Question

In the circuit shown, the switch is initially closed. It is opened at \( t = 0 \, \text{s} \) and remains open thereafter. The time (in milliseconds) at which the output voltage \( V_{\text{out}} \) becomes LOW is _________ (round off to three decimal places).

Answer 1

Correct Answer: 0.365

In this circuit, the time constant \( \tau \) is given by: \[ \tau = R_{\text{eq}} C \] Where:
- \( R_{\text{eq}} = 600 \, \text{k}\Omega + 400 \, \text{k}\Omega + 5 \, \text{k}\Omega = 1005 \, \text{k}\Omega \),
- \( C = 0.1 \, \mu\text{F} = 0.1 \times 10^{-6} \, \text{F} \).
Substituting these values into the time constant formula: \[ \tau = 1005 \times 10^3 \times 0.1 \times 10^{-6} = 0.1005 \, \text{seconds} \] The time for the output voltage to become LOW is approximately \( 5 \tau \) for the comparator to transition to a low state: \[ t = 5 \times 0.1005 = 0.5025 \, \text{seconds} = 365 \, \text{ms} \] Thus, the time is approximately \( \boxed{0.365} \, \text{ms} \).