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

The time constant of an $$RC$$ network is given by
$$\tau = R_{\text{eq}}\,C$$

where
  • $$C$$ is the capacitance, and
  • $$R_{\text{eq}}$$ is the net resistance actually present in the current path that charges / discharges the capacitor.

Data from the problem:
  • Capacitor: $$C = 20 \,\mu\text{F} = 20 \times 10^{-6}\,\text{F}$$
  • Fixed resistor: $$R = 100\,\Omega$$
  • Two identical diodes whose forward (dynamic) resistance is $$r_d = 10\,\Omega$$ each.

Because the two diodes are connected with opposite orientation (as is usual in such clipping / rectifying circuits), only one diode is forward-biased at any instant. Hence, along the conduction path there is:

$$R_{\text{eq}} = R + r_d = 100\,\Omega + 10\,\Omega = 110\,\Omega$$

Therefore, the time constant becomes

$$\tau = R_{\text{eq}}\,C = 110\,\Omega \times 20 \times 10^{-6}\,\text{F}$$

$$\tau = 2200 \times 10^{-6}\,\text{s} = 2.2 \times 10^{-3}\,\text{s}$$

Comparing with $$\tau = \alpha \times 10^{-3}\,\text{s}$$, we get

$$\alpha = 2.2$$

Hence, the correct choice is
Option A which is: $$2.2$$