Maple TA Quiz 20-3

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Maple TA Quiz 20-3

Post  DJ on Sat Mar 07, 2009 4:46 pm

20-3 here.

DJ
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Question 1

Post  DJ on Sat Mar 07, 2009 5:27 pm

So, we are given several values: V_m, f and R. We want to eventually find I_rms.

First, we need to find v_rms:

V_rms = V_m / (sqrt(2))

Then, let's find the average power by using this equation and the value of V_rms we just calculated:

P = (V_rms/R) * V_rms

Now, we can use this equation to find I_rms, which is our answer:

P = I_rms * V_rms

I_rms is the answer, and it is negative.

DJ
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Question 2

Post  DJ on Sat Mar 07, 2009 5:32 pm

There's a bunch of steps to this question as well.

So, we are given several values: V_rms, f, R, X_C (capacitor reactance), and X_L (inductive reactance). We are trying to find I_m.

First, let's find the value of Z by using this equation:

Z = sqrt(R^2 + (X_L - X_C)^2)

Then, let's find the value of V_m, which we will need in the end:

V_rms = V_m / (sqrt(2))

Finally, use this equation to find I_m, which is our answer:

V_m = I_m * Z

I_m is the answer, and it is positive.

DJ
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Question 3

Post  DJ on Sat Mar 07, 2009 5:36 pm

We are given the following values: V_rms, f, R, X_C, and X_L. We are looking for P average.

First, use this equation like above to find Z:

Z = sqrt(R^2 + (X_L - X_C)^2)

Then, find I_rms using this equation:

I_rms = V_rms / Z

Finally, use this equation to find Z, which is the answer:

P = I_rms^2 * R

P is the answer (average P) and it is negative.

DJ
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Re: Maple TA Quiz 20-3

Post  phystude on Sat Mar 07, 2009 6:43 pm

#1 above isn't working out for me. can u tell me wuts wrong with it plz?

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Re: Maple TA Quiz 20-3

Post  phystude on Sat Mar 07, 2009 6:46 pm

forget it i figured it out. i was converting the resistance wrong.

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Question 1

Post  frusty s on Sun Mar 08, 2009 12:50 am

hey..i really dont understand what you did in question one...and wat about th frequency????

frusty s
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RE: Question 1

Post  DJ on Sun Mar 08, 2009 2:47 am

I intuitively solved the question in that manner. MapleTA says it's the correct answer.

There must be another method which utilizes the frequency value.

DJ
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q4

Post  sjames on Sun Mar 08, 2009 8:56 am

Any ideas on #4?

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#4

Post  Guestttt on Sun Mar 08, 2009 4:39 pm

1= 4*pi^2*f^2*C*L

Just solve for f.
Answer is negative.

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RE: Question 4

Post  DJ on Sun Mar 08, 2009 5:05 pm

I'm still unable to calculate the correct answer. Do we just ignore the given values of R and V_rms?

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q4

Post  sck105 on Sun Mar 08, 2009 7:10 pm

I cant figure out number 4 plz help?

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question 4

Post  sjames on Sun Mar 08, 2009 8:23 pm

I am still having trouble with number 4 I cant figure it out can someone please post a more detailed description of how to do the problem? please?

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number 4

Post  cted on Sun Mar 08, 2009 9:23 pm

use f=1/(2pi*sqrt(LC))

cted
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question #1

Post  guest111 on Sun Mar 08, 2009 9:24 pm

i have trouble doing #1, do we have to convert anything in Question #1, and if we do what do we have to convert. i tried converting kilo ohms to ohms by multiplying by 0.001 but my answer is incorrect...pls help.

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Question #1

Post  guest111 on Sun Mar 08, 2009 9:32 pm

forget it ...i got it..

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question 4

Post  student0 on Sun Mar 08, 2009 9:49 pm

Question 4
First we need to rearrange the formula previously given to solve for f:
1= 4*(pi^2)*(f^2)*C*L -->
f= sqrt(1/(4*(pi^2)*C*L))

My values were:
C= 122.89 microF (convert to F by multiplying times 10^-6)
L= 51.59 microH (convert to H by multiplying times 10^-6)

Solution:
f=sqrt(1/(4*(pi^2)*(122.89E-6)*(51.59E-6))
f= 1998.84
Answer is negative.

The other values are given so we can answer the positive/negative question, just ignore them.
Hope this helps!

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Question 1

Post  need hel on Mon Mar 09, 2009 12:26 am

Guest 111 can you please explain how you got question 1?
For that matter can someone please elaborate more on what has already been posted on doing question 1? Please use numbers... Thanks

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probs 1 2 3

Post  user x on Mon Mar 09, 2009 2:04 am

i'm still having a problem with questions 1, 2, and 3
why are we not using the frequencies in any of the three questions?
also, can someone put up actual numbers cause idk what i'm doing wrong but i triple checked my calculations and i'm getting the same incorrect numbers...

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Re: Maple TA Quiz 20-3

Post  Justin on Mon Mar 09, 2009 1:00 pm

For #1 I just found the Peak current and divided by sqrt(2)

The power supply in an AC circuit with a 9.464 resistor delivers a peak voltage of 280 V at a frequency of 35.43 Hz. What is the rms current (see sheet 16)? Indicate with a positive (negative) sign whether the resistance depends (does not depend) on the frequency.

280 / 9464 = .029586
.029586 * (1/sqrt(2)) = .0292
Answer is negative.

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Questions 2 and 3

Post  green on Mon Mar 09, 2009 11:16 pm

Can someone please post the question and the work for questions 2 and 3. I tried DJ's solutions but they didn't work. Thanks!

green

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Conversions

Post  DJ on Tue Mar 10, 2009 12:30 pm

Don't forget to convert certain given values.

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Question 2 and 3

Post  Guest786 on Tue Mar 10, 2009 6:15 pm

2) The AC voltage source in an RLC circuit delivers 100 V rms voltage at a frequency of 250 s-1. The resistance of the resistor is 53.42 , the capacitive reactance is 41.17 and the inductive reactance is 136.1. What is the peak current (see sheet 27,31)? Indicate with a positive (negative) sign whether the minimum impedance at any frequency is equal to (larger than) the resistance of the resistor.

V_rms= 100
Frequency= 150 s^-1
Resistance of resistor (R)= 53.42 ohms
Capacitive reactance (X_C)= 41.17 ohms
Inductive reactance (X_L)= 136.1 ohms

First use this formula: to find Z:
Z = sqrt(R^2 + (X_L - X_C)^2)

Z= sqrt(53.42^2 + (136.1-41.17)^2)= 108.9284228

Then use this formula V_m= V_rms*sqrt(2)
V_m= 100*sqrt(2)= 141.4213562

Then finally use this formula: to find I_m:
I_m= V_m/Z

so then:
141.4213562/108.9284228= YOUR ANSWER ( In Positive)


3) The AC voltage source in a series RLC circuit delivers 148 V rms voltage at a frequency of 60 Hz. The resistance in the circuit is 86.23 , the capacitive reactance is 65 and the inductive reactance is 10.41 . What is the average power dissipated in the circuit (see sheet 36) ? Indicate with a positive (negative) sign whether the capacitor and the inductance (do not) dissipate power.

V_rms= 148
frequency= 60Hz
Resistance (R)= 86.23 ohms
Capacitive Reactance (X_c)= 65 ohms
Inductive Reactance (X_L)= 10.41 ohms

First use this formula: to find Z

Z = sqrt(R^2 + (X_L - X_C)^2)

Now plug the numbers in:
Z= sqrt(86.23^2 + (10.41-65)^2)= 102.0572437

Then use this formula: to find I_rms:
I_rms = V_rms / Z

Now plug in numbers:
I_rms= 148/102.0572437= 1.45016654

Finally use this formula: to find P:
P= I_rms^2*R

Plug in numbers:
P= (1.45016654)^2*86.23= 181.3402235 (MAKE SURE ANSWER IS IN NEGATIVE)

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THANK YOU

Post  thankies on Tue Mar 10, 2009 7:46 pm

ty so much!!!

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Re: Maple TA Quiz 20-3

Post  Guest111 on Wed Mar 11, 2009 12:11 am

Thank you DJ and everyone else for writing out the equations Very Happy!!!

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Re: Maple TA Quiz 20-3

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