Code Ques.

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curtnobel said:
First I guessed and
took 120/240 Ex., which I knew was 208. Did my math 208/240=.86666. Or 87 percent.
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Isn't it amazing that is also the cosine of 30? ? Or even better, it is 1/2 the square root of three...Although I don't think 120/240 is 208, more like .500
 
Res. Ques.

Res. Ques.

I did draw it out. Placed thing's where I felt they should be. If I had a amount of Res. that is on
each Res., that would make easer. I don't why I don't get.
 
curtnobel said:
I did draw it out. Placed thing's where I felt they should be. If I had a amount of Res. that is on
each Res., that would make easer. I don't why I don't get.
Click to expand...

You don't need to know resistance to answer the question. This one is simpler than you think.
 
The key is knowing that in series connected resistors, the current through each is the same and the overall resistance is the sum of the resistors. The voltage across a given resistor will be E = IR. If you know the voltage, then the resistance R = E/I. Just rearrange ohms law to get what you need.

They give you the voltage across R3 as 40V. So the resistance of R3 = 40/I.
The current in each resistor I = 180/(R1+R2+R3).
You also know that R2+R3 = 100/I
and R1+R2 = 140/I

Now it should just be an algebra problem. You have 4 equations and 4 unknowns, so it should be solvable.
 
You don't even have to find the current to solve this. Just some simple algbra and the information that has been given. It would involve 3 equations, but you really only need to use two of them to get the answer.
 
I did not

I did not

I did not have algbra in high school, the way you do it is different to me. But I will try to figure it
out.
 
jumper said:
I would highly suggest trying a course if there is a local community college. I took night classes for 6 years, it was a PITA, but worth it.
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You don't even need any "advanced" algebra. This is very basic math even though it is algebra. Look at drawing and tell us the voltage across R1 and voltage across R2. You probably don't even need a calculator for this one. All information that was given in the question is entered in the drawing.

rescalc.jpg
 
jumper said:
It was only a polite suggestion, I found night courses very valuable.

As far as your diagram, I may look at it later, but I already passed my exams.:)
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I wouldn't expect it to take you more than a minute or so to find the solution to the problem, that is if you are kind of slow with mathematics. If you are below a math wizard but yet still pretty fair - you can solve this problem in less than 15-20 seconds after looking at that diagram I would think.

I could re draw the same thing but instead of using volts change the unit of measurement to inches and then make the resistors to proper scale according to length instead of voltage dropped - I don't know if I could make the solution much more obvious than drawing it that way.
 
kwired said:
I wouldn't expect it to take you more than a minute or so to find the solution to the problem, that is if you are kind of slow with mathematics. If you are below a math wizard but yet still pretty fair - you can solve this problem in less than 15-20 seconds after looking at that diagram I would think.

I could re draw the same thing but instead of using volts change the unit of measurement to inches and then make the resistors to proper scale according to length instead of voltage dropped - I don't know if I could make the solution much more obvious than drawing it that way.
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It was 11:00PM, I was tired. But as you wish:


VR1=40V

VR2= 100V

VR1=VR3

Series circuit

R2 is 2.5 times the resistance of R1 or R2.

Ps. I never suggested an "advanced" algebra course.
 
jumper said:
It was 11:00PM, I was tired. But as you wish:


VR1=40V

VR2= 100V

VR1=VR3

Series circuit

R2 is 2.5 times the resistance of R1 or R2.

Ps. I never suggested an "advanced" algebra course.
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Those are not my answers - at least one of has to be wrong - I suppose it could be both of us. Was kind of hoping OP would give it one more try before anyone answers.
 
Nice Job

Nice Job

Nice job on the Dia.. You know how to work the PC. alot better than I do. That is somewhat how they had it on Asw.key also. Alot of people say that it is easy. It is a good thing that the Pastor sermon this
morning, was how somtimes we feel alone. Well I feel that way on this one. I did much better on
220.55. Sorry I don't see it.
 
curtnobel said:
Nice job on the Dia.. You know how to work the PC. alot better than I do. That is somewhat how they had it on Asw.key also. Alot of people say that it is easy. It is a good thing that the Pastor sermon this
morning, was how somtimes we feel alone. Well I feel that way on this one. I did much better on
220.55. Sorry I don't see it.
Click to expand...

I quickly drew something with pencil and had figured the problem out then decided to draw it on the computer and post it. I still think you possibly had a hard time figuring out just what to draw to help solve the problem. BTW what was your answer?

here is mine:
rescalc2.jpg
 
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I didn't

I didn't

I didn't have one. I have Doc. Appt., jury duty, and all these thing's hapening at once. Mom's
set for knee replace menet on Tues.. I have to much going on now. Still don't get it but, I have to deal with Mom.
Thank's
 
curtnobel said:
I didn't have one. I have Doc. Appt., jury duty, and all these thing's hapening at once. Mom's
set for knee replace menet on Tues.. I have to much going on now. Still don't get it but, I have to deal with Mom.
Thank's
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Curt, the sum of the voltage drops across the loads in a series circuit must =0V. Does help or make sense?

180V-80V-60V-40V=0V.
 
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