Take the breakdown voltage of the
zener diode used in the given circuit as 6V.
For the input voltage shown in figure below,
the time variation of the output voltage is:
(Graphs drawn are schematic and not to scale)

Solution

Physics sir · 3 steps

Back-to-back zeners: in each half one conducts forward, the other breaks down at 6 V, so both peaks clip.

1. Read the circuit. The two zeners are in series with their cathodes joined in the middle, so they are back to back. The output is taken across this pair, and R₁ takes up the extra voltage.

   Input	Top zener	Bottom zener	Output Vₒ
   |Vᵢₙ| < 6 V	one is forward biased	the other is reverse biased, below breakdown	no current, so Vₒ = Vᵢₙ and follows the sine
   Vᵢₙ > +6 V	forward biased, about 0 V	reverse biased, in breakdown	held at +6 V
   Vᵢₙ < -6 V	reverse biased, in breakdown	forward biased, about 0 V	held at -6 V

2. Shape of output. The input swings to ± 10 V, so the output follows the sine until it reaches ± 6 V. It stays flat until the input falls back below that level. The result has a flat top and a flat bottom in every cycle.
3. So. The output is clipped on both halves, which is the trapezoid-like wave in option (c).

The idea to keep. A single zener clips one side at V_Z and the other side near 0. Two zeners back to back clip both sides at about ± V_Z: one is forward biased while the other is in breakdown.

Answer from the NTA final answer key. Solution by PrepWiser.