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Question: the maximum efficiency of the full wave rectifier is (A) \(\dfrac{4}{{{\pi ^2}}} \times 100\% \) ...

the maximum efficiency of the full wave rectifier is
(A) 4π2×100%\dfrac{4}{{{\pi ^2}}} \times 100\%
(B) 8π2×100%\dfrac{8}{{{\pi ^2}}} \times 100\%
(C) 4040%
(D) 8080%

Explanation

Solution

Hint
The efficiency is defined as the ratio of DC power to the AC power i.e.
η=PdcPac\eta = \dfrac{{{P_{dc}}}}{{{P_{ac}}}}
As, Pac=Irms2(rf+Rl){P_{ac}} = I_{rms}^2\left( {{r_f} + {R_l}} \right) and Pdc=Idc2Rl{P_{dc}} = I_{dc}^2{R_l}, on substituting the values we will get the efficiency of full wave rectifier.

Complete step by step solution
The efficiency is the ratio of output power to the input power and the rectifier is a device which converts AC signal into DC signal. So we can say that efficiency is the ratio of DC power to the AC power. So, it can be written as
η=PdcPac…(1)\eta = \dfrac{{{P_{dc}}}}{{{P_{ac}}}} … (1)
Now, we also know that power is the product of voltage applied and the resistance.
Therefore, DC power can be written as Pdc=Idc2Rl{P_{dc}} = I_{dc}^2{R_l}
Where, RIR_I is the load resistance.
As, Idc=2Imπ{I_{dc}} = \dfrac{{2{I_m}}}{\pi }
Therefore, Pdc=(2Imπ)2Rl…(2){P_{dc}} = {\left( {\dfrac{{2{I_m}}}{\pi }} \right)^2}{R_l} … (2)
And AC power is Pac=Irms2(rf+Rl){P_{ac}} = I_{rms}^2\left( {{r_f} + {R_l}} \right)
RfR_f is the forward resistance of diode
And we also know that Irms=Im2{I_{rms}} = \dfrac{{{I_m}}}{{\sqrt 2 }}
Therefore, Pac=(Im2)2(rf+Rl)…(3){P_{ac}} = {\left( {\dfrac{{{I_m}}}{{\sqrt 2 }}} \right)^2}\left( {{r_f} + {R_l}} \right) … (3)
Substituting the values of equation (2) and (3) in equation (1), we get
⇒η=(2Imπ)2Rl(Im2)2(rf+Rl)\Rightarrow \eta = \dfrac{{{{\left( {\dfrac{{2{I_m}}}{\pi }} \right)}^2}{R_l}}}{{{{\left( {\dfrac{{{I_m}}}{{\sqrt 2 }}} \right)}^2}\left( {{r_f} + {R_l}} \right)}}
As, is less than then the efficiency becomes maximum therefore, above equation becomes
⇒η=8π2\Rightarrow \eta = \dfrac{8}{{{\pi ^2}}}
In percentage form it can be written as
⇒η=8π2×100%\Rightarrow \eta = \dfrac{8}{{{\pi ^2}}} \times 100\%
Hence, option (B) is correct.

Note
It must be noticed that the efficiency is maximum when the forward resistance of the diode is less than the load resistance across which the output is obtained. We can also say the maximum efficiency of the full wave rectifier is 81.2%.