Figure shows a typical circuit for low-pass filter. An `AC` input `V_i = 10 mV` is applied at the left end and the output `V_0` is received at the right end. Find the output voltages for `v = 10 kHz, 100 kHz, 1.0 MHz and 10.0 MHz`. Note that as the frequency is increased the output decreases and hence the name low-pass filter.

Explanation

Here `V_1 = 10 xx 10^-3V`,
`R=1 xx 10^3 Omega`,
`C=10 xx 10(-9) F`
(a) `X_C=(1)/(omegaC) = (1)/(2pifC)`
`=(1)/(2pi xx 10 xx 10^3 xx 10 xx 10^-9)
`=(1)/(2pi xx 10^-4) = 10^4/(2pi)=5000/pi`
`Z= sqrt(R^2 + X_c^2)`
`=sqrt((1+10^3) +(500//pi)^2)`
`=sqrt(10^6 + (5000//pi)^2)`
`I_0 = E_0/Z = V_1/Z`
`=(10 xx 10^-3)/(sqrt(10^6 + (5000//pi)^2)`
`V_0=I_0X_c`
`=10^-2/(sqrt(10^5 + (5000//h^2))) xx 5000/pi`
`= 0.00846V=8.46=mV =8.5mV`
`X_c=(1)/(omegaC) = (1)/(2pifc)`
`=(1)/(2pi xx 10^5 xx 10 xx 10^-9)`
`(1)/(2pi10^-3)`
`=(10^3)/(2pi) = (1000)/(2pi) = (500)/(pi)`
`Z=sqrt(R^2 + X_c^2)`
`=sqrt((10^3)^2 + (500//pi)^2)`
`I_0 = E_0/Z = V_1/Z`
`=(10 xx 10^3)/(sqrt(10^5 + (500//pi)^2)`
`V_0=I_0 X_c`
`=(10^2)/(sqrt(10^5 + (50//pi)^2)) xx (500)/(pi)`
`=1.6124 V= 1.6 mV`
(c) `f=1 MHz = 10^6 Hz`
`X_c = (1)/(omegaC)=(1)/(2pifC)`
`=(1)/(2pi xx 10^6 xx 10^-9 xx 10)`
`=1/(2pi xx 10^-2) = 100/ (2pi) = 500 /pi`
`Z=sqrt(R^2 + X^2C)`
`=sqrt((10^3)^2 +(50//pi)^2)`
`I_0 = E_0/Z = V_1/Z`
`=(10 xx 10^-3)/sqrt(10^6 +(50//pi)^2)`
`V_0 = I_0X_C`
`=(10^-2)/(sqrt(10^5 +(50//pi)^2)) xx 50/pi`
`=0.16 mV`
(d) `f=10 MHz`
`=10 xx 10^6 Hz = 10^7 Hz`
`X_c=(1)/(omegaC)= (1)/(2pifC)`
`=(1)/(2pi xx 10 ^7 xx 10 xx 10 ^-9) = 5/pi`
`Z=sqrt(R^2 + X_c^2)`
`=sqrt((10^3)^2 + (5//pi)^2)`
`i_0 = E_0/2 = V_1/Z`
`=(10 xx 10^-3)/sqrt(10^6 + (5//pi)^2)`
`V_0 = I_0X_C`
`=(10^-2)/(sqrt(10^6 +(5//pi)^2)) xx 5/pi = 16 mu V`.