Question

For the given circuit below it is given that
$V_{in}\left(t\right)=10\sqrt{2}\cos \left(2\pi 60t\right)V$ and $V_0\left(t\right)=10\sqrt{2}\cos (2\pi 60t-{120}^{\circ })V$.
(Assuming the OPAMP to be ideal)
The possible value of R is

• A. 35.94 k$\Omega$
• B. 40.00 k$\Omega$
• C. 44.94 k$\Omega$
• D. 45.94 k$\Omega$

Answer 1

The correct option is D 45.94 k$\Omega$

$V_P=\frac{\frac{1}{sC}}{\frac{1}{sC}+R}.V_{in}=\frac{1}{1+sRC}.V_{in}$Due to virtual ground,$V_N=V_P=\frac{1}{1+sRC}.V_{in}$

Apply KCL,

$\frac{V_{in}b-V_N}{100k}=\frac{V_N-V_0}{100k}$

$\Rightarrow$ $V_0=2V_N-V_{in}$

$V_0=\frac{2}{1+sRC}.V_{in}-V_{in}$

$\frac{V_0}{V_{in}}=\frac{1-sRC}{1+sRC}$

Put $(s=j\omega )$

$\frac{V_0\left(j\omega \right)}{V_{in}\left(j\omega \right)}=\frac{1-j\omega RC}{1+j\omega RC}$

$\angle \frac{V_0\left(j\omega \right)}{V_{in}\left(j\omega \right)}=-ta{n}^{-1}\omega RC-ta{n}^{-1}\omega RC=-2ta{n}^{-1}\omega RC$

$\angle \frac{V_0\left(j\omega \right)}{V_{in}\left(j\omega \right)}=-{120}^{\circ }=-2ta{n}^{-1}\omega RC$

$\omega RC=tan60$

$R=\frac{tan60}{2\pi \times 60\times 0.1\times {10}^{-6}}=45.94k\Omega$