Complex circuits help!

[GoldenBear]

Click here.

On my physics final all the currents went DOWN and I was like WHAT?? This and the other type of circuits were about the only thing I learned this semester, and I was bound to get those two right..but is it possible to figure it out when they're all going down?

 

[JohnnyKnoxville]

You got 2 batteries in series,1 parallel and 2 resistors in series and you call this a complex circuit?

 

[GoldenBear]

Okay help me with my simple circuit Mr. Einstein Hawking all rolled into one..

(at least you didn't type in bold)

 

[iamfried]

No, I don't think so. Depending on how you want to look at current going, you either have up in the middle and down at the sides or, up on the sides and down in the middle. Either way, the sides are going to be in the same direction and the middle is going to be opposite.

 

[RossGr]

I see three loops that need to be considered, all loops would have the current going "down" in the center but up on the sides. Actually how you draw the current to solve the problem doesn't matter as long as you are careful with polarity of the various voltage drops.

 

[JohnnyKnoxville]

OK since you shown the proper respect I will try to help.Please bear with since I learned this crap a very long time ago.Sooo first I have to understand what you are having difficulty with?
The currents going down?Electrons flow from negative to positive .Does that help?

 

[GoldenBear]

Okay, so I forgot exactly how I did it, but I was hoping for two positive answers and one negative, which is the typical resultant for the stuff we do in class..so basically I just did a bunch of loops and put them into my TI-89 solving feature hoping for solutions like that, and the closest I got was:

I1 = 1.43A
I2 = 1.7A
I3 = -.31A

It'd be quite the miracle if I got that..

 

[iamfried]

I was refering to whether he was taught hole flow or electron flow, hence the different points of view.

 

[Passions]

The flow of current doesnt matter. You can reverse the polarity and switch signs. The outer edges should all flow in one direction with the middle going reverse.

 

[Whitecloak]

GoldenBear
its been a long time since i have used my electrical fundaes(6years in fact) but i will do my best to help you.
what is your exact prob?i used Kirchoff's law on the circuits and it seems to be fine. i mean the currents are not cancelling out etc. could be that you have made a mistake in putting the things together or maybe your ammeter is faulty. i dont wanna rub it in but like JohnnyKnoxville said it is a pretty simple circuit.

 

[JohnnyKnoxville]

I believe you are correct sir,i tried the math (with pen&paper) and got the same figure for |1 as you did.20 volts divided by 14 ohms = 1.43Amps

 

[OS]

<< You got 2 batteries in series,1 parallel and 2 resistors in series and you call this a complex circuit? >>



hahahaha, no kidding, I was expecting like a cascaded transistor amplifier or something.

Just write loop equations for the two loops and solve with simultaneous equations.

 

[Passions]

how are u getting 1.7A for i2? I am getting 1.5A

 

[RossGr]

Most of you seem to be neglecting the outside loop. It is driven by a net 4V through 30Ohms for a current of .133A . This current flows ccw arround the outside of the circiut. It OPPOSSES the current in the left side box ( 20V though 14Ohms for 1.43A ) and reinforces the right side loop (24V through 16Ohms for 1.5A) so from this I get

I1 = 1.3A
I2 = 1.63A
I3 = 2.93A

This is consistent, if you look at the either of the junctions you can see that I1+I2=I3.

 

[Ameesh]

this is why i changed from computer engineering to computer science, i hate anolog circuits.