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 _____

From the given circuit diagram, the battery is connected such that its positive terminal is upwards. This means the current will flow in the clockwise direction during the charging of the capacitor.

In the parallel combination of the two diodes, the left diode is pointing downwards (in the direction of the current) and will be forward-biased. The right diode is pointing upwards (opposite to the current) and will be reverse-biased.

The reverse-biased diode offers infinite resistance and acts as an open circuit. The forward-biased diode offers a resistance of \(10\ \Omega\).

The total equivalent resistance of the circuit is the sum of the resistor \(R\) and the resistance of the forward-biased diode:
\(R_{eq} = 100\ \Omega + 10\ \Omega = 110\ \Omega\)

The time constant \(\tau\) of the \(RC\) circuit is given by:
\(\tau = R_{eq}C\)
\(\tau = 110 \times 20 \times 10^{-6}\ \text{s}\)
\(\tau = 2200 \times 10^{-6}\ \text{s} = 2.2 \times 10^{-3}\ \text{s}\)

Comparing this with the given time constant \(\alpha \times 10^{-3}\ \text{s}\), we get:
\(\alpha = 2.2\)

Answer: \(2.2\)