Question

Consider the op-amp circuit shown in the figure below.

$\text{The source}{V}_i\text{is a sinusoidal signal with input frequency of 50 Hz and rms value of 10 V. Then the magnitude of source current}{I}_i\text{is equal to}$

• A. 18.224 mA
• B. 36.666 mA
• C. 43.145 mA
• D. 31.415 mA

Answer 1

The correct option is D 31.415 mA


Applying KCL at inverting terminal.
$\frac{V^{-}-0}{1/sC}+\frac{V^{-}-V_0\left(s\right)}{1/sC}=0[\because V^{+}=V^{-}=V_i(s\left)\right]$
$\frac{V_0\left(s\right)}{V_i\left(s\right)}=(1+RCs)$
$\therefore \text{The current}{I}_i=\frac{V_i\left(s\right)-V_0\left(s\right)}{R}$
$\therefore I_i=\frac{-sRC}{R}.V_i\left(s\right)=sC{V}_i\left(s\right)$
$I_i=-j\omega C{V}_i\left(s\right)$
$|I_i|=2\pi f\times 10\times {10}^{-6}\times 10$
$=10\pi \times {10}^{-3}=31.415\text{mA}$