Frequency response 4

Consider the circuit on the image with a voltage source V = 10 2 s i n ( ω t ) V = 10\sqrt{2} sin(\omega t) . What is the effective voltage across the capacitor as the frequency of my voltage source goes to infinity?


The answer is 0.

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1 solution

João Areias
Apr 9, 2018

The resistor with the capacitor are a voltage divider. Let's call Z c Z_c the capacitor's impedance and R R the resistor's impedance, we can calculate the voltage across the capacitor as

V o = V i ( Z c R + Z c ) V_o = V_i \cdot \left(\frac{Z_c}{R + Z_c}\right)

We can calculate Z c Z_c in function of the frequency as Z c = j 2 π f C Z_c = \frac{-j}{2\pi f C} with j = 1 j = \sqrt{-1} . It's easy to see that the capacitor's impedance will go to infinity as the frequency goes to zero, which means that

lim Z c ( Z c R + Z c ) = 0 R = 0 \lim_{Z_c \to \infty}\left(\frac{Z_c}{R + Z_c}\right) = \frac{0}{R} = 0 so V o = 0 V_o = 0

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