I refer to my previous answers on this exercise:
Make electronics - charles pratt - experiment 11
PUT Base Resistors Question
The resistors aren't exactly in series; they're constructing a voltage divider, putting the PUT's base voltage at $$ 27/(15+27) * 6V $$
This is the programmable part of a PUT which sets the on voltage for it. In circuits 2 and 3 the gate is connected to the negative rail and is therefore at 0V, preventing it from ever turning on.
As in my other answers, PUTs are an unusual near-obsolete part and I wouldn't bother learning about them.
There are many errors in the diagrams. How you connect these transistor, is called Common Emitter, and is one of the three ways in which you can connect a transistor.
For common emitter configuration, it is usually considered that the load is connected in the collector circuit. To light a led, the basic circuit would be:
![schematic](https://i.stack.imgur.com/TrQ6Y.png)
simulate this circuit – Schematic created using CircuitLab
When the source \$V_b\$ is applied, the LED ligths on.
The LED is connected in the output circuit. The LKV for this:
$$
V_{CC} = I_C\,R_C + V_{D1} + V_{CEsat}
$$
For example, if \$V_{D1} = 1.2\,\mathrm{V}\$ (like common red LED), \$V_{CEsat}= 0.8\,\mathrm{V}\$ (Collector-emmiter saturation voltage) and \$V_{CC}=9\,\mathrm{V}\$
$$
I_C\,R_C = 7\,\mathrm{V}
$$
for a common red LED, we can suppose 15 mA, then
$$
R_C = \dfrac{7\,\mathrm{V}}{15\,\mathrm{mA}} = 466.66\,\Omega\approx 470\,\Omega
$$
The input circuit is on the base terminal. For a current collector of 15 mA and a factor \$h_{FE}=100\$ (typical)
$$
I_B = \dfrac{I_C}{h_{FE}} = \dfrac{15\,\mathrm{mA}}{100} = 150\,\mu\mathrm{A}
$$
If \$V_b = 3.3\,\mathrm{V}\$ (typical for a \$\mu\$C) and \$V_{BE}=0.7\,\mathrm{V}\$ (Base-emmiter voltage for a silicon transistor, typical)
$$
R_{b} = \dfrac{V_b - V_{BE}}{I_B} = \dfrac{3.3 - 0.7}{150\times 10^{-6}} = 17.333\,k\Omega\approx 18\,k\Omega
$$
Best Answer
It wasn't even true back then. Well, maybe that's why Dawkins is a biologist and not an engineer. :-)
Today's processors pack billions of transistors on a die a few square cm in area and less than a mm high. There would fit hundreds of them in a skull, maybe \$10^{12}\$ transistors.
Even if you look at discrete transistors there would fit more than just a few hundred. I guess SOT-23 already existed in 1989, and then you would get \$10^5\$-\$10^6\$ of them in a skull.
edit (2011-06-13)
I own a copy of The Selfish Gene, and was curious what Dawkins had in mind, so I looked into it. Her's more from that paragraph:
Somebody must have told Dawkins that a transistor has 3 pins :-).
Anyway, he doesn't only compare the numbers of neurons (or neurones, BE?) to transistors, but also points out that the neuron is a lot more complex, partly because of its thousands of connections. My guesstimate is that you'd need \$10^5\$ to \$10^6\$ transistors to emulate one such neuron (maybe as an analog instead digital computer?). Which means that a skull stuffed with GPUs wouldn't still come close to the processing power of a brain.
And then there's the problem of all these connections. They're the real power, not just the large number of neurons. We don't have the technology to build such complex systems, and IMO won't for a long time. And then I'm not even talking about the dynamic nature of these connections: they can rearrange themselves, making new connections and breaking others.
To put all these AI suckers in perspective, take a look at our vision system. In a second we can process a stereoscopic image of \$10^8\$ pixels, create a virtual 3D model of the scene and identify objects in detail. Move half a meter to the right and you add lots of new data. There's still a long way to go...