Calculate current of resistor with Norton's Theorem

Question

I have a problem with the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

I have to calculate the following data using Norton's theorem:

• \$R_n\$
• \$A_n\$
• \$I_{R_2}\$

For the first I calculate the following circuit:

^{simulate this circuit}

That is:

$$R_n=\frac{R_4\cdot(R_2+R_1+R_3)}{R_4+R_2+R_1+R_3}=\frac{2,5\Omega\cdot(2,5\Omega+5\Omega+2,5\Omega)}{2,5\Omega+2,5\Omega+5\Omega+2,5\Omega}=\frac{25\Omega}{12,5\Omega}=2\Omega$$

To calcultate \$A_n\$ I used superposition: for \$A_4\$ is simple because of current con \$A-B\$ is the same of \$A_4\$, so $$I_{AB}'=5A$$

Considering \$A_1\$ I calculate the voltage of real current source:

$$V_1=A_1\cdot R_1=15A\cdot 2,5\Omega=37,5V$$

And then the current:

$$I_{AB}''= \frac{V_1}{R_2+R_4+R_3}=\frac{37,5V}{10}=3,75A$$

At the end: $$I_{AB}=5A+3,75A=8,75A$$

Now my problem come: how can I calculate \$I_{R_2}\$??

I know the result is \$2 A\$ and I tried all I can but without success....

Answer 1

Since you know \$I_{AB} = 8.75A\$ and \$R_n = 2 \Omega\$, calculate \$V_{AB} = 8.75*2 = 17.5V\$. From this, we get \$I_{R4} = V_{AB}/R_4 = 17.5/2.5 = 7A\$.
If 7A is flowing in \$R_{4}\$ and 5A in \$A_4\$. Rest 2A is coming from \$R_2\$ from KCL.