calculate currents class ab amplifier
Chicken &amp; egg is an accurate way to describe what happens in this circuit... or any circuit that is enclosed in a feedback loop.

Here, you have the operating point - and therefore the output voltage at node 2 determined by the node 2 voltage fed back through bias resistor R6 to the input of your amplifier. This overall feedback is negative because there is only one inverting stage, Q1. Most circuit analysis is performed based on achieving a specified output.

So, to begin, assume output voltage (node2) = 1/2 V1 = 6 volts. This will allow the maximum peak-to-peak output voltage swing under signal conditions.

Choose a quiescent emitter current for Q2; let's say 10 milliamps.

Calculate the voltage drop across R1 = 0.33 volts.

Calculate voltage at the base of Q2 = 6.0 + 0.7 + 0.33 = 7.03 volts.

Calculate the voltage drop across R3 = 12 - 7.03 = 4.97 volts.

Calculate current through R3 = 4.97V/1000 = 4.97 milliamps.

Calculate collector voltage of Q1 = 7.03 - 0.7 - 0.7 = 5.63 volts.

Calculate the emitter voltage of Q3 = 5.63 + 0.7 = 6.33 volts.

Calculate the voltage drop across R2 = 6.33 - 6.0 = 0.33 volts.

Calculate the current through R2 = 0.33 volts / 33 ohms = 10 milliamps.

We now know that Q2 and Q3 emitter currents are ~ equal. 

Assume Q1, Q2 &amp; Q3 beta = 500.

Calculate Q2 &amp; Q3 base currents = 10 milliamps / 500 = 20 microamps.

Calculate Q1 collector current = 4.97 milliamps - 0.02 milliamps - 0.02 milliamps = 4.93 milliamps.

Calculate Q1 base current = 4.93 / 500 = 10 microamps.

Calculate Q1 emitter current = 4.93 milliamps + 0.01 milliamps = 4.94 milliamps.

Calculate voltage across R4 = 100 * 4.94 milliamps = 0.494 volts.

Calculate Q1 base voltage = 0.7 + 0.494 = 1.194 volts.

Calculate current through R5 = 1.194 volts / 1500 ohms = 0.796 milliamps.

Calculate current through R6 = (6 volts - 1.194 volts) / 5000 ohms = 961 microamps. 

Now you could reiterate, knowing that about a milliamp of the 10 milliamp emitter current of Q2 is flowing out of "Node 2" back to the base of Q1. Q2 &amp; Q3 are dissipating about 60 milliwatts each, which might be closer to Class AB+, but you can see that lowering the quiescent power dissipation in the output transistors is as simple as choosing a lower value for quiescent emitter current in Q2 and Q3.

In practice, the operating point is often set by a resistor in parallel with diodes D1 &amp; D2.

Hope this helps. Have fun!