Simple Circuit Question

Windburn

I am working my way through Make: Electronics.
Experiment 9: Time and Capacitors asks you to construct the following circuit:


The purpose is to demonstrate that a capacitor can allow a signal to pass through it. By pressing switch A, the capacitor charges and briefly illuminates the LED. Once "fully" charged, the LED no longer illuminates regardless of pressing switch A. To reset the capacitor, pressing switch B discharges the capacitor and allows you to illuminate the LED again with switch A.

Now, my question is why is the 10k ohm resistor necessary? Shouldn't a simple short circuit discharge the capacitor?
I templated this in circuits.io and it doesn't work, which is expected. But why? If the resistor is necessary to safely discharge the capacitor, why must it be connected to the negative bus? Why does it not work to put the resistor in the short circuit with switch B?

Thanks.

romanqwerty

The LED is a diode which means current only goes one way through it. The 10k resistor is there so that when charging the capacitor, current moves through the led while still having an available pathway to discharge the cap. If you put the 10k right next to B either:

There is a short circuit pathway to charge the capacitor (no lighting of LED) or there is no way to discharge the capacitor (remember current can't flow the wrong way through a led).

Windburn

Thank you for the response.

The following circuit illustrates what I'm imagining. After the capacitor has been charged, why doesn't pressing switch B create the short circuit path in yellow to discharge the capacitor?



If I had to guess, I suspect my confusion lies in a misunderstanding of the way capacitors work.

romanqwerty

It will equalize the difference across it but won't necessarily discharge it. You have a way to pull current out of that yellow square but no way to put some back in. Also there will be a voltage drop (or raise depending on how you look at it) across the diode. Essentially you're floating your capacitor. You're setting both sides to the same potential but that is not really pinned anything elsewhere in the circuit.

It's generally bad practice to directly short a capacitor (with some types that can explode them) and they also tend to be directional (one side should be charged positive and the other negative).

The short of it is that your system is complicated and difficult to predict what will happen, which floating components tends to do. The book example is simpler to analyze and the real world components will more closely match how theory dictates they should.