0
$\begingroup$

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.

textbook example problem

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!

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

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.