For a half-wave rectifier, the load resistanc | vedclass

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(C) The efficiency of a half-wave rectifier is given by the formula: $\eta = 40.6 \left( \frac{R_L}{R_d + R_L} \right) \%$. Given: $R_L = 2 \, k\Omega

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$20.3$

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(C) The efficiency of a half-wave rectifier is given by the formula: $\eta = 40.6 \left( \frac{R_L}{R_d + R_L} ight) \%$. Given: $R_L = 2 \, k\Omega$ and $R_d = 2 \, k\Omega$. Substituting the values into the formula: $\eta = 40.6 \left( \frac{2 \, k\Omega}{2 \, k\Omega + 2 \, k\Omega} ight) \%$. $\eta = 40.6 \left( \frac{2}{4} ight) \%$. $\eta = 40.6 	imes 0.5 = 20.3 \%$. Thus, the rectification efficiency is $20.3 \%$.

For a half-wave rectifier, the load resistance is $R_L = 2 \, k\Omega$ and the $P-N$ junction diode resistance is $R_d = 2 \, k\Omega$. The rectification efficiency is .... $\%$.

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