Maple TA Quiz 203
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Question 1
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.
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 Guest
Question 2
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.
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 Guest
Question 3
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.
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 Guest
Re: Maple TA Quiz 203
#1 above isn't working out for me. can u tell me wuts wrong with it plz?
phystude Guest
Re: Maple TA Quiz 203
forget it i figured it out. i was converting the resistance wrong.
phystude Guest
Question 1
hey..i really dont understand what you did in question one...and wat about th frequency????
frusty s Guest
RE: Question 1
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.
There must be another method which utilizes the frequency value.
DJ Guest
RE: Question 4
I'm still unable to calculate the correct answer. Do we just ignore the given values of R and V_rms?
DJ Guest
question 4
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?
sjames Guest
question #1
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.
guest111 Guest
question 4
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)*(122.89E6)*(51.59E6))
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!
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)*(122.89E6)*(51.59E6))
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!
student0 Guest
Question 1
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
For that matter can someone please elaborate more on what has already been posted on doing question 1? Please use numbers... Thanks
need hel Guest
probs 1 2 3
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...
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...
user x Guest
Re: Maple TA Quiz 203
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.
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.
Justin Guest
Questions 2 and 3
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 Posts : 4
Join date : 20080922
Question 2 and 3
2) The AC voltage source in an RLC circuit delivers 100 V rms voltage at a frequency of 250 s1. 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 + (136.141.17)^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 + (10.4165)^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)
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 + (136.141.17)^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 + (10.4165)^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)
Guest786 Guest
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