# What's wrong with this solution to freshman textbook example for Ohm's law and Kirchhoff's current law?

I'm a freshman taking an introductory course on circuit theory and I've run into a bit of trouble working through a textbook example practice problem. Unfortunately, it only lists the answer and not the steps involved. In the above problem, it is given that $i_x=3\,A$ and the leftmost current is $8\,A$.

From what I understand, the three parallel branches (all but that with the $v_x$ source) should have identical voltage drops; this suggests $(6\,\Omega)i_x=18\,V$ which is consistent with the given $i_x=3\,A$ -- so far, so good.

I run into trouble when trying to apply Kirchhoff's current law to the top node connecting these three branches, however. Since the voltage drop across each branch is $18\,V$ it follows the current through the $R_a$ branch ought to be $(18\,V)/R_a$. As the other two branches are specified to be $8\,A$ and $3\,A$ it follows we have:$$13\, A=3\, A+\frac{18\,V}{R_a}+8\,A$$... from which I conclude $R_a=9\,\Omega$, rather than $R_a=1\,\Omega$.

Try as I might I've been unable to figure out exactly where I've gone wrong. Which steps were incorrect?

Any and all help is greatly appreciated. Thank you!

• Think carefully about the 8A current delivered by the voltage source. Does it enter or leave the top node? – Alfred Centauri Sep 5 '14 at 2:18
• wow... I can't believe it. Thanks @AlfredCentauri, I understand now! Would you post an answer for me to accept? – oldrinb Sep 5 '14 at 2:20
• I modified my comment. – oldrinb Sep 5 '14 at 2:23

Would you post an answer for me to accept?

It is stipulated that the 18V voltage delivers 8A which I interpret to mean that the 8A leaves the positive terminal of the source to enter the top node thus, the currents entering the top node sum to

$$8A + 13A$$

The currents leaving the top node sum to

$$\frac{18V}{R_A} + 3A$$

Setting both sums equal yields

$$8A + 13A = \frac{18V}{R_A} + 3A$$