Does nodal analysis work with alternating currents if every step is treated like DC

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 :)

The circuit as shown is not viable - or you could analyze it in two phases, if you must:

Phase 1:

• Each 1n4148 diode is rated for 200 mA continuous, 450 mA peak repetitive current.
• When wired as indicated, each diode will drop approximately 1 to 1.5 Volts (Fig.3 in datasheet) before the current exceeds absolute maximum rating
• As the supply voltage is 5 Volts, this far exceeds the maximum 3 Volts noted above, so one of the two diodes will burn out.

Phase 2.a: If D2 burns out and becomes an open circuit:

• There will be no voltage at V_out as D2 is now an open circuit
• Result: V_out = 0 Volts

Phase 2.b: If D1 burns out and becomes an open circuit:

• V_out = V1 - V_D2 = ~ 4.4 Volts

Then there are the possibilities of D1 or D2 burning out to become short. That resultant analysis is left for you to do :-)