Electronic – meant by dc drop

AFAIK, you can always solve any linear circuit the 'brute force' way using nodal analysis:

1. Write Kirchoff's Current Equations on all nodes except ground
2. For every circuit component, (i.e. resistors, capacitors etc.), write down their behaviour (for instance, ohm's law for a resistance, i = c dV/dt for a capacitance and so on)
3. At this point, we'll have a handful of equations with us. We can also try to eliminate as many equations from them as possible using any info we have; however in the end, we need to be left with N simultaneous equations in N unknowns. Solve them and we'll get all the node voltages and branch currents.

Coming to the circuit above, let's define the current through V2 as I2, and the ones through R_n as I_n. Let me also call the node at the top as V_a, the node between the CCCS and R5 as V_c, the one between the R_7 and R_8 as V_b and the node in the middle as V_e. Now, writing Kirchoff's Current Law on these nodes will leave us with $$ I_2 = I_7 + I_5\\ I_7 = I_8 + I_x\\ I_x + I_5 = I_6 $$ respectively. Writing down the 'behaviour' of R6, R5, R7, R8, V2, V3 and the CCCS will respectively yield $$ V_E = I_6 R_6 \\ V_A - V_C = I_5 R_5 \\ V_A - V_B = I_7 R_7 \\ V_B = I_8 R_8 \\ V_A = V_2 \\ V_B = V_E + 0.7\\ I_5 = 180 I_x $$

That's 10 linear equations in 10 unknowns. Solve them, and we'll find all I_x as 88.18uAmps...

Of course, 10 equations is a bit too much to solve by hand (I generally use Gauss-Jordan elimination to do this part), but as far as I've seen, this method works in situations where the usual 'text-book' approach using nodal and mesh analyses fail. Furthermore, we don't have to deal with the painful Thevenin equivalent/Super-mesh workarounds here...

On the downside, I'm not quite sure if this approach works with every possible circuit (so far I haven't seen any where it fails), so any negative feedback on this part is welcome :)

I don't know how to proceed with this question or where to start.

If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:

$$v_+ = v_-$$

Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.

This should get you started.

@AlfredCentauri I still don't see the bottom loop, do you mean the loop v+ connected to VB then connected to the voltage source and then the resistor and finally VA. Is that considered a loop even with the op-amp? And when I do I still don't get your equation.

^{simulate this circuit – Schematic created using CircuitLab}

This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:

$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$

rearranging yields

$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$

If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.