Question

Consider the ideal op-amp based circuit shown below:
then,

• A. $\text{the value of current}{I}_L\text{is equal to}2\cos \left(\omega t\right)\text{mA}$
• B. $\text{the value of current}{I}_{\text{out}}\text{is equal to}-6\cos \left(\omega t\right)\text{mA}$
• C. $\text{the value of current}I\text{is equal to}-4\cos \left(\omega t\right)\text{mA}$
• D. $\text{the value of current voltage}{V}_0\text{is equal to}-20\cos \left(\omega t\right)\text{V}$

Answer 1

Answer: (B), (D)

$\text{the value of current voltage}{V}_0\text{is equal to}-20\cos \left(\omega t\right)\text{V}$


$I=\frac{2\cos \left(\omega t\right)-0}{1\text{k}\Omega }=2\cos \left(\omega t\right)\text{mA}$

$I_L=\frac{V_0}{R_L}$

$\text{and}{V}_0=-\frac{10}{1}\times 2\cos \left(\omega t\right)=-20\cos \left(\omega t\right)V$

$\therefore I_L=-\frac{20}{5}\times \cos \left(\omega t\right)\times {10}^{-3}A=-4\cos \left(\omega t\right)\text{mA}$

$\therefore I_{\text{out}}+I=I_L$

$I_{\text{out}}=[-4\cos (\omega t)-2\cos (\omega t\left)\right]\text{mA}=-6\cos \left(\omega t\right)\text{mA}$