JEE Main 2026 · 4 Apr · Shift 1 · Physics Q42

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Consider a circuit consisting of a capacitor (20 μF), resistor (100 Ω) and two identical diodes as shown in figure. The resistance of diode under forward biasing condition is 10 Ω. The time constant of the circuit is $\alpha\times10^{-3}$ s. The value of $\alpha$ is ________.

Series loop: battery V, capacitor C = 20 μF, resistor R = 100 Ω, and a pair of identical diodes connected in parallel with opposite orientations (one points down, the other up), so one of them always conducts.

2.2

2.0

2.1

2.4

Answer and worked solution

Official answer: A

The diode pair. The two diodes are antiparallel, so one is always forward-biased and adds 10 Ω in series.

Time constant. $\tau=(100+10)\times20\times10^{-6}=2.2\times10^{-3}$ s.

(A)

Topics:RC circuits Semiconductors and p-n junction

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 42