17_3 maple quiz
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17_3 maple quiz
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.
blahhh Guest
Question 1,2,&4
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...
guest001 Guest
Question 2
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:
V11/2XI_1rI_2V_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
V11/2XI_1rI_2V2 = 0
19.25.5(I_1)rI_212.53 = 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
Basically, you start off with these equations:
I_1 = I_2+1/2
And then following what they did to find the left loop:
V11/2XI_1rI_2V_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
V11/2XI_1rI_2V2 = 0
19.25.5(I_1)rI_212.53 = 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 Posts : 42
Join date : 20081218
Q 3 and 4
I've been trying to figure this one out using the CD, but I'm stuck .. PLEASE HELP!! Thanks!
Anon Guest
Question 3:
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.
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.
Kathleen Posts : 42
Join date : 20081218
QUEST 1
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
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
SMARTALL Guest
Question 4
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!
Anon Guest
Q4
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
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
almost Guest
Re: 17_3 maple quiz
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
To find R I used a different equation
R=(R_{1}*R_{2})/(R_{1}+R_{2})
Guest01 Posts : 133
Join date : 20080919
#4
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
guest101 Guest
Re: 17_3 maple quiz
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!
(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!
help me Guest
Re: 17_3 maple quiz
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
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
Betta Guest
Question 4
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.
Qs4 Guest
Re: 17_3 maple quiz
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.
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.
Betta2 Guest
Q#4
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!
Hope it helps!
phy122 Guest
#4
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,
jsyz Posts : 7
Join date : 20081124
Re: 17_3 maple quiz
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.
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.
Kathleen Posts : 42
Join date : 20081218
Question 4
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
Friend Guest
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