Find the resistance through a resistor

[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.