Question

Consider the ideal op-amp circuit shown in the figure below:



An input voltage signal is applied to the op-amp where $V_{in}\left(t\right)$= 6sin($\omega t$) mV, then the value of output current $I_0$ from the op-amp is equal to

Answer 1

From the figure node A will be at ground due to virtual grounding.
Thus,

$I_1=\frac{V_i}{1\times {10}^3}=6\sin \left(\omega t\right)\mu A$

$V_o=-\left(\frac{10}{1}\right)V_i=-10V_i$

$\therefore V_o=-60\sin \left(\omega t\right)mV$

$Thus,I_2=\frac{V_0}{4\times {10}^3}=-15\sin \left(\omega t\right)\mu A$

Applying KCL at node $V_0,$ we get

$I_0+I_1=I_2$

$I_0=I_2-I_1$

$=[-15\sin (\omega t)-6\sin (\omega t\left)\right]\omega A$

$=-21\sin \left(\omega t\right)\mu A=-0.021\sin \left(\omega t\right)mA$