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JEE Main 2026 · 2 Apr · Shift 1, Physics question 37

For the given circuit (shown in part (A)) the time dependent input voltage vin(t)v_{in}(t) and corresponding output vo(t)v_o(t) are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y.

(A) Circuit used: an AC source v_in drives, in parallel, (i) a branch with an ideal diode (pointing down) in series with a Zener diode of V_Z = 5 V (pointing up), (ii) an unknown branch between points X (top) and Y (bottom), and (iii) the output resistor across which v_o(t) is taken. (B) Input voltage: one sine cycle of amplitude 20 V over t = 0 to 2 min. (C) Output voltage: +5 V for the first half-cycle and −5 V for the second, flat-topped (clipped at ±5 V).
Choose one option
Answer and worked solution

Official answer: B

Given branch. The ideal diode in series with the 5 V Zener conducts for one polarity and holds vov_o at 5 V (forward diode + Zener breakdown), clipping the positive half at +5 V.

Missing branch. The output is also clipped at −5 V in the negative half, so the X-Y branch must be the mirror image: an ideal diode and a Zener with both directions reversed relative to the given branch. Option (B) is that pair.

A branch with a resistor (A, C) would not hold the output at a fixed −5 V. In (D) the diode and Zener point the same way, so the branch conducts in the same half-cycle as the given one and nothing clips the negative half.

(B)

Topics:Special diodes (LED, Zener, photodiode, solar cell)Diode as rectifier
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