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DC circuit problems

by Dx
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Dx
#1
Jul8-03, 11:55 AM
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a) I must determine the current in 8 [ome] resistor? Current in the 4[ome] ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.

What have I done wrong? how do attach my diagram in here?
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HallsofIvy
#2
Jul8-03, 12:01 PM
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It's hard to tell what you did wrong when we don't know what you did. How about at least telling us exactly what the problem was? For one thing where did you get the "12 volts" and "9 volts" that suddenly appeared in your answer?
Dx
#3
Jul8-03, 12:43 PM
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OK! I think I got it posted remember that the batteries are from neg to pos so according to kirchoffs law I believe we add them together. I just don't know how to solve for these.


a) I must determine the current in 8 resistor? Current in the 4 ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.


Sorry took so long.
Dx
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File Type: doc circuit.doc (24.0 KB, 10 views)

Dave
#4
Jul8-03, 05:08 PM
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DC circuit problems

Please post the question up and I can try it.
At the moment the problem doesn't have enough information.
Tom Mattson
#5
Jul8-03, 05:34 PM
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Originally posted by Dx
how do attach my diagram in here?
When you make a post, look below the "Your Reply" and "Options" sections, and you'll see the "Attach File" section. You can either do it that way, or upload it to your personal webspace if you have one.
Dx
#6
Jul8-03, 06:15 PM
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Anyways picuture this:

a 12V battery connected in series loop with 7 and 4 ohms, ok. Also a 9V battery is in this loop connected in series with a 8 and the same 4 ohm resistor. Now the battery connects both points in the same direction like so. neg to pos wire conected to battery neg to pos, follow me.

a) I must determine the current in 8 resistor? Current in the 4 ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.

Dx
Attached Files
File Type: doc circuit.doc (24.0 KB, 10 views)
Tom Mattson
#7
Jul8-03, 06:18 PM
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Originally posted by Dx
OK! I think I got it posted
No, I still don't see it.
Dx
#8
Jul8-03, 06:22 PM
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Originally posted by Dx
Anyways picuture this:

a 12V battery connected in series loop with 7 and 4 ohms, ok. Also a 9V battery is in this loop connected in series with a 8 and the same 4 ohm resistor. Now the battery connects both points in the same direction like so. neg to pos wire conected to battery neg to pos, follow me.

a) I must determine the current in 8 resistor? Current in the 4 ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.

Dx
Can you picture this?
Tom Mattson
#9
Jul8-03, 06:25 PM
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Originally posted by Dx
Can you picture this?
It sounds like everything is connected in series. Is that it?
Dx
#10
Jul8-03, 06:36 PM
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Originally posted by Tom
It sounds like everything is connected in series. Is that it?
There's more than one way to skin a cat, Tom! Its like this!

   |--- 7 [ome]-------- 8 [ome]-------------| 
   |               |                 |
   |               |                 |
   |               4[ome]                |
   |               |                 |
   |       |       |          |      | 
   ------| |----------------| |------|
           |                  |

         12 Volts            9 Volts
And I must find:

a) I must determine the current in 8 resistor? Current in the 4 ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.

Hope this is better.
Dx
Tom Mattson
#11
Jul8-03, 06:56 PM
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Originally posted by Dx
                   A
   |--- 7 [ome]--------*---8 [ome]-----------| 
   |   ----->      |  ----->         |
   |     I1      | |      I3         |
   |          I2 | 4[ome]                |
   |             | |                 |
   |       |     V |          |      | 
   ------| |----------------| |------|
           |                  |

         12 Volts            9 Volts
OK, that's great. I've modified your drawing a little bit so that it includes current arrows and so that a node "A" is labeled.

a) I must determine the current in 8 resistor? Current in the 4 ?

a) I started out using ohms law. E=IR, ok. I took the difference between the Voltages eg.12-9 = 3V then multiply by 11 ohms total resistance.

b) Did pretty much the same way except that 4 ohms is where most current is drawn.
I don't know why you did that, but you have to use Kirchhoff's laws here. You have 3 unknown currents (I1,I2, and I3), and you have to solve for them simultaneously.

Try this:

1. Write down Kirchhoff's current law for Node A.
2. Write down Kirchoff's voltage law for the left loop going clockwise.
3. Write down Kirchoff's voltage law for the right loop going clockwise.

Do that, and we'll go from there.

edit: fixed diagram
Dx
#12
Jul8-03, 07:19 PM
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1) I_3 = I_1 + I_2
2) I_2 = I_3 - I_1
3) I_1 + I_3 - I_2

But I am having problems determining should I seperate the loops to find the total resistance or what? I donno what to do really, ive read that section on this and the example makes not a whole lotta sense to me, plz help me?
Dx
Tom Mattson
#13
Jul8-03, 07:33 PM
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Originally posted by Dx
1) I_3 = I_1 + I_2
2) I_2 = I_3 - I_1
3) I_1 + I_3 - I_2
No, you got them all wrong.

Kirchoff's current law (KCL) states that the sum of currents at a node equals zero. Take currents entering the node to be positive and currents leaving the node to be negative.

Kirchoff's voltage law (KVL) states that the sum of voltage rises/drops around a closed loop is zero. For instance, going through a 12 V battery from the + to - terminal is a voltage drop of 12 V. Going through a 7Ω resistor against the direction of current I1 is a voltage rise of 7I1 V.

All this should have been explained to you in class and in the book. Try writing down the equations again, using those rules.

But I am having problems determining should I seperate the loops to find the total resistance or what?
You do not need the total resistance at all. You need to learn how to use KVL.
Doc
#14
Jul15-03, 12:45 PM
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Basically the laws state that the sum of the currents in all three resistors = 0. You know the value of the resistors and are given some voltages. Simple algebra should do it for you. Set an equation up involving everything you are given. One side of the equation will be 0. You will also need ohms law figured into your equation. I am not sure you have a handle on ohms law though DX, because in your first post you are talking about multiplying a resistance value by the voltage to get current. Do you see something wrong there? OHMS multiplied by AMPS gets you VOLTS.

NOT: OHMS multiplied by VOLTS gets you AMPS.


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