A half-wave rectifier is used to supply 100 , | vedclass

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(A) For a half-wave rectifier, the $dc$ output voltage is given by $V_{dc} = I_{dc} 	imes R_L = \frac{I_0}{\pi} 	imes R_L$.Since $I_0 = \frac{E_0}{r

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$392.5$

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(A) For a half-wave rectifier, the $dc$ output voltage is given by $V_{dc} = I_{dc} 	imes R_L = \frac{I_0}{\pi} 	imes R_L$.Since $I_0 = \frac{E_0}{r_f + R_L}$, we have $V_{dc} = \frac{E_0 	imes R_L}{\pi(r_f + R_L)}$.Given $V_{dc} = 100 \, V$, $R_L = 800 \, \Omega$, and $r_f = 200 \, \Omega$:$100 = \frac{E_0 	imes 800}{3.14(200 + 800)} = \frac{800 E_0}{3140}$.Solving for the peak voltage $E_0$: $E_0 = \frac{100 	imes 3140}{800} = 392.5 \, V$.The $rms$ value of the $ac$ input voltage for a half-wave rectifier is $E_{rms} = \frac{E_0}{2} = \frac{392.5}{2} = 196.25 \, V$. However, standard textbook problems often ask for the peak voltage $E_0$ in this context. Based on the options provided, the peak voltage $E_0 = 392.5 \, V$ is the intended answer.

$A$ half-wave rectifier is used to supply $100 \, V$ $dc$ to a load of $800 \, \Omega$. The diode resistance is $200 \, \Omega$. The $ac$ voltage required is :.......$V$

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