RLC Circuit Frequencies Help

[EROC]

Homework Statement

In each of the four circuits shown,
an AC input voltage, ##V_0 cos(\omega t)## is applied on the
left and the output voltage is measured on the
right. Treat the input frequency as a variable.

(a) For each case, state whether high or low
frequencies are passed or suppressed in the
output and explain your answer briefly.

(b) You have a 0.47 µF capacitor. Select the
circuit that allows high frequencies to pass.
Calculate the resistor value such that ##V_(out) / V_(in)## is
1/ sqrt(2) at 200 Hz. Such a filter could be used to reduce 60 Hz “hum” in an audio circuit. What is ##V_(out) / V_(in)## at 60 Hz?

(c) You have a 3 mH inductor. Select the circuit that allows high frequencies to pass. Calculate the
resistor value such that ##V_(out) / V_(in)## is 1/ sqrt(2) at 200 Hz. Optional: comment on whether you anticipate
any issues implementing this circuit with real components

**Cannot use high-pass / low-pass filters**
This is mainly an RLC circuit question in a basic physics: electrostatics and magnetism course, so it is focused on basic circuit ideas.

Homework Equations

##V=IZ##

##X_L = \omega L##

##X_C = 1 / (\omega C)##

The Attempt at a Solution

I don't know what I am supposed to be doing or observing for the first one. An upperclassman I know can explain it in circuit theory, but I do not know circuit theory and will not be learning it in this course.

Any help would be great, and an explanation please!

 

[EROC]

Here are the diagrams for the problem

 

[NascentOxygen]

These are all LP or HP filters.

To get a feel of what happens as frequency changes here, you can think of capacitors and inductors as frequency-variable resistors. So for each two-element circuit you are looking at a potential divider arrangement, but where the impedance of one of the components changes as the signal frequency rises.

Capacitors drop in ohms as frequency of the signal increases, but with an inductance it's the opposite. At low frequencies an inductor has low ohms (you could think of it as almost a short circuit), whereas at higher frequencies the inductor exhibits lots of ohms.

 

[rude man]

Without basic circuit theory you are really forced to define capacitance C and inductance L as
i= C dV/dt and
V = L di/dt,
where V is the voltage across the device and i is the current thru it.
Then you need to solve the differential equations for the given circuits and excitation functions. This is mandatory if you want quantitative results, since you can't simply assume that capacitors and inductors are frequency-determined resistors.

 

[HalfLostAndSearching]

I understand that you may not have a background in circuit theory and may not be familiar with the concepts being discussed in this question. However, as a scientist, it is important to have a basic understanding of various scientific fields and to be able to learn and apply new concepts when needed.

In this question, we are dealing with RLC circuits, which are circuits that contain a resistor (R), an inductor (L), and a capacitor (C). These components have different properties that affect the behavior of the circuit at different frequencies.

(a) In each of the four circuits, the input voltage is an AC voltage, meaning it varies with time. The input voltage is given by ##V_0 cos(\omega t)##, where ##V_0## is the amplitude of the voltage and ##\omega## is the angular frequency. The angular frequency is related to the frequency (f) by the equation ##\omega = 2\pi f##.

In the first two circuits (a and b), the output voltage is measured across the capacitor. In the third and fourth circuits (c and d), the output voltage is measured across the resistor. The behavior of the circuit at different frequencies can be understood by considering the properties of the components.

Resistors have a constant resistance and do not depend on frequency. Therefore, they do not affect the output voltage.

Inductors have a property called inductive reactance (X_L), which is given by the equation ##X_L = \omega L##. This means that the inductor will have a higher resistance to high frequencies compared to low frequencies. Therefore, in circuits c and d, the output voltage will be lower for high frequencies compared to low frequencies.

Capacitors have a property called capacitive reactance (X_C), which is given by the equation ##X_C = 1 / (\omega C)##. This means that the capacitor will have a higher resistance to low frequencies compared to high frequencies. Therefore, in circuits a and b, the output voltage will be lower for low frequencies compared to high frequencies.

(b) To select the circuit that allows high frequencies to pass, we need to look at circuits c and d, as they have the output measured across the resistor. We can see that circuit c has a higher resistance to high frequencies compared to circuit d, as it has a larger inductor (3mH compared to 1mH). Therefore, circuit c is