Solve for the Following

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Below is Diagram of a RCL Series Circuit along with the Formulas for filling in each of the circuits unknown values. What is the least number of formulas you can use to fill in all the unknown values?

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This is not a test nor any such thing, just wanted to show the instructors a lil something in case they have students who are not good at memorizing Formulas or finding these types of problems a problem.


Will post later on my solution to this and how many formulas I actually used to solve all missing variables.


~FyE~
 

crossman

Senior Member
Location
Southeast Texas
These are the formulas you need to know:

VA = I(t) x E(t)
xVARS = I(t) x E(x)
PF = cos theta
C =1/2PiFXc
L = Xl/2PiF
Ohm's Law

in conjunction with:

The impedance vectors for series LCR circuits
Pythagorean's Theorem
Basic Trigonometry

To me, the student is much better off knowing the theory behind what is happening in the circuit, and this comes from understanding the vectors.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
080728-201% EST

By inspection all the currents are the same and in phase with each other because this is a single closed loop.

The sum of the voltages around the loop are zero. Thus,
Vsource = R*I + jXl*I -jXc*I = ( R+j(Xl-Xc) )*I or
I = Vsource / ( R+j(Xl-Xc) )
Because Xl is greater than Xc the current will lag the input voltage by
Phi = arctan ( (Xl-Xc)/R )


What is the circuit called when Xl = Xc?

Moderators note: Edited at Gar's request

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nakulak

Senior Member
my recollection is that for a system of linear equations, the number of equations required to provide a solution will be equal to the number of unknowns. in this case unknowns = 4, so I believe you need 4 linear equations to solve the problem (assuming the system has a valid solution (ie is consistent))
 

cadpoint

Senior Member
Location
Durham, NC
nakulak said:
my recollection is that for a system of linear equations, the number of equations required to provide a solution will be equal to the number of unknowns. in this case unknowns = 4, so I believe you need 4 linear equations to solve the problem (assuming the system has a valid solution (ie is consistent))

Man, that sounds like Quatum Physics !

If you notice most equations come down to A = B / C, if one can remember what the variables are they can get to the correct equation.
 

coulter

Senior Member
1st -

I'm missing one piece. The figure has a "Z" notation next to the source, which would lead one to think the source has an impedance and we don't know what it is.

carl​
 
Yes Z is the total Impedance of this circuit ( you have resistance, Inductance and Capacitance) and can be computed using the proper formula from the tables given.

My whole concept here was to show any insructors an easy forumla to help out there students who may be at this part of an electrical course instead of trying to remember all the different formula. This is just one practice problem and every problem only gives you 3 or 4 knowns and you have figure out the rest if you can remember the forumla.

In our class most eveyone was having problems tring to remember the right forumla to use to solve for specific unknown as you can see there are quite a few forumla's to put under one belt.

Folks dont beat ur brains out on this, this is just for any instructors who may want a different approach to obtaining the unkowns with just combining 85% of all the formula into 1 formula. :wink:

After spending hrs and hrs practicing these types of problems I noticed something and made a lil chart that wipes of most of the forumla listed. After solving these problems alot quicker using the lil chart, I then showed my class buddy whom I tutoring at night. He then show another class mate and next thing I know the two instructors drew the circuit above with only 4 known values and told me to come up infront of the class and show them and the rest of my class mates what all the hype was about. Lets put it this way, on exam day everyone passed with flying colors :grin: :D

I was nervous that the teachers may consider it a cheat but it all there in the book just a lil hidden. But they was shocked on how easy all this can be made and was given "Nice job" by the instructors after it was explained.

Again this is just a lil help if a student is having difficulties, it is not replacing on how, why or what happen in these types of circuits just an easier method of remembering the right forumla. If you like to work it out, I can post the answer key to the unknowns.


~FyE~
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
080729-0628 EST

coulter:

This is a series resonant circuit and with Xc = Xl it is at the resonant frequency and the lowest impedance, looks like a resistance. At this point the voltage across the capacitor or inductor may be larger than the source voltage. Depends on the ratio of reactance to resistance.

If it were a parallel resonant circuit, R L and C in parallel, then when Xc = Xl it is at resonance and at its highest impedance, looks like a resistance, and would be driven from a current source instead of a voltage source to exhibit a maximum voltage at resonance.


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crossman

Senior Member
Location
Southeast Texas
1st, let's see what you got.

From my point of view, for the series circuit:

1. Ohm's law for each type of component must be memorized. (easy enough)

2. XL = 2PiFL and XC = 1/2PiFC must be memorized. (easy enough)

3. Basic power formulas must be memorized. VA = It x Et and the various permutations for watts and VARs.

4. Power Factor = Cosine angle must be memorized.

5. Impedance vectors and how to manipulate them must be memorized. (easy enough)

6. You need some basic trig.

Now, the cool thing is, this also solves parallel RLC circuits if you substitute the current vectors for the impedance vectors.
 
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