Make: Electronics (2nd edition) experiment 9 - capacitive coupling

[Not posting an image of the circuit because it's copyright]

The answer is in the text:

• when the capacitor is discharged (initial state), and you press button A, current will flow through the capacitor, charging it, and through the 10K resistor, and the LED and 470 resistor in parallel with the 10K resistor.  If the LED did not have the 470 resistor in series then the current through it may well blow it.  The 470 is really in the circuit to protect the LED and has a coincidental small role to play in charging the capacitor - but that isn't its real purpose in this experiment.
• Once the capacitor is charged the LED will turn off because the capacitor will not pass DC current: the current's only path is via button B which is open thus nothing flows.
• In order to see the LED flash, the speed of charging the capacitor has to be slowed down.  This is achieved with the 10K resistor in series with it.  If you tried a small value resistor here, you may well not see the LED flash at all; conversely, trying a larger resistor you may see the LED light for longer.
• The 470 and 10K resistor are in parallel with each other so, yes it does somewhat impact the charging rate, and that rate can be calculated by determining the actual resistance (you should be able to calculate total resistance from resistors in parallel by this stage) and thus the time constant. 

What he's teaching you here is that even with DC, a capacitor will allow some current through whilst it is charging; what he's reinforcing though is that you should always current limit your LEDs to stop blowing them.  Note that when he moves on to replace the LED with the DMM, that 470 resistor isn't needed.

If you have plenty of LEDs to hand you could experiment to try and make it blow: remove the 470 resistor and see what happens; try substituting larger capacitors.  I expect at some time you'll see it flash and then no more.

Don't forget to discharge that capacitor using button B before removing it from the circuit!  Especially if you try larger ones.

Hope that helps but feel free to ask more questions.

[Not posting an image of the circuit because it's copyright]

The answer is in the text:

• when the capacitor is discharged (initial state), and you press button A, current will flow through the capacitor, charging it, and through the 10K resistor, and the LED and 470 resistor in parallel with the 10K resistor.  If the LED did not have the 470 resistor in series then the current through it may well blow it.  The 470 is really in the circuit to protect the LED and has a coincidental small role to play in charging the capacitor - but that isn't its real purpose in this experiment.
• Once the capacitor is charged the LED will turn off because the capacitor will not pass DC current: the current's only path is via button B which is open thus nothing flows.
• In order to see the LED flash, the speed of charging the capacitor has to be slowed down.  This is achieved with the 10K resistor in series with it.  If you tried a small value resistor here, you may well not see the LED flash at all; conversely, trying a larger resistor you may see the LED light for longer.
• The 470 and 10K resistor are in parallel with each other so, yes it does somewhat impact the charging rate, and that rate can be calculated by determining the actual resistance (you should be able to calculate total resistance from resistors in parallel by this stage) and thus the time constant. 

What he's teaching you here is that even with DC, a capacitor will allow some current through whilst it is charging; what he's reinforcing though is that you should always current limit your LEDs to stop blowing them.  Note that when he moves on to replace the LED with the DMM, that 470 resistor isn't needed.

If you have plenty of LEDs to hand you could experiment to try and make it blow: remove the 470 resistor and see what happens; try substituting larger capacitors.  I expect at some time you'll see it flash and then no more.

Don't forget to discharge that capacitor using button B before removing it from the circuit!  Especially if you try larger ones.

Hope that helps but feel free to ask more questions.

Are the bullet points yours or from the book? I can't find that written anywhere in it.

Thank you

They are mine - just trying to rephrase what is written there to explain it in a different way.  

thanks, that's very helpful