Electronic – Circuit analysis with known current

First off, mesh analysis would be simpler because you would only have one unknown (the current in the right mesh).

As you know, in nodal analysis you are summing currents at the nodes. Current into the nodes positive and current out of the nodes negative. The way you have defined the problem you should have 3 equations and three unknowns. Your first equations seems to do this correctly for node B. It kind of falls apart after that. For example, node A equation should just be .005 = (Va-Vb)/3000. You already know that Vc = 16 volts.

So simplifying the node A equation gives you what Andy aka said, then just substitute it back into your node B equation.

The first thing you should notice is that from a small-signal preceptive point of view \$Q_1\$ \$r_{o1}\$ act just like the emitter resistance for \$Q_2\$

And the small signal mode will look like this

^{simulate this circuit – Schematic created using CircuitLab}

The small-signal parameters are:

\$\large g_{m2} = \frac{0.1\textrm{mA}}{25\textrm{mV}} = 4\textrm{mS}\$

\$\large r_{\pi 2} = \frac{\beta}{g_{m2}} = 25\textrm{k}\Omega\$

\$\large r_{o2} = r_{o1} = \frac{V_A}{\textrm{I}} =\frac{5 \textrm{V}}{0.1\textrm{mA}} = 50 \$

Hence the output resistance will be the same as I show here:

BJT common-base output resistance derivation

finding output resistance of CB amplifier with ro

$$R_{OUT} = \frac{V_X}{I_X} = r_{o2}+\left(1+ g_{m2} r_{o2}\right)\cdot r_{o1}||r_{\pi2} = 3.4\textrm{M}\Omega $$

Also, remember that in the small-signal analysis you can replace PNP with the NPN small-signal model.

^{simulate this circuit}

^{simulate this circuit}