# Is this a parallel circuit? How can I tell whether it's parallel or not?

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

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.)

• 1.3 plus 2.6 is? Jun 10 '14 at 3:22
• the resistance and voltage is correct. 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.

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,

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

• I did try current dividing. Io = [R_Ix / (R_Ix + R_Io)] * (4V) --> Io = (1/3)*4 = 4/3 = 1.3333... Jun 10 '14 at 3:36
• 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? Jun 10 '14 at 3:40
• sorry thats wrong. your current source is 6A not 4A. Jun 10 '14 at 3:41
• @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 $I_o = V_o/2\Omega$ 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.

• @Johnathan That equation you just used is wrong. You will never get current by multiplying a voltage with a resistance ratio. Jun 10 '14 at 3:55
• @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 $I_x = \frac{V_o}{1\;[\Omega]}.$ Nov 5 '15 at 18:51
• @Pål-KristianEngstad I said $i\ne v \times \frac{R_1}{R_2}$. I was replying to jonathan's comment which he deleted. Nov 6 '15 at 9:43