How Does Current Divide in a Series Circuit with Different Resistors?

[fish]

what is the current through the 2 ohm resistor?
what is the current through the 8 ohm resistor?


8 V
...! |...
. empty space .
. empty space . I
. empty space .
...2 ohm Res...
. empty space .
. empty space 8 ohm Res
. empty space . I_2
. empty space .
...! |...
16 V

! = (-) side of battery
| = (+) side of battery

I_1 is on the 2 ohm branch (current going left to right)
I current going down (clockwise)
I_2 current going down (clockwise)

Is this setup right with correct sign conventions?

I_2=I+I_1

2 ohm res:
+8V-2I_1=0
8=2I_1
I_1=4 A

8 ohm res:
-8V-8I_2+16V=0
8V=8I_2
I_2=1 A

 

[redrogue]

Originally posted by fish
what is the current through the 2 ohm resistor?
what is the current through the 8 ohm resistor?


8 V
...! |...
. empty space .
. empty space . I
. empty space .
...2 ohm Res...
. empty space .
. empty space 8 ohm Res
. empty space . I_2
. empty space .
...! |...
16 V

! = (-) side of battery
| = (+) side of battery

I_1 is on the 2 ohm branch (current going left to right)
I current going down (clockwise)
I_2 current going down (clockwise)

Is this setup right with correct sign conventions?

I_2=I+I_1

This is correct by Kirchoff's junction rule.

2 ohm res:
+8V-2I_1=0
8=2I_1
I_1=4 A

8 ohm res:
-8V-8I_2+16V=0
8V=8I_2
I_2=1 A

Yes, this is correct. The potential difference in a circuit loop is equal to zero.
However a more correct way to present your answer would be:

8V-I_1(2\Omega)=0
8V=I_1(2\Omega)
I_1=4A


-8V-I_2(8\Omega)+16V=0
8V=I_2(8\Omega)
I_2=1A

It's good practice to keep units in your work.

 

[fish]

thanks redrogue