Assuming the ideal diode, draw the output waveform for the circuit given in fig. (a), explain the waveform

Explanation

When the input voltage is equal to or less than `5V`, diode will be revers biased. It will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above `5V`.
If input voltage is more than `+5 V`, diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond `5V` as the voltage beyond `+5V` will appear across R.
When input voltage is negative, there will be oppositive to `5V` battery in `p - n` junction input voltage becomes more than `- 5V`, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on the output terminals.