Is this a parallel circuit? Can I collapse it and make one equivalent resistor of \$2/3\Omega\$?

enter image description here

Also, will someone confirm whether or not I got the right answers for the unknown voltages and currents? \$I_o = 1.333\dots\text{A}\$, \$I_x=2.6666\dots\text{A}\$, \$V_o = 4\text{V}\$.

I found \$I_o\$ with current division: \$(1/3)\times(4) = 4/3 = 1.333\text{A}\$ and \$I_x = (2/3)\times 4 = 8/3 = 2.6666\text{A}\$

(Update: I see my mistake. I was multiplying by 4V instead of 6A when current dividing.)

  • 2
    \$\begingroup\$ 1.3 plus 2.6 is? \$\endgroup\$ Commented Jun 10, 2014 at 3:22
  • \$\begingroup\$ the resistance and voltage is correct. \$\endgroup\$ Commented Jun 10, 2014 at 3:28

2 Answers 2


Starting from one terminal of source, if current has more than one path to reach the other terminal, then those two paths are parallel. I see two such paths in your circuit.

You didn't say how you calculated \$I_x\$ and \$I_o\$. The answers you got are wrong. Try Current dividision.

EDIT: You used 4A instead of 6A in your calculations.

If you want to find the current by dividing voltage across resistance by resistance value, you have to find the voltage \$V_o\$ first. $$V_o = 6A\times (1\Omega || 2\Omega) = 4V$$ now,

$$I_x = V_o/1\Omega = 4A$$ $$I_o = V_o/2\Omega = 2A$$

  • \$\begingroup\$ I did try current dividing. Io = [R_Ix / (R_Ix + R_Io)] * (4V) --> Io = (1/3)*4 = 4/3 = 1.3333... \$\endgroup\$
    – Johnathan
    Commented Jun 10, 2014 at 3:36
  • \$\begingroup\$ But above ^^^ someone pointed out that 1.3333.... + 2.666.... = 4A, but the current source has a total of 6 amps, and KCL doesn't work at that node (V_o) --- 6A in, 4A out.... What did I do wrong? \$\endgroup\$
    – Johnathan
    Commented Jun 10, 2014 at 3:40
  • \$\begingroup\$ sorry thats wrong. your current source is 6A not 4A. \$\endgroup\$
    – nidhin
    Commented Jun 10, 2014 at 3:41
  • \$\begingroup\$ @Johnathan You can do. find equivalent resistance (\$1\Omega || 2\Omega\$) and multiply with current (6A) to get \$V_o\$. Then you can calculate \$I_x = V_o/1\Omega\$ and \$I_o = V_o/2\Omega\$ \$\endgroup\$
    – nidhin
    Commented Jun 10, 2014 at 3:50

Look at the current going into node 1 (labeled \$V_{o}\$). Then remember KCL: everything going into the node must come out.

node 1: \$6 - I_{o} - I_{x} = 0 \$

and from inspection: \$I_{o} = \frac{V_{o}}{2}, I_{x} = \frac{V_{o}}{1}\$

That is everything you need.

  • \$\begingroup\$ @Johnathan That equation you just used is wrong. You will never get current by multiplying a voltage with a resistance ratio. \$\endgroup\$
    – nidhin
    Commented Jun 10, 2014 at 3:55
  • \$\begingroup\$ @nidhin I'm not sure what you mean, did Ohm's Law suddenly become debatable? Across a resistor, if \$v\$ is the voltage across it, \$i\$ is the current, and \$R\$ is the resistance, then surely Ohm's Law holds: \$v = i\cdot R.\$ Applied to this example, \$I_0 = \frac{V_0}{2\;[\Omega]},\$ and \$I_x = \frac{V_o}{1\;[\Omega]}.\$ \$\endgroup\$ Commented Nov 5, 2015 at 18:51
  • \$\begingroup\$ @Pål-KristianEngstad I said \$i\ne v \times \frac{R_1}{R_2}\$. I was replying to jonathan's comment which he deleted. \$\endgroup\$
    – nidhin
    Commented Nov 6, 2015 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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