Question

High-pass filter is the one shown in Fig 26.45

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (C), (D)

The correct options are
C C
D D

$Z_R=R;Z_L=jL\omega ;Z_C=1/jC\omega$
impedance of circuit =$Z=Z_R+Z_L+Z_C$
current in circuit =$\frac{V_{in}}{Z}$
output in fig.c : $V_o=$current in circuit$\times Z_L=\left|V_{in}\frac{L\omega }{R+j(L\omega -1/C\omega )}\right|$
output in fig.d : $V_o=$current in circuit$\times (Z_R+Z_L)=\left|V_{in}\frac{R+jL\omega }{R+j(L\omega -1/C\omega )}\right|$
At small frequency $1/C\omega$ increases and $L\omega$ decreases. In both the above outputs, numerator decreases and the denominator increases as frequency is reduced. Therefore output tends to zero as frequency decreases.
Both circuit reject small frequency and allow output at high frequencies.

At high frequency, $1/C\omega$ decreases and $L\omega$ increases. In bot the above outputs, numerator ans the denominator approach the same value as $1/C\omega$ approaches zero. therefore output tends to be equal to the input as the fraction of impedances reaches unity.