Question

Consider the circuit shown in the figure below:


If the op-amp is ideal, then the average value of the voltage across $100\Omega$ resistor is equal to

• A. 3 V
• B. 0 V
• C. 5 V
• D. 10 V

Answer 1

The correct option is B 0 V



$\begin{array}{cc}{I}_i& =C\frac{d}{dt}\left[V_{in}\right(t)-0]\\ I_i& =C\frac{d{v}_i}{dt}\\ \text{now}{v}_0& =-R{I}_i\\ v_0& =-RC\frac{d{v}_i}{dt}\\ v_0& =-{10}^{-3}\frac{d{v}_i}{dt}\\ v_{in}\left(t\right)& =\{\begin{array}{cc}5t& 0<i<1s\\ 5& 1s<5<2s\\ -10t+25& 2s<t<3s\\ 5t-20& 3s<5<4s\end{array}\\ \therefore v_0\left(t\right)& =\{\begin{array}{cc}-5\times {10}^{-3}& 0<i<1s\\ 0& 1s<5<2s\\ +{10}^{-2}& 2s<t<3s\\ -5\times {10}^{-3}& 3s<5<4s\end{array}\end{array}$


$\therefore v_0\left(t\right){|}_{average}={\int }_{-\infty }^{\infty }{v}_0\left(t\right)dt=0$