Diagram shows a circuit consisting of some elements. The input and output of the circuit is shown. Choose the correct option that shows the component in box. (Zener diode has a breakdown potential of 5V)

The correct option is Option 4 (as shown in the fourth image, consisting of a Zener diode pointing from X to Y in series with a standard diode pointing from Y to X).

Step-by-Step Analysis:

1. Input and Output Waveforms:
The input voltage is given as:
$V_{\mathrm{in}}=20\sin \left(\omega t\right)$
This is an alternating voltage source with a peak value of 20 V. The output waveform is symmetric and clipped at exactly +5 V and -5 V. The Zener breakdown potential is given as $V_Z=5\text{V}$.

2. Operation in the Positive Half-Cycle ($V_{\mathrm{in}}>0$):
During the positive half-cycle, terminal X is at a higher potential than terminal Y ($V_X>V_Y$).
Looking at the middle branch containing two back-to-back Zener diodes: - The upper Zener diode (pointing down) is forward-biased.
- The lower Zener diode (pointing up) is reverse-biased.
Once the potential difference exceeds 5 V, the lower Zener diode undergoes breakdown, clamping the voltage across terminals X and Y at +5 V (assuming ideal diode characteristics where the forward voltage drop is negligible).
For the box component in Option 4: the Zener diode (pointing from X to Y) is forward-biased, and the standard diode (pointing from Y to X) is reverse-biased. Thus, the box acts as an open circuit and does not affect the output voltage, which remains regulated at +5 V by the middle branch.

3. Operation in the Negative Half-Cycle ($V_{\mathrm{in}}<0$):
During the negative half-cycle, terminal Y is at a higher potential than terminal X ($V_Y>V_X$).
For the component in the box (Option 4): - The standard diode (pointing from Y to X) is forward-biased and conducts.
- The Zener diode (pointing from X to Y) is reverse-biased.
Once the potential difference $V_Y-V_X$ reaches the breakdown voltage of 5 V, the Zener diode in the box undergoes breakdown, clamping the output voltage at -5 V.
This ensures that the negative peak is also clipped at exactly -5 V.