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

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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\$ Jun 10 '14 at 3:22
  • \$\begingroup\$ the resistance and voltage is correct. \$\endgroup\$ Jun 10 '14 at 3:28

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
    Jun 10 '14 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
    Jun 10 '14 at 3:40
  • \$\begingroup\$ sorry thats wrong. your current source is 6A not 4A. \$\endgroup\$
    – nidhin
    Jun 10 '14 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
    Jun 10 '14 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
    Jun 10 '14 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\$ Nov 5 '15 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
    Nov 6 '15 at 9:43

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