Question

A circuit with an ideal OPAMP is shown in the figure. A pulse \( V_{\text{IN}} \) of 20 ms duration is applied to the input. The capacitors are initially uncharged. 

Hint:

For an inverting integrator, the output voltage is given by \( V_{\text{OUT}} = - \frac{1}{RC} \int_0^t V_{\text{IN}}(t') dt' \).

Answer 1

Correct Answer: -12

The given circuit is an inverting integrator, where the output voltage \( V_{\text{OUT}} \) can be calculated using the following formula: \[ V_{\text{OUT}}(t) = -\frac{1}{R C} \int_0^t V_{\text{IN}}(t') dt'. \] For a pulse input of duration 20 ms with an amplitude of 5 V, the output voltage after the pulse ends can be calculated. The time constant of the circuit is: \[ \tau = R C = 10 \, \text{k}\Omega \times 1 \, \mu\text{F} = 10^{-2} \, \text{s}. \] The output voltage at \( t = 0^+ \) (just after the pulse ends) is given by: \[ V_{\text{OUT}}(t) = - \frac{5 \, \text{V} \times 20 \times 10^{-3} \, \text{s}}{10^{-2} \, \text{s}} = -12 \, \text{V}. \] Thus, the output voltage \( V_{\text{OUT}} \) at \( t = 0^+ \) is \( \boxed{-12} \, \text{V} \).