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Calculate the \$I_3\$ value,i know i can calculate the \$I_3\$ easily with transforming norton to thevenin circuit and KVL ,and i can know \$I_3=2A\$

schematic

simulate this circuit – Schematic created using CircuitLab

However,i ask myself "if i don't want to transform norton to thevenin circuit,can i still calculate the \$I_3\$?" so here is my thinking,however ,i found that i can't calculate the \$I_3=2A\$

schematic

simulate this circuit

\$6+2=I_2+I_3 =>8=I_2+I_3\$,so it seems like \$8A\$ current is divided across two parallel resistors,

so i think \$I_2=8\times \frac{4}{(3+2)+4}=3.55\$,and \$3.55=2+I_3\$,so \$I_3\$ should be \$1.55\$.where am i wrong in this method?

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    \$\begingroup\$ Now is the time to use superposition. \$\endgroup\$
    – Andy aka
    Apr 26, 2020 at 10:28

2 Answers 2

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You need nodal or mesh equations which is basically KCL or KVL applied a grand scale. Note the plural EQUATIONS, not singular. They are more systematic than whatever it is you are trying to do by applying KCL piecemeal and trying to get away with just one equation.

\$6+2=I_2+I_3 =>8=I_2+I_3\$

You are just going to ignore that I_1 exists at that 4-way node and not I_3? This is because you are trying to use just one equation and shortcut to I_3. I_3 splits up into I_1 and I1 so by using I_3 and I1 in the same node equation you are counting I_3 twice while also ignoring I_1. By the way, I1 and I_1 are terrible naming schemes.

Try an equation for every 4 way node and 3 way node. Three equations for three unknowns I_1, I_2, I_3. Then solve simultaneously. When everything affects everything else you cannot solve for a small part without solving for everything.

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  • \$\begingroup\$ so you mean??i can't get the point from your reply \$\endgroup\$
    – shineele
    Apr 26, 2020 at 4:13
  • \$\begingroup\$ Look up a nodal analysis example. You need to learn some basics. \$\endgroup\$
    – DKNguyen
    Apr 26, 2020 at 4:14
  • \$\begingroup\$ i know the nodal analysis ,indeed it can know the \$I_3=2\$ easily too,but you just provide another method called "nodal analysis".The question i ask is why is the method i use can't calculate the correct answer.Nodal analysis do present the current with resistance and voltage,if i can know the \$I_3\$ by calculating resistance and voltage,why can't i just know the \$I_3\$ by calculating current? \$\endgroup\$
    – shineele
    Apr 26, 2020 at 4:32
  • \$\begingroup\$ is my $I_2$ formula wrong?or? \$\endgroup\$
    – shineele
    Apr 26, 2020 at 4:33
  • \$\begingroup\$ Your equation is simply wrong. I already explained why it is wrong. Where in that circuit does 6A, 2A, I_2, and I_3 all meet up? Nowhere. Regardless of the reasoning you used to come up with your equation, that's what you equation ultimately says and it is clearly not the case. The current divider thing is also nonsense because there is no current divider from parallel resistors. R2 is not parallel with R1 + R3? Why? Because I1 is in parallel with R1. You can't just pretend I1 is not there. \$\endgroup\$
    – DKNguyen
    Apr 26, 2020 at 5:21
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You are right, in a way. There is \$8\:\text{A}\$ arriving into the shared node of \$I_1\$ and \$I_2\$. And if you imagine that the sum of the currents in \$R_1\$ and \$R_2\$ must also be \$8\:\text{A}\$ then you'd be right about that, too. However, you write, "...is divided across two parallel resistors." And this is, I think, at least one point where you make an error. \$R_1\$ and \$R_2\$ are not in parallel.

Before I continue, let's note first that you are always allowed to designate exactly one node as \$0\:\text{V}\$ (or "ground.") So I want to redraw the schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

That said, you can observe that the sum of the currents in \$R_1\$ and \$R_3\$ (you know the direction, I assume) must also be the same as \$I_1\$ or \$6\:\text{A}\$. From simple inspection, we can say:

$$\begin{align*} I_{\text{R}_1}+I_{\text{R}_2}&=8\:\text{A}\\ 6\:\text{A}&=I_{\text{R}_1}+I_{\text{R}_3}\\ 2\:\text{A}+I_{\text{R}_3}&=I_{\text{R}_2}\\ I_{\text{R}_2}\cdot R_2 &= I_{\text{R}_1}\cdot R_1-I_{\text{R}_3}\cdot R_3 \end{align*}$$

That's a little "over-specified" with four equations when you only need two of the first three plus the last one. Regardless, you could wrestle with the above and get the answers you want for all three currents, now.

Or use just two equations and two unknowns using KCL:

$$\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_2}&=I_1+I_2+\frac{V_2}{R_2}\\\\ \frac{V_2}{R_2}+\frac{V_2}{R_3}+I_2&=\frac{V_1}{R_2} \end{align*}$$

When done, you know the current in \$R_3\$ is \$I_3=I_{\text{R}_3}=\frac{V_2}{R_3}\$.

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