Trick Problem

Discussion in 'Homework Help' started by shannonm1, Apr 13, 2008.

Apr 13, 2008
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Hello, I am taking an electronics course. My instructor gave us a trick problem to try and figure out. We are not being graded on this, but I am stumped. Would anyone be able to point me in the right direction to figure it out? The knowns are listed, need to fill in the unknowns for Amps, Volts, Resistance, Watts. Thanks for any help anyone can provide!
Shannon

2. wijendra14 New Member

Apr 13, 2008
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I didnt quite get your question. Could you please tell me exactly what we need to find out?

Apr 13, 2008
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I need to fill in Voltage, Amps, Resistance, and Watts for each resistor.

4. wijendra14 New Member

Apr 13, 2008
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Oh i see. Ok you know that there is 2V across 1 resistor. Therefore, there must be 8V across the other 2 resistors. The current in each resistor is same because it is a series circuit. You can find this current. Say the voltage across the 2ohm resistor is V1 and the voltage across the 2watt resistor is V2.
V1+V2=8
P=VI => 2= IV2
V1=2I
You can solve these equations!
Hope this helps!

Apr 13, 2008
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I believe I was able to figure out the problem.... mostly by luck. I'm not fully understanding the equations. Would someone be able to explain in more detail please? Thank you!!

6. hgmjr Retired Moderator

Jan 28, 2005
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So what value did you get for the current flowing in the loop of your "trick" problem?

hgmjr

Apr 13, 2008
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I came up with approx 267.5 mA

8. hgmjr Retired Moderator

Jan 28, 2005
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218
I'm afraid that answer is not correct.

If you can scan and post your worksheet so that we may see it, maybe we can assist you in locating where you went off the track.

hgmjr

9. SgtWookie Expert

Jul 17, 2007
22,201
1,807
The only resistance shown is more than twice as much as the other two in series.
The only power shown is by far the least power dissipated by the other resistors.
The only voltage drop shown is the 2nd largest voltage drop.

Apr 13, 2008
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Thanks for the help so far. The last answer I got was pretty much through trial and error. I really need to brush up on my algebra It's been 15 years since I've been in school and my math is a little fuzzy. I'll try working on this some more and I'll post what I come up with. If there any other hints to point me in the right direction, thank you. I'm not asking for any answers, just guidance. Thank you everyone.

11. MusicTech Active Member

Apr 4, 2008
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well if this helps (suppose you want to find x)
x+a=b
subtract a frome either side to get x=b-a

x-a=b

xa=b
divide by a x=b/a

x/a=b multiply by a x=ab

If you need anything more than that you are probably doing something wrong, hope it helps

by the way, you probably combined total circuit and individual resitor values, which led you to wrong values, I make that mistake sometimes too. be careful

12. MusicTech Active Member

Apr 4, 2008
144
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when somebody gets this, which one of you probably has, please tell me, I have been fiddling with systems for about 25 minutes now and got nothing

13. SgtWookie Expert

Jul 17, 2007
22,201
1,807
Algebra bites

But in electronics, it's reality. Until you get into time. Then triginometry comes into play. Until you get into high frequencies where you need other stuff.

It gets worse, lots worse.

I'm admittedly math-handicapped. Yet I love electronics and computers, which are all math-based. It's this wierd love-hate relationship.

I've achieved great things by having math/physics gurus sitting next to me.
They do the math. I write the programs. Magic happens.

14. hgmjr Retired Moderator

Jan 28, 2005
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218
Have you made any progress toward a solution?

hgmjr

15. Ron H AAC Fanatic!

Apr 14, 2005
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Has anyone else realized that this problem has two sets of solutions? The equation for current is quadratic, and both solutions are positive real numbers. The answer our OP came up with (267.5mA) is pretty close to one of the solutions, but he came up with it by trial and error.

16. JoeJester AAC Fanatic!

Apr 26, 2005
4,055
1,774
All my numbers came out as integers.

17. hgmjr Retired Moderator

Jan 28, 2005
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I recognized early on that the expression was second-order.

hgmjr

Apr 13, 2008
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I appreciate the help, but the math is a little beyond me. Like I said this was only a challange for us, so I'll wait and see Thursday evening how it is figured out. I'm not too bad with algebra. But like I said, this is a bit out of my grasp at the moment. Thank You!

19. Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
I've got a graphic of the method (with one key omission) and the solution for the current all made up. Do you want me to post it?