Kirchoff's Laws, nodes and current

[Alt+F4]

http://img157.imageshack.us/img157/2147/showmegf4.gif








a) What is the current I1 which flows through R1?

I1 =

(b) What is the current I3 that flows through R3?









3 Equations

I1-I3 = 0

I1R14 + I1R2 - E1 =0

I3R2 + I3R3 +E2 = 0



is that even right?

 

[FrogPad]

http://img157.imageshack.us/img157/2147/showmegf4.gif








a) What is the current I1 which flows through R1?

I1 =

(b) What is the current I3 that flows through R3?









3 Equations

I1-I3 = 0

I1R14 + I1R2 - E1 =0

I3R2 + I3R3 +E2 = 0



is that even right?

Not quite.
Label the node in between R1, R3, and R2. What can you say about this node in terms of current?

From kirchhoffs law you know that
$$\sum i_{in} = \sum i_{out}$$

The trouble you are having is that there is another current. Can you see why?

Looking at that node we talked about.

$$I_1 = I_2 + I_3$$
where $I_2$ is defined as leaving the node.

 

[Alt+F4]

Not quite.
Label the node in between R1, R3, and R2. What can you say about this node in terms of current?

From kirchhoffs law you know that
$$\sum i_{in} = \sum i_{out}$$

The trouble you are having is that there is another current. Can you see why?

Looking at that node we talked about.

$$I_1 = I_2 + I_3$$
where $I_2$ is defined as leaving the node.

so is this how?

I1 = I2 + I3
I1R14 + I2R2 - E1 =0
I2R2 + I3R3 +E2 = 0

 

[silver-rose]

For the last equation, you want to make sure you keep the signs right. You can indicate a drop of voltage as a result of resistors as NEGATIVE, while an increase in PD as a result of a cell as POSITIVE or vice versa, just keep it CONSISTENT

 

[Alt+F4]

For the last equation, you want to make sure you keep the signs right. You can indicate a drop of voltage as a result of resistors as NEGATIVE, while an increase in PD as a result of a cell as POSITIVE or vice versa, just keep it CONSISTENT

So

I1 = I2 + I3
I1R14 + I2R2 - E1 =0
-I2R2 - I3R3 +E2 = 0


i tried that and it is not the right answer

 

[Alt+F4]

Ok so I am guessin it is something like that
I1 = I2 + I3
-E1 + I1R1 + I2R2 + E2 + I1R4 = 0

wat bout the 3rd equation? Thanks

 

[Valhalla]

wat bout the 3rd equation? Thanks

try writing KVL for the other loop

 

[Alt+F4]

Actually
I1 = I2 + I3
-E1 + I1R1 + I2R2 + E2 + I1R4 = 0
E2+R2I2+R3I3

 

[Alt+F4]

Actually
I1 = I2 + I3
-E1 + I1R1 + I2R2 + E2 + I1R4 = 0
E2+R2I2+R3I3

HW is due in 2 hrs and i tried everything any help is appreciated

 

[FrogPad]

HW is due in 2 hrs and i tried everything any help is appreciated

The two loop equations are:

$$-\xi_1 +I_1(R_1) +I_3(R_2) +\xi_2 +I_1(R_1) = 0$$
$$-\xi_2 -I_3(R_2) +I_2(R_3) = 0$$

Note: That is two equations with THREE unknowns. So you need to come up with another equation to solve.

 

[Alt+F4]

Thanks Alot

 

[FrogPad]

Try to do this when you are coming up with the equations.

1) First draw the currents (you can choose whatever direction you want).
- Remember that current does not change when passing elements in a series. - It only changes when it reaches elements in parallel.

2) Traverse a branch (ie make a loop).
- When you get to an element (such as a resistor) if the current direction is pointing into the element then write that voltage term (ie V=IR, where I = current going into the element, and R = the resistance of the element) as POSITIVE.
- If you are going around the loop and you reach an element but the current is pointing in the opposite direction of the element, then write that term as NEGATIVE.
- Set all of these terms equal to 0. Take note that the voltage supplied (such as from a voltage source) is equal to the voltage drops (such as a resistor takes away a certain amount of voltage).

You really need to do a few examples for all of this to sink in.

 

[Alt+F4]

The two loop equations are:

$$-\xi_1 +I_1(R_1) +I_3(R_2) +\xi_2 +I_1(R_1) = 0$$
$$-\xi_2 -I_3(R_2) +I_2(R_3) = 0$$

Note: That is two equations with THREE unknowns. So you need to come up with another equation to solve.

ok sorry but this equation for some reason did not work, i know one equation should be


I1 = I2 + I3
-E1 + I1R14 + I2R2 + E2 = 0



I just still can't get the third equation for the left loop

i was thinking it was

-E2 + R2I2 + R2I3 + R3I3

any help, thanks

 

[FrogPad]

ok sorry but this equation for some reason did not work, i know one equation should be


I1 = I2 + I3
-E1 + I1R14 + I2R2 + E2 = 0



I just still can't get the third equation for the left loop

i was thinking it was

-E2 + R2I2 + R2I3 + R3I3

any help, thanks

Woops sorry buddy. I interchanged I2 and I3.

The correct equations for this network are:

-E1 +I1R1 +I2R2 +E2 +I1R4 = 0
-E2 -I2R2 +I3R3 = 0
I1 = I2 +I3

 

[Alt+F4]

Woops sorry buddy. I interchanged I2 and I3.

The correct equations for this network are:

-E1 +I1R1 +I2R2 +E2 +I1R4 = 0
-E2 -I2R2 +I3R3 = 0
I1 = I2 +I3

lol :( nope not it i have the answer, it is the 2nd equation that is giving me trouble to get. i need to understand this for coming exam. I asked my TA and he said it is not a straight forward thing

 

[FrogPad]

lol :( nope not it i have the answer, it is the 2nd equation that is giving me trouble to get. i need to understand this for coming exam. I asked my TA and he said it is not a straight forward thing

Are you saying that what I posed (the last one) is not the right answer?

If that's the case, let me know what the right one is.


It is VERY straightforward... I have no idea why your TA is saying that. You basically have to remember two things.

1) the voltage around a loop is equal to zero.
2) current in equals current out.

 

[Alt+F4]

lol :( nope not it i have the answer, it is the 2nd equation that is giving me trouble to get. i need to understand this for coming exam. I asked my TA and he said it is not a straight forward thing

Never mind, for some reason it wouldn't accept my answer but i kept clickin ok and all of a sudden it took it. I had the equation from the start way before i did this thread. It wouldn't just accept my answer. Damn computers. Thanks for ur help

 

[FrogPad]

I'm almost positive the last answer I put down is correct... if I'm wrong, I'm either having a huge space out (I did already with the switching of I2 and I3) or I have no idea what I'm doing... (which I hope is not the case ;)

oh...
so what is the answer?

 

[Alt+F4]

I'm almost positive the last answer I put down is correct... if I'm wrong, I'm either having a huge space out (I did already with the switching of I2 and I3) or I have no idea what I'm doing... (which I hope is not the case ;)

oh...
so what is the answer?

Answer was .031 but i had .03064435 even though it says on top to keep all sig figs and to aviod rounding

 

[Alt+F4]

ur equations were right

 

[FrogPad]

ur equations were right

Good. Otherwise I'd be freaking out.


Let me know if you still need help with those circuits. Trust me. It's straightforward, I don't know why the TA is trying to make it sound like it is so difficult to you.

It's not easy until you've done hundreds of them, but even then if you are not careful you can make careless mistakes. The hard part is just about watching your signs.