RCL Circuit: 2 Resistors, 1 Capacitor, 2 Inductors

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SUMMARY

The discussion focuses on solving an RCL circuit problem involving two resistors (50 Ohms each), two inductors (20 mH and 2.5 mH), and a capacitor (50 µF) with a voltage source of V = (45v)sin(80πt). Key tasks include calculating the power dissipated by each resistor, determining the frequency at which both resistors dissipate equal power, and analyzing the impact of swapping the inductors. Participants emphasize the importance of using RMS current for power calculations and suggest setting up equations based on Kirchhoff's Voltage Law to simplify the analysis.

PREREQUISITES
  • Understanding of RCL circuit analysis
  • Familiarity with Kirchhoff's Voltage Law
  • Knowledge of RMS current and power calculations
  • Basic algebra involving complex numbers
NEXT STEPS
  • Learn how to calculate average power in RCL circuits using RMS values
  • Study the effects of frequency on impedance in RCL circuits
  • Explore the application of Kirchhoff's laws in circuit analysis
  • Investigate the relationship between reactance and frequency in inductors and capacitors
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Electrical engineering students, circuit designers, and anyone involved in analyzing RCL circuits and optimizing power dissipation in electrical systems.

B. Vane
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Homework Statement


NOTE: Image uploaded with thread shows problem in clearest possible form.

V = (45v)sin(80[pi] t)
Resistors 1 and 2 = 50 Ohms
Inductor 1 = 20 mH
Inductor 2 = 2.5 mH
Capacitor = 50 uF

a) Find the power dissipated by each resistor.
b) At what frequency will both resistors dissipate the same power?
c) What would be the answer to part (b) if you swapped the inductors?
d) Justify your answers to parts (b) and (c) with a graph of the impedance of each branch as function of frequency.
d) What is the current delivered by the power supply at very low frequencies?

Homework Equations


They are in the second uploaded document. They are all the equations I believe for RCL Circuits.

The Attempt at a Solution


I'm having trouble knowing where exactly to start.
I believe I have to try and condense the Resistors and Inductors into 1 loop but I'm not sure how to.
Plus all the other parts of the question are intimidating to say the least for me.
I can do simpler problems but we never covered an example like this in the course and I've searched high and low online to no avail. Please help, it will be much appreciated.
 

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Hello B. Vane,

Welcome to Physics Forums! :smile:

B. Vane said:

Homework Statement


NOTE: Image uploaded with thread shows problem in clearest possible form.

V = (45v)sin(80[pi] t)
Resistors 1 and 2 = 50 Ohms
Inductor 1 = 20 mH
Inductor 2 = 2.5 mH
Capacitor = 50 uF

a) Find the power dissipated by each resistor.
b) At what frequency will both resistors dissipate the same power?
c) What would be the answer to part (b) if you swapped the inductors?
d) Justify your answers to parts (b) and (c) with a graph of the impedance of each branch as function of frequency.
d) What is the current delivered by the power supply at very low frequencies?

Homework Equations


They are in the second uploaded document. They are all the equations I believe for RCL Circuits.

The Attempt at a Solution


I'm having trouble knowing where exactly to start.
I'd start with part a).

Seriously though, start with finding the current through each resistor. Your list of impedances for different types of components should be useful. Prepare for a considerable bit of algebra involving complex numbers.

Once you have the current through a given resistor, you should then find the expression for the power dissipated by that resistor.

I advise coming up with an expressions for the power (or at least the current through each resistor) in terms of variables, and plug the numbers into the expression as a final step. That way, you can "reuse" much of your work between steps (part a and c for example). Otherwise you might find yourself having to start from the beginning each time something changes.

I believe I have to try and condense the Resistors and Inductors into 1 loop but I'm not sure how to.
I'm not sure what you mean.

There are two loops. If you set the loops up smartly, you should be able to work with each loop individually for parts a), c), and d).

(Well, the first d that is. The second d should be pretty easy though after some thought [it's one of the easier steps].)

Plus all the other parts of the question are intimidating to say the least for me.
This problem is long, containing many parts. Give yourself time. This isn't the type of problem that you can complete in a couple minutes. Just give yourself a good chunk of time to work on it, and work through each part one step at a time. Think of it as a project.

I can do simpler problems but we never covered an example like this in the course and I've searched high and low online to no avail. Please help, it will be much appreciated.
I have a feeling that this problem combines many of the simpler problems that you already know into one bigger problem (with many steps). Just be patient, have perseverance, and work through it step by step. :smile:
 
Last edited:
B. Vane said:
I believe I have to try and condense the Resistors and Inductors into 1 loop but I'm not sure how to.

You don't have to set up equations for loops ( Kirchhoffs Voltage Law). For example:

Upper path: Voltage over resistor+capacitor+inductor = 45V.
Middle path: Voltage over resistor+inductor = 45V.

It's as simple as that.
 
Last edited:
collinsmark said:
Hello B. Vane,

Welcome to Physics Forums! :smile:I'd start with part a).

Seriously though, start with finding the current through each resistor. Your list of impedances for different types of components should be useful. Prepare for a considerable bit of algebra involving complex numbers.

Once you have the current through a given resistor, you should then find the expression for the power dissipated by that resistor.

I advise coming up with an expressions for the power (or at least the current through each resistor) in terms of variables, and plug the numbers into the expression as a final step. That way, you can "reuse" much of your work between steps (part a and c for example). Otherwise you might find yourself having to start from the beginning each time something changes.I'm not sure what you mean.

There are two loops. If you set the loops up smartly, you should be able to work with each loop individually for parts a), c), and d).

(Well, the first d that is. The second d should be pretty easy though after some thought [it's one of the easier steps].)This problem is long, containing many parts. Give yourself time. This isn't the type of problem that you can complete in a couple minutes. Just give yourself a good chunk of time to work on it, and work through each part one step at a time. Think of it as a project.I have a feeling that this problem combines many of the simpler problems that you already know into one bigger problem (with many steps). Just be patient, have perseverance, and work through it step by step. :smile:

I've been working on it and here Is what I've got so far... not really sure any of it is correct though. (Sorry its so big, can't/ don't know how to resize it)
dWKCs6D.jpg

I don't really have an answer for much. I guess the power disapated in R1 and R2 but I'm pretty sure my current calculated running though R2 is wrong.

Any more guidance will be greatly appreciated!
 
Last edited by a moderator:
Hello B. Vane,

You have the right idea up to a point.

In the first branch (the one that has the capacitor, inductor and resistor) things look pretty good up until the point where you calculate the current at time t = 0. It is not necessary to calculate the instantaneous current in the circuit. I think the problem as asking for the "average" power dissipated by each resistor, not the instantaneous.

Just use the rms current, when calculating the power.

In the second branch (the one that contains an inductor and resistor), again, use the rms current, rather than the peak current, when calculating average power.
 
As collinsmark said, when you use ##Power=I^2R## that formula needs the RMS voltage.

For (b) you have to repeat the calculations, this time retaining ##\omega## in the reactances because you'll end up solving for that value of ##\omega## that makes the resistors' powers equal.
 

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