Question

Determine the corner frequency $\left(f_c\right)$ in Hz, for the below given high pass filter.

• A. 318.4

Answer 1

The correct option is A 318.4

$\frac{V_0\left(t\right)}{V_i\left(t\right)}=\frac{Z_L}{Z_R+Z_L}$

$=\frac{j\omega L}{R+j\omega L}$

$=\frac{j\omega \frac{L}{R}}{1+j\omega \frac{L}{R}}$


From here, we can write

${\omega }_c=\frac{R}{L}$
$and{f}_c=\frac{{\omega }_c}{2\pi }=\frac{\frac{R}{L}}{2\pi }$

$\Rightarrow f_c=\frac{R}{2\pi L}=\frac{200}{2\pi \times 0.1}$

$\Rightarrow f_c=\frac{200}{0.2\pi }=318.4Hz.$