# Norton's equivalent circuit with independent sources

I need to find the current flowing through $$\R_{5}\$$ by using Norton's theorem. What I tried to do was:

1. Open-circuit the part of the circuit where there is $$\R_{5}\$$.

2. Calculate $$\R_{n}\$$ by open-circuiting the independent current sources (that is, $$\I_{1}\$$) and short-circuiting the voltage sources ($$\V_{1}\$$).

3. Calculate $$\I_{n}\$$ by short-circuiting the part where there is $$\R_{5}\$$.

So,

2. $$R_{n}=R_{4}+\frac{R_{2}R_{3}}{R_{2}+R_{3}}=7.5ohms$$

3. I used KCL to find $$\V_{C}\$$:

$$\frac{V_{1}-V_{C}}{R_{3}}+\frac{0-V_{C}}{R_{2}}+\frac{0-V_{C}}{R_{1}}-I_{1}=0$$

$$V_{C}=4.18V$$

Then,

$$I_{n}=\frac{V_{C}}{R_{4}}=\frac{4.18}{6}=0.696A$$

Apparently, $$\I_{n}\$$ is wrong, but I can't find any mistake in my reasoning. Any idea?

• Try to write a proper KCL for Vc node. Because now your KCL is not consistent. – G36 Nov 13 '18 at 17:31
• Current through R4 does not feature in your first node equation. Also some signs are wrong. – Chu Nov 13 '18 at 17:44
• @G36 Okay, I checked my KCL, but I still get the wrong result. – Arnau Nov 13 '18 at 17:55
• What current direction you have assumed to be "positive" and what "negative"? Do you understand my question? – G36 Nov 13 '18 at 17:59
• @G36 I'm assuming all the currents are coming into the node, and I always leave Vc at the end of the subtraction (v1-vc, 0-vc, etc.). That's what our professor told us. Is it okay? – Arnau Nov 13 '18 at 18:05

I'm assuming all the currents are coming into the node, and I always leave Vc at the end of the subtraction (v1-vc, 0-vc, etc.)

For sure you can do that. But notice that the $$\I_1\$$ current is leaving the $$\V_C\$$ node.

So the $$\I_1\$$ current needs to have a different sign.

I will show you two cases:

First case:

I will give a minus sign if the current entering the node (coming into the node).

And the plus sign if the current is leaving the node.

$$-\frac{V_1 - V_C}{R_3} - \frac{0 - V_C}{R_2} - \frac{0 - V_C}{R_4} + I_1 =0$$

Notice that I give $$\I_1\$$ the plus sign because $$\I_1\$$ is leaving the node (current source).

Second case:

Plus sign to all the currents that are entering the node.

$$\frac{V_1 - V_C}{R_3} + \frac{0 - V_C}{R_2} + \frac{0 - V_C}{R_4} - I_1 =0$$

Do you know why I give a minus sign to $$\ I_1\$$?

So what is important here is that you always must be consistent with the sign convention you have chosen when you are writing nodal equations.

And of course, those two "method" will give the same answer.

And we do not take $$\R_1\$$ resistor into account because $$\R_1\$$ is in series with the constant current source. Hence, it won't change the current in this branch.

• But I include R4 in my KCL. – G36 Nov 13 '18 at 18:40
• Okay, I understand now. I have a question, though. The branch where R1 is, is it going to flow only I1 through it? Or it will flow I1 + the diverged current to that branch? – Arnau Nov 13 '18 at 18:41
• Only I1 current is flowing in the I1+R1 branch. In series circuit only one current can flow – G36 Nov 13 '18 at 18:47
• Okay, thank you very much. Everything makes sense now. – Arnau Nov 13 '18 at 18:50