Collector current for common-emitter audio amplifier
For \$3W\$ and \$8\Omega\$ loudspeaker, you will need \$\sqrt{\frac{3W}{8\Omega}} = 0.61A\$ of RMS current (\$0.86A\$ peak)

So, the collector current should be larger than this current because the  maximum positive current at the load is equal to:  

$$I_{L\; max} = I_C \frac{R_C}{R_C + R_L}$$  

Additional if we assume \$I_C = \frac{1/2V_{CC}}{R_C}\$ 

Hence \$ I_C &gt; \frac{V_{CC}*I_{Lmax}}{V_{CC} -2I_{Lmax}R_L}=\frac{24V*0.86A}{24V - 2*0.86A*8\Omega} =2A\$ and \$Rc = \frac{12V}{2A} = 6\Omega \$ 

The power dissipation in the \$R_C\$ resistor is \$24W\$ 

And in the BJT the power dissipation is also equal to \$24W\$

As you can see the circuit efficiency is very pure 

$$\frac{3W}{48W} * 100 = 6.25\% $$

But this is normal for class A amplifiers. 

This is why people don't use class A as a "output stage". 

We are preferring the class AB push-pull amplifier.
How does this Push-Pull amplifier work?