Peak Reverse Voltage Diode

Peak value of the input to a half-wave rectifier bein 10V. what reverse voltage must the diode withstand?

I just need to know how to work the problem

10 volts

if there is a capacitor across the output, then the PIV should be 20 volts.

How? look at waveforms for input and output and see what the voltage difference is, and where it is highest.

In this case, when the input is at -10 volts and output is at zero.

With a cap across the output, max occurs when the output is at +10, from the charge stored on the cap, and the input is at -10

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A full wave p-n junction diode rsctifier uses load resister of 1500-ohms with a forward resistance of 10-ohms?

A full wave p-n junction diode rectifier uses load resister of 1500-ohms with a forward resistance of 10-ohms and reverse resistance of alpha.the wave voltage applied to each diode has an amplitude of 30v and frequency of 50Hz.calculate the (i)peak,the d.c and R.m.s. load current.(ii)d.c. power output.(iii)efficiency of the rectifier.

Assume the 30 Volts rms is a sine wave applied to each diode by the opposite ends of the secondary of a 100% eefficient center tapped transformer. I have assumed the reverse resistance alpha to be large enough to be considered an open circuit and therefore dissipate zero power during the time the diodes are reverse biased.

(i) Peak load current = (30V/.707) / 1510 Ohms = 28,1 mA
DC load current and rms ac input current = 30V / 1510 Ohms = 19.87 mA

(ii) DC power out = [(.01987A) x (.01987A) x 1500 Ohms = 592mW

(iii) Efficiency of the rectifier = output Watts / input Watts =.592W / [(30V x .01987A) = 99.3%