Question

Consider the circuit shown in the figure below.

The source $I_s=5u\left(t\right)$ mA is connected to an RC circuit which in turn is connected to an ideal op-amp circuit.
The value of output voltage $V_0$ at t=5 sec will be ______V.

• A. 0.6

Answer 1

The correct option is A 0.6



$V_{out}=V^{+}$
Since op-amp is used as a voltage buffer $V_0=V^{+}$
Thus $V^{+}\left(0\right)$ can be given as



$i_1=\frac{250}{250+1000}\times 5\times {10}^{-3}A=\frac{250}{1250}\times 5\times {10}^{-3}$
$i_1=1mA$ $\therefore V^{+}\left(0^{+}\right)=1\times {10}^{-3}\times 1\times {10}^3=1V$
At $t=\infty$, we have



Thus, $V^{+}\left(\infty \right)=0V$
Time constant, $\tau =R_{eq}C$
$R_{eq}=250+1000=1250\Omega$
$\tau =1250\times 8\times {10}^{-3}sec$
$=10sec$
$V^{+}\left(t\right)=V^{+}\left(\infty \right)+\left(V^{+}\right(\left(0^{+}\right)-V^{+}\left(\infty \right))e^{-t/\tau }=V^{+}(0\left)\right)e^{-t/\tau }=e^{-t/10}volts$
for t=5sec
$V^{+}\left(5\right)=e^{-\frac{5}{10}}Volts=e^{-\frac{1}{2}}=e^{(-0.5)}$
$V_0=V^{+}\left(5\right)=0.6Volts$