Code Ques.
- Thread starter Flight987
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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
kwired
Electron manager
- Location
- NE Nebraska
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.
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.
- Location
- Illinois
- Occupation
- retired electrician
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.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
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.
kwired
Electron manager
- Location
- NE Nebraska
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.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
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.
kwired
Electron manager
- Location
- NE Nebraska
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.
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.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
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.
kwired
Electron manager
- Location
- NE Nebraska
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.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Actually I was wrong.. Misread numbers. Old eyes.
I saw 140V twice, not the 100V on top.
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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.
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.
kwired
Electron manager
- Location
- NE Nebraska
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:
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jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
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.
97catintenn
Senior Member
- Location
- Columbia, TN
Sure, just that sum means addition
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