# Find the following information from the circuit provided

#### leejohnson222

Joined Jan 11, 2023
57
Determine the following parameters using ac theory

See the following circuit

Vs = 10 V @ 1.5 kHz, R = 15 Ω, L = 2.7 mH and C = 1.8 μF

Find
1.Inductive reactance, XL
2.Capacitive reactance, XC .
3.Circuit impedance, Z
4.Current, IS
5.Resistor current, IR
6. Inductor current, IL
7.Capacitor current, IC.

#### leejohnson222

Joined Jan 11, 2023
57
i assume i just use the formulas for each one and slowly get through the maths, XL = 2pifl
XC = 1/pifc

#### Alec_t

Joined Sep 17, 2013
14,327

#### leejohnson222

Joined Jan 11, 2023
57
Xc = 1/ 2pifc my error then i would use

Xl = 2pifl
Xc = 1/2pifc
Z = treat impedance like Resistance formula, series or parallel so here 1/R + 1/Xl + 1/Xl = Z
Is = IR + IL + IC
Ir = Vo / R
IL = ??
Ic = Vo / Sq XL
Not sure on Inductor Current IL

Last edited:

#### leejohnson222

Joined Jan 11, 2023
57
Just unsure of the way to work out inductor current here??

#### Alec_t

Joined Sep 17, 2013
14,327
Are you being asked to determine RMS current or instantaneous current at time t or peak current?

#### leejohnson222

Joined Jan 11, 2023
57
Are you being asked to determine RMS current or instantaneous current at time t or peak current?
good question it doesnt specify, it just says find IL

#### leejohnson222

Joined Jan 11, 2023
57
so far i have
Xl = 25.43ohms
Xc = 58.94 ohms
Z = 71.5 ohms
Current = V/Z = 0.280A
Resistor voltage = I x R = 2.8V
Would i multiply current by In XL for inductor voltage = 13.2A
and the same for Capacitor voltage ?

#### WBahn

Joined Mar 31, 2012
30,071
Xc = 1/ 2pifc my error then i would use

Xl = 2pifl
Xc = 1/2pifc
Z = treat impedance like Resistance formula, series or parallel so here 1/R + 1/Xl + 1/Xl = Z
Is = IR + IL + IC
Ir = Vo / R
IL = ??
Ic = Vo / Sq XL
Not sure on Inductor Current IL
1/2pifc is (1/2)*(pi)*(fc)

Remember order of operations. You want

1/(2pifc) = 1/(2*pi*fc)

This sloppiness will bite you. You will enter an equation into a program or a spreadsheet and you will do it as 1/2*pi()*freq because you are used to interpreting the '/' as a horizontal line separating numerator from denominator. But computers are very literal minded, so you need to get in the habit of writing equations with an appreciation for proper order of operations.

#### WBahn

Joined Mar 31, 2012
30,071
good question it doesnt specify, it just says find IL
The problem just says that the voltage source is a 10 V sinusoidal source. That could be the RMS voltage, or it could be the amplitude. It could, conceivably, be the peak to peak voltage, but that's a stretch.

The best you can do is answer in the same terms. Use the 10 V and if your current come out to be, say 2 A, then if the 10 V is amplitude, the 2 A is amplitude. If it is RMS, then the answer is RMS.

Actually, that's second best. The best you can do is state your assumption about what it is, with a note of how to correct the results if the assumption is wrong. That not only shows that you understand these concepts, but that you are paying enough attention to take note of the ambiguity in the problem statement, but even further, that you are able to proceed to come up with a solution that deals with it.

#### WBahn

Joined Mar 31, 2012
30,071
Z = treat impedance like Resistance formula, series or parallel so here 1/R + 1/Xl + 1/Xl = Z
You can't add resistances and reactances this easily, because they have different phase relationships.

Even if you could, that equation would be wrong because you have reciprocals of impedance on the left and an impedance on the right.

#### WBahn

Joined Mar 31, 2012
30,071
so far i have
Xl = 25.43ohms
Xc = 58.94 ohms
Z = 71.5 ohms
Current = V/Z = 0.280A
Resistor voltage = I x R = 2.8V
Would i multiply current by In XL for inductor voltage = 13.2A
and the same for Capacitor voltage ?
You aren't showing enough work for me to figure out the mistakes you are making.

Your inductive reactance value is correct.

Your capacitive reactance value is probably correct. Capacitive reactance is properly a negative value, but I'm guessing that this is incorporated into the equations you have been given. If you see things like (Xl - Xc) in equations, that subtraction is there because the people that developed much of the military technical school training materials nearly a century ago decided that students couldn't handle the concept of a capacitor having a negative reactance and so instead jury rigged a bunch of equations that imply that we have to treat it as a distinct and separate thing from inductive reactance. The result is that you have an explosion of equations to handle a relatively small number of simple situations and they can't easily be generalized to arbitrary circuits.

What a shame. But, it's probably the situation you have to live with.

Your impedance value is clearly wrong. What do you know about the total resistance of a 15 Ω resistor placed in parallel with anything else?

Also, particularly since you are treating both inductive and capacitive reactance as positive values, you need to indicate what the 'flavor' of your impedance is.

Since your have both resistance and reactance, your total impedance is going to be a combination of the two. How have you been taught to represent that? Commonly you would indicate it with an angle. Does that sound familiar? Let us know how you have been taught, and we will do our best to conform to that.

How are you coming up with V/Z being 0.280 A? What are you using for V? What are you using for Z?

What are you using for I and R to get I x R = 2.8 V?

It seems like you are just grabbing some I and some R and throwing them into Ohm's Law because they happen to be handy.

Ohm's Law uses the voltage across THAT resistor and the current through THAT resistor, along with the resistance of THAT resistor.

Look at your original schematic. Without doing any calculations at all, you should be able to tell me immediately what the voltage is across each and every one of those components.

In your next line, it impossible to tell exactly what you mean, because you talk about multiplying a current by a reactance to get a voltage but then indicate that the result is, out of nowhere, 13.2 A. Where is that 13.2 A coming from?

You REALLY need to show your work, not just provide the answers. That forces us to have to become mind readers to divine what we think you might have done to get the answers you provided. Instead, show the actual equations that you set up and solved to get those answers. Then we can better see what you actually did and go from there.

#### leejohnson222

Joined Jan 11, 2023
57
ok here are what my workings out are, i can see as the circuit is parallel all voltages will be the same for each branch

XL = 2*Pi*f*L = 2 * 3.14 * 1.5khz * 2.7mH. = 25.43ohms
XC = 1/ 2*Pi*f*C. = 58.94ohms

to find total impedance i need to find total current
Resistor current 10/15 = 0.66A
Inductor current V/XL = 10 / 25.43. = 0.39A
Capacitor Current = V/ Xc = 10 / 58.976 = 0.169

Total current IS = IR + IL + IC =
0.666 + 0.39 angle -90 + 0.169 angle 90
IS = 0.701 angle - 18.357A

Z = V/IS = 10 / 0.701 Angle -18.357 = 14.14 Angle 18.357 ohms

i think i had mixed up 2 questions and mixed values by mistake from my rough notes and workings out

#### WBahn

Joined Mar 31, 2012
30,071
ok here are what my workings out are, i can see as the circuit is parallel all voltages will be the same for each branch

XL = 2*Pi*f*L = 2 * 3.14 * 1.5khz * 2.7mH. = 25.43ohms
XC = 1/ 2*Pi*f*C. = 58.94ohms
You are still being sloppy with order of operations. What you wrote is neither what you meant nor what you did.

What you wrote:

XC = 1/ 2*Pi*f*C

This is

XC = (((1/ 2)*Pi)*f)*C

What you meant and did:

XC = 1/ (2*Pi*f*C)

Trust me, this sloppiness is going to get you in trouble. You need to deal with it now when that trouble is nothing more than some lost points on an assignment, and not down the road with the trouble is damaged/destroyed equipment and/or injured/dead people.

[QUOTE[
Resistor current 10/15 = 0.66A
Inductor current V/XL = 10 / 25.43. = 0.39A
Capacitor Current = V/ Xc = 10 / 58.976 = 0.169
[/QUOTE]

You also need to start properly tracking your units instead of just tacking on some units to the answer that you think you would like the answer to have.

10 V / 15 Ω = 0.667 A

The A on the answer isn't there because I want the answer to be in amps, it is there because 1 V = 1 A/Ω and 1 A/Ω multiplied by 1 Ω is 1 A. I put the units there that the math dictates, and THEN consider whether the units agree with what I expected them to be. We all make silly math errors. Most (not all) such mistakes will mess up the units, allowing us to detect that a mistake has been made and to then track it down and correct it before wasting a bunch of time and effort pursuing an answer that is now guaranteed to be wrong.

Also, note that these currents are not complete because only the magnitude is indicated. The phase angle is a critical part of the description of the current. The results should be

Resistor current: IR = V/R = 10 V / 15 Ω = 0.6667A
Inductor current: IL = V/XL @ 90° = 10 V / 25.43 Ω = 0.3932 A @ 90°
Capacitor Current: = IC = V/Xc @ -90° = 10 V / 58.976 = 0.16956 @ -90°

Unless there is a problem-specific reason to do otherwise, the usual convention in engineering is to report answers to three significant figures. Intermediate results that are used in subsequent calculations should be reported with one or two additional sig figs so that the resulting round-off errors have little to no effect on the final result. It is also common to not count a leading 1 when counting sig figs.

Total current IS = IR + IL + IC =
0.666 + 0.39 angle -90 + 0.169 angle 90
IS = 0.701 angle - 18.357A
Hopefully you can see why it is important/useful to include the angle with your currents. You needed them in this calculation, but your intermediate results didn't include them. So now, out of nowhere, you have a bunch of angles pop up in your work as if by magic. Having to always keep track of when and where to throw in what angles is a recipe for disaster, so keep it to an absolute minimum.

Z = V/IS = 10 / 0.701 Angle -18.357 = 14.14 Angle 18.357 ohms

i think i had mixed up 2 questions and mixed values by mistake from my rough notes and workings out
I think you have some of that roundoff creeping in. I get that the total current is

I_S = 703 mA @ -18.52°
Z = 14.22 Ω @ 18.52°

You need to indicate that your angles are in degrees, if that's what they are, because without units, the angle is in radians.

#### leejohnson222

Joined Jan 11, 2023
57
right thats fair enough to pick me up on the lazy units and angles, i agree that it would be extremely important in a real world scenario, so it is better to clean it up and stop doing it now, while this is just theory etc.
thank you for taking to time to look at this and correct my mistakes

#### leejohnson222

Joined Jan 11, 2023
57
i need to look at creating the same circuit in a sim and create a resonance curve, i am not sure how to do this
i can create the circuit but clearly it needs to be changed slightly to create the curve

#### WBahn

Joined Mar 31, 2012
30,071
I assume that a "resonance curve" is simply referring to "something" as a function of frequency. The AC Sweep is generally the tool that you want for that, though I'm not familiar with your simulator (can't even tell which one it is).

#### MrAl

Joined Jun 17, 2014
11,489
Determine the following parameters using ac theory

See the following circuit

View attachment 306059

Vs = 10 V @ 1.5 kHz, R = 15 Ω, L = 2.7 mH and C = 1.8 μF

Find
1.Inductive reactance, XL
2.Capacitive reactance, XC .
3.Circuit impedance, Z
4.Current, IS
5.Resistor current, IR
6. Inductor current, IL
7.Capacitor current, IC.
Hi,

Your approach to this will depend on what you have learned in the past. From the other posts I read it seems that you never did an AC circuit like this before where you had reactive components like L and C as well as R. You would need to learn that first.

The most general method is to use the concept of imaginary numbers. If you can learn that you can solve any of these problems.
Even better, work in the Laplace domain, although you would have to learn how to deal with imaginary numbers also.
Next is the phasor concept.
Last, and definitely least, is using learned formulas, but I do not recommend that method because as soon as you get handed a different network you will not know how to solve it.

The method I have to recommend is to use the concept of imaginary numbers. It's not that much harder than regular math.