# RLC Circuit Frequencies Help

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1. Nov 20, 2014

### EROC

1. The problem statement, all variables and given/known data

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

(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.

2. Relevant equations

$V=IZ$

$X_L = \omega L$

$X_C = 1 / (\omega C)$

3. 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!

2. Nov 20, 2014

### EROC

Here are the diagrams for the problem

#### Attached Files:

• ###### Physics HW 11.png
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3. Nov 20, 2014

### Staff: Mentor

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.

Last edited: Nov 21, 2014
4. Nov 21, 2014

### 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.