Find the resistance through a resistor

[gneill]

that's where I'm starting to get confused. so if my current through the top half from a to B is 2.089(yes 8V/R is how I arrived at that.) but that's the current through 2 resistors, so I can't use that as my I value?

2.087A is the total current flowing from A to B. After passing through the 1 Ohm resistor it splits into two paths, an upper path that has two resistors in series (2 and 3 ohms), and the bottom path containing three resistors in a slightly more complicated configuration. You want to find out how much of that original 2.087A is flowing via this bottom path.

You have calculated the total resistances of the two individual paths. You have the total current. You have the current divider rule. You want the current in the lower path. I don't know how I can be more explicit.

 

[Resmo112]

ok so I'm looking for my I2 if 2.087A is my I value then I need to put the resistor values into the R2/R2+R1 part of the equation.

I'm sorry dude I'm really trying I promise I am I really appreciate the patience, there's no reason this should be taking you this long.

is it just I2=2.087*(6.54/5+6.54)=1.18? the problem I have there is I don't understand why I'm using the top half of the circuit.

 

[gneill]

ok so I'm looking for my I2 if 2.087A is my I value then I need to put the resistor values into the R2/R2+R1 part of the equation.

I'm sorry dude I'm really trying I promise I am I really appreciate the patience, there's no reason this should be taking you this long.

is it just I2=2.087*(6.54/5+6.54)=1.18? the problem I have there is I don't understand why I'm using the top half of the circuit.

The current divider rule determines the fraction of the total current, I, that splits off and flows through a given path. If you look closely at the rule, it states that the fraction of the current that flows through a given path is equal to the resistance of the *other* path divided by the sum of the resistances of both paths.

By placing the resistance of the lower path in the numerator of the ratio, you've actually calculated the current flowing through the upper path rather than current in the lower path. So replace the 6.54 quantity in the numerator with 5, the resistance of the upper path. this will give you the current in the lower path. What value do you calculate?

 

[Resmo112]

ok so I got .904.
I did 2.087 * 5/6.54+5 and that = .904

 

[Resmo112]

but that's not the right answer either.

 

[gneill]

ok so I got .904.
I did 2.087 * 5/6.54+5 and that = .904

Good. We're not quite done, but almost.

Now take a close look at the lower path. You can see that the current in the lower path also splits within that path -- some going through the 4.00 Ohm resistor and the rest through the 17 Ohm resistor. You're interested in the current in the 17 Ohm resistor. So apply the divider rule again, this time using these two resistances and the current value that you have for the lower path.

 

[Resmo112]

ok SO THEN I take .904 * (4/17+4) and get .164

 

[gneill]

ok SO THEN I take .904 * (4/17+4) and get .164

Your method and numbers are correct, but you seem to have slipped on the calculator keypad! Try running the calculation again.

P.S. you might want to make use of parentheses to make it clear that the "17 + 4" is all denominator.

 

[Resmo112]

ahhh .172, yeah that had to have just been a finger slip probably hit like .804 rather than .904. if this is right and works I will be naming my first child Gneili

 

[Resmo112]

just making sure before I input this, there are NO other steps I'm missing? and is there any way this could be a negative number?

 

[gneill]

ahhh .172, yeah that had to have just been a finger slip probably hit like .804 rather than .904. if this is right and works I will be naming my first child Gneili

Heh. I won't hold you to that!

The value looks alright to me.

 

[Resmo112]

THANK YOU SO SO MUCH! it was right! I see where the other guy was trying to guide me too, which makes a lot more sense NOW! Sorry this had to have been really frustrating for you. I'm a bit slow at times (especially when it comes to physics)

 

[gneill]

just making sure before I input this, there are NO other steps I'm missing? and is there any way this could be a negative number?

There are no internal source in the network that could make current flow "against the grain". So everything should be flowing from left to right (assuming the current enters at A and proceeds to B). So no, no negative numbers. The calculation is complete. the current through R2 is 0.172 A flowing from left to right.

 

[gneill]

THANK YOU SO SO MUCH! it was right! I see where the other guy was trying to guide me too, which makes a lot more sense NOW! Sorry this had to have been really frustrating for you. I'm a bit slow at times (especially when it comes to physics)

No worries. Glad things got sorted in the end.