Electronic – Thevenin’s theorem

I think you're misreading the problem. If I understand you right, you're breaking the circuit on the left side of R3 and R6, then finding the Thevenin equivalent of E, R1, and R2. But what you're supposed to do is break the circuit above and below R2, then find the Thevenin equivalent of E, R1, and R3-R6. You're "looking into" the same terminals that R2 is connected to.

Also, is R3 really 1000 ohms? That looks like a typo.

Edit: I can't go into much more detail without just giving you the answer. You already know the basic steps:

1. Disconnect R2 from the circuit.
2. Determine the voltage between nodes A and B. This is Vth
3. Deactivate E, then determine the resistance between nodes A and B. This is Rth.
4. Now you have a Thevenin equivalent circuit. Attach R2 to it, then compute the current.

Most of the actual work is combining series and parallel resistors, which you should already know how to do.

Norton's theorem says that any voltage source in series with a resistor can be converted to a current source in parallel with a resistor:

^{simulate this circuit – Schematic created using CircuitLab}

If you'd like to switch between the two circuits, the relationship is \$V_{TH} = I_NR\$; the resistor doesn't change.

In your case, the circuit simplifies to:

^{simulate this circuit}

(I'll let you figure out the values of the components)

With this new circuit, it's pretty easy to tell that the total current split between the three resistors is \$I_1\$ + \$I_2\$. Using that total current, you can figure out the current (and then the voltage and power) by treating the circuit as a current divider.