# Homework Help: LRC circuit help

1. Apr 26, 2017

### Tyler R

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

I built a multiloop circuit for a project. The 4.2v is from mini-USB. The 3.7V is a lipo battery. The circuits are A/V Transmitters (LC circuit) which have an impedance of 75 ohms. The resistor is a 470-ohm resistor. I don't think I'm calculating KVL properly and would like some advice on where to move next. I want to calculate the current throughout the circuit without using my ammeter. The two large ovals are copper plates that act as positive and negative terminals
http://imgur.com/a/j1qgc

2. Relevant equations
KVL & Ohms Law.

3. The attempt at a solution
I'm not sure how to incorporate a charging battery as well as the LC circuits. Do I create a loop for each separate circuit? Thank you for any help

http://imgur.com/a/j1qgc

Last edited by a moderator: Apr 26, 2017
2. Apr 26, 2017

### Staff: Mentor

Can you provide more details about the transmitter circuits? It's not clear what kind of load they actually represent. Is your circuit only providing power to them? Maybe you have power specs for them? Operating current (load/no load)?

Your 470 Ω resistor is only going to allow about 1 mA to flow from the 4.2 V USB source to the 3.7 V battery and load.

3. Apr 26, 2017

### Tyler R

Thank you for the reply gneill :). The transmitter circuits are as follows: My circuit provides power to them at 3.73V measured with a voltmeter
http://imgur.com/a/i2eRQ

From what I can tell, the transmitters use about 90mA each

Last edited: Apr 26, 2017
4. Apr 26, 2017

### Tyler R

We can find i1 by Applying KVL across the first loop. 4.2-470∗i_1-3.7=0. i_1=1.06mA.

5. Apr 26, 2017

### Staff: Mentor

According the spec sheet they draw 90 mA @ 5 V typically. Depending upon how clever their design they may draw more if the supply voltage is lowered below 5 V if they are to maintain their transmit power spec.

So, supposing that their power requirement is fixed, use the datasheet specs to determine the DC power. Then see what resistance that represents at your measured operating voltage (it'll be significantly less than 70 Ω).

After that the circuit can be analyzed. For completeness you might want to include the internal resistance of the Lipo battery, although it'll be pretty small (a few milliohms).

6. Apr 26, 2017

### Tyler R

I appreciate all your feedback. I'm currently unsure how I would go about determining the resistance at 3.73V if I'm following correctly. The circuits might have been designed cleverly because the signal is very strong when connected. Let's say the power requirement is fixed. Would I be using V=IR to determine the current from the data sheet.
5V / 75 Ohms=6.66E-2,
Then using the current to determine the resistance? If so I'm getting around 0.2487 Ohms

7. Apr 26, 2017

### Staff: Mentor

The 75 Ω doesn't pertain to the DC power load. It refers to the video connection interface.

You want to use the typical supply voltage and supply current to find the supply power consumption of the circuit. So

P = (90 mA)(5 V). So about 0.45 W. Probably closer to 0.5 W if the supply voltage degrades from 5 V, but let's ignore that for now.

Do you know an expression that relates power to voltage and resistance? Use it to find the load resistance when the voltage is your measured value.

By the way, was your load voltage measured with all four circuits connected, or just one?

8. Apr 26, 2017

### Tyler R

Yes! P= I^2*R = v^2 / R.

So p=(90 mA)(3.73 V) = .3337 W = 3.732 / R

.3337 W = 13.9129 / R

R = 41.7 Ohms

The 3.73V was measured while the entire load was being applied. When just one circuit was connected I was getting ~3.9V

9. Apr 26, 2017

### Staff: Mentor

Ah. Not quite. Use the spec sheet supply current and voltage values to calculate the required power. Assume that that power must be drawn for the circuit to do its job.

Then using that power value and your measured voltage, find the resistance. You'll want to use the 3.9 V value for one load if you use the power for one load in your calculation.

10. Apr 26, 2017

### Tyler R

Alright, so taking your original P = (90 mA)(5 V). ~0.45 W I would

.45W=I2 / R
.45=3.932 / R
.45=15.4449 / R
.45*R=15.449
R=34.322 Ohms

Then, I can use KVL on the second loop. 3.93V-34.322*I2-3.7=0 to find I2 which is distributed to all four circuits?

11. Apr 26, 2017

### Tyler R

I see what you're saying. I could also calculate the total resistance of all four circuits.

.45*4=3.732 / R
1.8=13.9129 / R
R*1.8 = 13.9129
R=7.73 Ohms because they are parallel?

3.7-7.33*i2=0 i2 is then i_2=0.479mA.

Last edited: Apr 26, 2017
12. Apr 26, 2017

### Staff: Mentor

Your 3.7 V value for the cell is probably not its actual value. It's going to be higher than that, and "pulled down" by the load current across its internal resistance. Note that the cell is effectively connected directly across your loads, so whatever voltage you measure across the load will also appear across the cell.

The EMF of the cell is probably closer to your 3.93 V value. Have you measured the unloaded cell voltage?

In parallel. So a total load of about 8 Ohms. That's a bit less than you'd predict from the single load result: 34/4 = 8.5 Ohms for four of them in parallel, which goes to show that the circuits are drawing more power to compensate for the drop in supply voltage.

Your circuit will look something like this:

With the load voltage and the estimated total load resistance of about 8 Ω you should be able to find the total load current easily enough with just Ohm's law.

13. Apr 26, 2017

### Tyler R

Wow!!! You are awesome. Using Ohms law I got the same value as I did using KVL

14. Apr 26, 2017

### Tyler R

I never measured it before I connected the circuits. Unfortunately, everything is soldered in. This was my first time creating any sort of circuit and I had a lot of fun. You have been immensely helpful. I will upload a picture of the device. It is a wi-fi jammer and each circuit runs on a specific channel.

15. Apr 26, 2017

### Staff: Mentor

I have to say that I'm not exactly thrilled to hear that that is its intended use . One hopes it won't be used to illegally interfere with commercial or private networks.

16. Apr 26, 2017

### Tyler R

Not at all. It is strictly educational. I made sure it was legal in my state before working on the project and my Professor said it was ok to create.

17. Apr 26, 2017

### Tyler R

The only person I've annoyed is my dad who I live with Lol, but he understands. I only had it on when I wanted to test it.

18. Apr 26, 2017

### Tyler R

I truly appreciate all your help. You helped me the first time I had a question on the Physics Forum. I hope you enjoy your evening :)

19. Apr 26, 2017

### Staff: Mentor

You're welcome, and thanks.