Question

Consider a circuit consisting of a capacitor (20 μF), resistor \((100 \Omega)\) 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 \(\alpha \times 10^{-3}\) s. The value of \(\alpha\) is ________.

• A. 2.2
• B. 2.0
• C. 2.1
• D. 2.4

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The time constant ($\tau$) of an RC circuit is given by $\tau = R_{net}C$. In a circuit with diodes, the net resistance depends on the direction of current flow. If the capacitor is charging or discharging through a specific path, only the forward-biased diode contributes its resistance.

Step 2: Key Formula or Approach:
1. $\tau = R_{eq} \times C$ 2. In forward bias, $R_{diode} = 10 \, \Omega$. 3. Total resistance $R_{eq} = R + R_{diode}$ (assuming they are in series in the active branch).

Step 3: Detailed Explanation:
1. Identifying the active branch: Usually, in such problems, the capacitor discharges through the resistor and one forward-biased diode. 2. $R_{eq} = 100 \, \Omega + 10 \, \Omega = 110 \, \Omega$. 3. $C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F}$. 4. $\tau = 110 \times 20 \times 10^{-6} = 2200 \times 10^{-6} = 2.2 \times 10^{-3} \, \text{s}$. 5. Comparing with $\alpha \times 10^{-3}$, we find $\alpha = 2.2$.

Step 4: Final Answer:
The value of $\alpha$ is 2.2.