T. Ment wrote:
On Wed, 26 Jun 2019 16:25:17 -0400, Paul wrote:

And this is the way to buy resistors. An "assortment pack"
gives resistors at from $0.03 to $0.05 a piece. This one
happens to be sold out, but it's just an illustration.

https://www.radioshack.com/collectio...nt=20332255301

You will find 30 each of 1k, 10k, and 100k;
ten each of 1, 10, 100, 120, 150, 220,
330, 470, 560, 1.5k, 2.2k,
3.3k, 4.7k, 15k, 22k, 47k,
220k, 470k, 1meg, and 10meg;

Five each of 2.2, 15, 22, 33, 39, 51, 68, 82,
180, 270, 390, 510, 680, 820, 1.2k,
1.8k, 2.7k, 3k, 3.9k, 5.1k, 5.6k,
6.8k, 8.2k, 12k, 18k, 27k, 33k, 39k,
51k, 56k, 68k, 82k, 120k, 150k, 180k,
270k, 330k, 1.5meg, 2.2meg, 3.3meg,
and 4.7meg.

Notice I'm getting 180, 680, 820, the usual kinds
of standard values.

$15 is more than I want to spend right now.

Ebay has cheap assortments, if you can wait 4-6 weeks for shipping from
China. USA assortment sellers are pricey, like Radio Shack.

But I found 1/4 watt packs, choose your ohms, shipping included, cheap:

https://www.ebay.com/itm/1-4W-25-Wat...SJJa BvQvMz9g

$1.55 is my kind of purchase. If 1k ohms is the wrong value, it won't
cost too much to buy different ohms. Even three wild guesses are less
than $5.

I prefer carbon composition (as they're flat to 1GHz), but for
this project, the metal film will suffice.

The metal film might be a bit more brittle as well.

If you need to bend the legs on a resistor,
you can use needle nose pliers, hold the leg
away from the body of the component, and bend
the leg ninety degrees, such that the stress
doesn't go onto the meniscus during the bending
operation.

The carbon composition ones, in my opinion,
can take a bit more stress.

The metal film ones use more "glassy" materials.

You can place resistors in series and in parallel,
to create values other than the "mono-value" you
get in a bulk bag. Two 1K in parallel is 500 ohms.
Three 1K in parallel is 333 ohms. I could combine
a composite 500 ohms and a composite 333 in series
and make a 833 ohm resistor. See the fun you can have ?

Paul