(1) Use metal film where possible. Fewer bad surprises. At 1 cents each either way the cost of bad surprises exceeds the component cost, even if the cost is only measured in frustration and wasted effort.
(2) Wouter (correctly (of course)) says "evenly spaced" but doesn't quite explain it. He means that the ratio between adjacent resistors should be about the same. You should aim to always include the powers of 10 values and then have as many as appropriate in between to fill in.
SO
1, 10, 100, 1000, 10000 ...
OK, that one was obvious.
But sqrt(10 ) = 3.16, so
- 3.16, 10, 31.6, 100, 316 ... :-)
BUT they don't make 3.16 etc in sensible standard ranges, so using the nearest "E12" values:
1, 3.3, 10, 33, 100, 330, 1000, 3k3, 10k, 33k ...
The "obvious" thing to do may be to use
1, 4.7, 10, 47, 100, 470 etc
BUT the ratio of 47/10 = 47 (of course) BUT the ratio of 100/47 = 2.13.
So, if you had a fixed voltage and were connecting successively higher value resistors to ground the change from 100 to 470 would decrease the current by a factor of 4.7, but the next step from 470 to 1000 would reduce the current by a ratio of 2.13. As you went up the currents would change by factors of 4.7, 2.13, 4.7, 2.13, 4.7 ...
You usually get more than 2 steps per decade.
The smallest sensible number has 12 steps per decade.
These are say 1, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2, 10 ...
If looked at by resistance difference the series seems uneven, The differences are.
0.2, 0.3, 0.3, 0.4, 0.5, ... 1.4, 1.8
BUT - when looked geometrically by ratio we see:
1.2/1 = 1.2
1.5/1.2 = 1.25
1.8/1.5 = 1.2
2.2/1.8 = 1.222
2.7/2.2 = 1.227
3.3/2.7 = 1.222
...
10/8.2 = 1.22
SO, within the resolution afforded by 2 significant digit numbers we see that the ratio of adjacent resistances is about 1.21152766 :-) .
I use that "strange" value as it is the twelfth root of 10. If you multiply a number by 1.21152766 twelve times you get a result 10 times larger.
So if you space twelve resistors across a decade range with each a factor of 10^(1/12) larger than the prior one you get resistors which increase in value "smoothly" from a current flow point of view.
E12 - 12 resistors per decade spaced in value by a ratio of the 12th root of 10 .
E24 - 24 resistors per decade spaced in value by a ratio of the 24th root of 10 .
E48 - 48 resistors per decade spaced in value by a ratio of the 48th root of 10 .
E96 ...
More anon maybe .... brake pads to change, darkness fallen ...
There is inductance and there is capacitance between the spiral turns and from end-to-end. It's generally not important until you get to VHF frequencies, but of course the effect is relatively larger on higher resistances for the capacitance and the inductance effect is relatively higher on very low resistances.
For example, if you use this calculator, a coil with 8 turns 2mm diameter and 7mm long would have an inductance of 0.04uH, so at 100MHz and Xl/R = 0.1, R < 250\$\Omega\$. Note that if you have a current sense resistor of very low value, even a fraction of a microhenry inductance will start to have a noticeable effect at moderate frequencies. A 10m\$\Omega\$ resistor with 40nH inductance would be affected similarly at only 4kHz.
Parasitic capacitance works similarly from the other end of the frequency scale- a 10M\$\Omega\$ resistor with 0.4 pF of end-to-end capacitance would be be affected similarly at only 4kHz.
Best Answer
Unfortunately answer is no. There is no quick test you can perform to figure out the wattage of the resistor. Based on the appearance you can make an educated guess, but that's all it is, a guess. The only way to be 100% sure is if it's marked or you happen to have the manufacturer's part number and can look it up in the datasheets.