Simple Circuit Question (KVL/KCL)

1. Oct 16, 2009

Novark

[Solved]

1. The problem statement, all variables and given/known data

2. Relevant equations

KCL, KVL, Ohms Law

3. The attempt at a solution

6 = i + (i/4) + (v/8)

I just can't get another equation that relates v & i.

Last edited: Oct 17, 2009
2. Oct 16, 2009

cepheid

Staff Emeritus
The branch with the 2 ohm resistor tells you that the voltage drop across it is given by i*(2 ohms), which, from the way the circuit is laid out, must also be equal to v.

3. Oct 16, 2009

arithmetix

we know that the equivalent resistance of the io/4 branch is 4 times that in the 2 ohm resistor, because 1/4 the current flows. So it's easy to redraw the circuit using an 8 ohm resistor in place of that branch, and that makes it all nice and easy.

4. Oct 16, 2009

Novark

If that's the case, then won't I get:

6 = i + (i/4) + (2i/8) = (5i/2)
Which would give: i = 12/5.

Am I missing something?

5. Oct 16, 2009

cepheid

Staff Emeritus

i = 4i/4

i/4 = i/4

2i/8 = i/4

so the sum is:

4i/4 + i/4 + i/4 = 6i/4 = 3i/2

if 3i/2 = 6, then 3i = 12, and i = 4.

6. Oct 16, 2009

Novark

Ah, that's right. Not sure how I made such a silly mistake.
Thank you.