17_3 maple quiz

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17_3 maple quiz

Post  blahhh on Sat Feb 14, 2009 12:15 pm

does anyone know how to do #4? i'm so stuck on it. i tried everything but its not working out for me. Please help.

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Question 1,2,&4

Post  guest001 on Sat Feb 14, 2009 1:49 pm

does anyone know how to do #1,2,and 4 because i looked at the lecture notes but i can't seem to get it right...

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Question 2

Post  Kathleen on Sun Feb 15, 2009 1:21 am

This is basically just following the example given in sheets 28&29 in the notes.

Basically, you start off with these equations:
I_1 = I_2+1/2
And then following what they did to find the left loop:
V1-1/2XI_1-rI_2-V_2 = 0
(I don't know if this is an actual equation or where it came from but I'm just substituting the letters for our problem in with the numbers they used)
For right loop:
V_2+rI_2-.5xR_3 = 0

In this question, the givens:
V1 = 19.25 V
V2 = 12.53 V
R3 = 28.14

Left loop
V1-1/2XI_1-rI_2-V2 = 0
19.25-.5(I_1)-rI_2-12.53 = 0 (combine the 19.25 and 12.53)
6.72 - .5(I_1)-rI_2 = 0

Right loop
V2 + rI2 - .5R_3 = 0
12.53 + rI_2 - .5(28.14) = 0
12.53+rI_2 = 14.07
rI_2 = 1.54

Go back to the equation we left off with from left loop
6.72-.5(I_1)-rI_2 = 0
plug in what we just found
6.72-.5(I_1) - 1.54 = 0
5.18 = .5(I_1)
10.36 = I_1

Back to original equation now
I_1 = I_2 + 1/2
10.36 = I_2 + .5
I_2 = 9.86

Now to solve for r:
rI_2 = 1.54
r(9.86) = 1.54
r = .1562

So, this is how you insert your answers:
I_1, I_2, r .... or 10.36, 9.86, .1562

Kathleen

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Q 3 and 4

Post  Anon on Sun Feb 15, 2009 1:23 am

I've been trying to figure this one out using the CD, but I'm stuck .. PLEASE HELP!! Thanks!

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Question 3:

Post  Kathleen on Sun Feb 15, 2009 1:38 am

This one is pretty easy,
We are asked to look at the left resistor and are given R and R1. For the sake of this problem though, and seeing as to how it relates to the equation, indicate the first value as R1 though and the second as R2. Seeing as these are parallel, we can use the equation:

(1/R) = (1/R1) + (1/R2)
(1/R) = (1/first given value) + (1/second given value)
Solve for R, and that is your answer. Refer to sheet 31 for help.

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Question 1 and 4

Post  dan on Sun Feb 15, 2009 1:57 am

does anyone know how to find question 1 and 4...pls help

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QUEST 1

Post  SMARTALL on Sun Feb 15, 2009 5:46 am

Question 1: (1 point)



Use the figure attached with R = 12.36 , V1 = 5.815 V, and V2 = 10.631 V, and calculate the current (see sheet 24,25,29 bottom) ? Indicate with a positive (negative) sign whether the current direction shown is (not) consistent with your solution.
HERE IS HOW I DID QUESTION 1...

YOU NEED TO PICK A STARTING POINT. I PICKED THE TOP LEFT CORNER. THE FIRST THING YOU COME ACROSS IS V1... THIS WILL BE NEGATIVE B/C IT IS A POTENTIAL DROP...NEXT YOU HIT THE GIVEN 20 WHICH IS THE R. THIS ALSO WILL BE NEGATIVE AND MUST BE MUTIPLIED BY THE CURRENT I. NEXT YOU HIT V2 WHICH THIS TIME IS POSITIVE...AND THEN FINALLY YOUR R WHICH AGAIN IS NEGATIVE...SO THE FINAL EQUATION FOR THIS IS....

-V1 -20I + V2 - RI=0

YOU SET EVERYTHING EQUAL TO ZERO BECAUSE OF THE LOOP RULE.

IN MY CASE: -5.815 -20I + 10.631 -12.36I=0

REARRANGE ALGERBAICALLY AND YOU GET -32.36I=-4.816
I=.1488
THE ANSWER IS NEGATIVE BECAUSE WHEN YOU GET NEGATIVE CURRENTS (-20I AND -12.36I) YOU HAVE SET THE CURRENT FLOW IN THE WRONG DIRECTION.... HOPE THIS HELPS

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Question 4

Post  Anon on Sun Feb 15, 2009 12:33 pm

For this question I've been using the 3 Rs given, plus the one in the diagram and using the equation: 1/R = (1/R1)+(1/R2)+(1/R3).... and the voltage given in the diagram (6V). After solving for R and using the equation V=RI to solve for I, I still can't get the right answer. Can someone please help? Thanks!

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Q4

Post  almost on Sun Feb 15, 2009 1:35 pm

ok guys here's what ive done so far and have gotten nowhere so maybe someone can work off of this becuz im about to throw my computer out the window soon

i used the 1/R=1/R1+1/R2+1/R3 then find R like in Q3 then plug into V=IR and that got me nowhere.
Next i tried I=V/R1+V/R2+V/R3 and got the same results so if anyone has any ideas that would be super

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Re: 17_3 maple quiz

Post  Guest01 on Sun Feb 15, 2009 2:16 pm

For Question 4 you want to first find the total resistance in the circuit, i.e. find R. Then calculate the current using Ohm's Law.

To find R I used a different equation
R=(R1*R2)/(R1+R2)

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Re: 17_3 maple quiz

Post  bigjkjhn on Sun Feb 15, 2009 3:47 pm

lik\\nnjhn hbnbuixvb h436g vbctgcb folllow it

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#4

Post  guest101 on Sun Feb 15, 2009 4:15 pm

treat the left loop and right loop as in series and then just add them up because the two loops are in parallel to each other and then use V=IR to find the I

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Re: 17_3 maple quiz

Post  help me on Sun Feb 15, 2009 5:32 pm

I keep having problems with question 4. here is what i'm doing:

(1/R)= (1/R1) + (1/R2) + (1/R3)
Then solving for R

Once I do that I plug it into V= RI

I am using 6V for V...and I keep solving it and getting it wrong...does anyone know what i'm doing wrong? if so please help! thanks!

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Re: 17_3 maple quiz

Post  Betta on Sun Feb 15, 2009 6:15 pm

For 4, you have to do
1/R= (1/R1) + (1/there is an R on the diagram, mine was 10)
and then
1/R= (1/R2)+(1/R3)
Add up the Rs.
Then use that for V=IR

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4

Post  STUDENT on Sun Feb 15, 2009 6:18 pm

OMG I CAN'T FIGURE #4 OUT, I WORKED MORE THAN 2 HOURS ON IT AND CAN'T FIGURE IT OUT.

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4

Post  STUDENT on Sun Feb 15, 2009 6:21 pm

BETTA

WHAT IS THE VALUE OF I

STUDENT
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Question 4

Post  Qs4 on Sun Feb 15, 2009 6:33 pm

I think that mapleTA has messed up the answer for that one. I put in the number I got for R and it accepted it.

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Re: 17_3 maple quiz

Post  Betta2 on Sun Feb 15, 2009 7:36 pm

R1 = 14.03 , R2 = 7.312 , and R3 = 23.52
were my numbers.
Then on the diagram, directly below your R1 there should be another value, __omega
Mine was 10.
So, first, you must do
(1/14.03)+(1/10)=1/R
R=5.8385

Then you must do it again for R2 and R3, so
(1/7.312)+(1/23.52)=1/R
R=5.5779

Now add the two Rs together:
So,
5.8385+5.5779=11.4164

Now for V=IR
The V is also on the diagram, mine was 6V.
So, 6=I(11.4164)
I= .52555

And it's negative. I cut the digits down but it should work. Very Happy

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Q#4

Post  phy122 on Sun Feb 15, 2009 8:22 pm

Although question #4 is asking for the Current (I) there seems to be some sort of problem with either the way the question is asked and the answer it expects you to give. Just follow the above steps WITHOUT using the V=RI equation, the answer is just the Rs ADDED UP and it is NEGATIVE.

Hope it helps!

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#4

Post  jsyz on Mon Feb 16, 2009 7:17 am

phy122 wrote:Although question #4 is asking for the Current (I) there seems to be some sort of problem with either the way the question is asked and the answer it expects you to give. Just follow the above steps WITHOUT using the V=RI equation, the answer is just the Rs ADDED UP and it is NEGATIVE.

Hope it helps!

the r is not the 2 rs add up, when you do 1/r1+1/r2 in your calculation you get a answer of .xxxx and then you have to do 1/.xxxx in order to get the r because 1/r1+1/r2 =1/r, then just use the formula V=IR
PS: always be careful with the calculation,

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Re: 17_3 maple quiz

Post  Kathleen on Mon Feb 16, 2009 9:53 am

What Phy122 said was right, it is just the R's added up, as in (1/R) = (1/R1)+(1/R2)
The professor posted a blog about this, saying that there was an error in question 4 where he forgot to do it so that the current was calculated, so if you started the quiz before a certain time then thats why you only have to add the R's and not use Ohm's law. If, however, you take the quiz a second time or started later in the weekend, you will have to do Ohm's law after solving for R.

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Question 4

Post  Friend on Mon Feb 16, 2009 8:46 pm

You have to use the Given resistance in the diagram too (the one under R1) calculate the parallel R using 1/R= 1/R1+1/R in the diagram solve for R by taking 1/(1/R) then solve 1/R=1/R2+ 1/R3 and take the 1/(1/R) and then add the two R's together to solve for the total R. then use I=V/R

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Re: 17_3 maple quiz

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