Question

Low-pass filter is the one shown in Figure

• A. a
• B. b
• C. c
• D. d

Answer 1

The correct option is A a

$Z_R=R;Z_L=jL\omega ;Z_C=1/jC\omega$
impedance of circuit =$Z=Z_R+Z_L+Z_C$
current in circuit =$\frac{V_{in}}{Z}$
$V_o=$ current in circuit$\times Z_C$
$V_o=|\frac{V_{in}}{R+j(L\omega -1/C\omega )}\times 1/C\omega |$
$\omega =2\pi f$
The reactance of a capacitor varies inversely with frequency, while the value of the resistor remains constant as the frequency changes. At low frequencies the capacitive reactance, ( Xc ) of the capacitor will be very large compared to the resistive value of the resistor, R and inductor. This means that the voltage potential, Vc across the capacitor will be much larger than the voltage drop, Vr developed across the resistor and inductor. At high frequencies the reverse is true with Vc being small and Vr being large due to the change in the capacitive reactance value
It allows low frequency. AT high frequency output is zero.