pricingresistors
Does anybody know of any reasons that the 560 Ohm resistor is so much cheaper than the other values? It does not appear to be a trend of price increasing or decreasing with resistance since there are resistors both above and below this value that are more expensive. This is not the only example of this line. Every now and then there will be a particular value that is half the price of all the others.
Mouser.ca screenshot, 2020-10-17
Like gbarry says. The next higher value you find in the Digikey offering is 2 kΩ, so you would need a 6386 Ω resistor in parallel. That is between one end and the wiper, not across the full resistance. 6386 Ω is also an odd value, so you may want to compose that using different standard values as well.
Now the 1523 Ω, how precise does that have to be? You represent it in 4 significant digits, that's 0.1 %. Potmeters are often 20 %, more precise ones can be pretty expensive. So our 2 kΩ can be anywhere between 1600 Ω and 2400 Ω. In the first case your parallel resistor should be 31650 Ω, in the second 4170 Ω. That's a wide range. You could use a 33 kΩ trimming potmeter. Use a multiturn is you want to get the 1523 Ω really precise.
Note that placing another resistor over the potmeter's wiper distorts its linear characteristic.
The greenish curve shows the resistance as a function of the rotation in % for a 2.2 kΩ resistor parallel to a 5 kΩ linear potmeter. The purple curve shows that resistance for the 2 kΩ potmeter plus the 6386 Ω, so that's not so bad after all. The rule for the most linear characteristic is to have the ratio resistor/potmeter as large as possible.
The equation is basic Ohms Law with parts replaced for power calculation.
The formula is: $$ P=\frac{V^2}{R} $$ As it's got 12V across the resistor, and the resistor measures 51Ω, that becomes: $$ P=\frac{12^2}{51} = 2.82W $$ So you need at least 3W to be safe.
Placing resistors in parallel results in lower resistance. The long form of the formula for that is: $$ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n} $$ So taking three 150Ω resistors in parallel gives you: $$ \frac{1}{R_T} = \frac{1}{150} + \frac{1}{150} + \frac{1}{150} $$ $$ \frac{1}{R_T} = 0.02 $$ $$ R_T = \frac{1}{0.02} = 50\Omega $$ If all the resistors are the same value then it is a simple as dividing the value by the number of resistors, and the power ratings just add together, 20 three 0.5W resistors would give 1.5W of power. Not enough for your application.
If you cannot get 1W resistors then you would be better with six 0.5W resistors of double the value - so 300Ω each.
Best Answer
Some values of resistors and capacitors are manufactured in much greater quantity, and thus cost less. For designs where that exact value of the resistance isn’t relevant (such as designs where the ratio of resistances determines the operation), this is perfectly fine and you can lower costs by selecting these values. If you have a very specific need, and can’t meet it with the more common resistor values, you may have to pay more to get specialized components.