^{2} }{Rl}$

$Vdc = \frac{Vₘ}{ π}$

where,

Vdc = DC voltage

Vₘ = voltage applied

Rl = resistence

Similarly,

AC power can be defined as -

$Pac = \frac{ {Vrms}^{2} }{Rl}$

$Vrms= \frac{Vₘ}{ \sqrt{2} }$

where,

Vrms = RMS voltage at load resistence

Vₘ = Voltage applied

Rl = Load resistence

Now, efficiency of the rectifier is the ratio of power of DC to power of AC

so,

$η= \frac{Pac}{Pdc}$

Putting the values of Pdc and Pac we get,

$η \: = \frac{Vdc^{2} }{Vrms^{2} }$

Now putting the value of Vdc and Vrms  we get

$η = \frac{ { \frac{2Vₘ}{\pi} }^{2} }{ { \frac{Vₘ}{ \sqrt{2} } }^{2} }$

$η = \frac{8}{ {\pi}^{2} } = 0.812$

or,

η = 0.812 × 100 = 81.2%

Hence, The maximum efficiency of a full wave rectifier is 81.2%

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