Is it typically advisable to use a voltage divider when trying to achieve a specific voltage in a power supply? I have a 15W +/-12V power supply and I need 10V, so what are the drawbacks of using a voltage divider with, say, 68K ohms and 340K ohms?
And yes, I could get by with a 7810 regulator, but I can't find a regulator such as a 7910 for negative voltage. So I'm looking for other options.
No, not typically advisable. The drawback is that with the resistance values you have chosen, even a small load will bring the voltage down to 0V, unless you buffer it somehow.
A voltage divider is not an appropriate replacement for a voltage regulator if you want to draw any significant amount of power. This is because the output is high impedance. The source impedance will be equivalent to the parallel combination of the two resistors. Using the values you list in your post, 68K and 340K, the output impedance would be around 56.6K. The maximum current you can draw from that is 10V/56.6K = 176 microamps...by shorting the output to ground. And if you want the output to be, say, within 1% of 10 volts, then you can only draw 1% of that current, or less than 2 microamps. If your circuit only needs 1 microamp, then it will work fine, although the divider is going to pass 12/(340+68) = 29 microamps directly to ground, resulting in a power efficiency of about 1*10/((29+1)*12) = 2.8%. And that efficiency will be the same if you use smaller resistors so you can draw more current - for instance if you drop down to 340 ohms and 68 ohms and draw 1 mA instead of 1 uA, the efficiency is still 2.8%.
So yeah, if you want to supply power, use a voltage regulator, that's what they're made for. Also, if you're trying to regulate 12V down to 10V, you'll want a regulator that can deal with that relatively small 2V difference. You'll have to get a relatively decent low-dropout regulator (LDO).
Voltage dividers should almost never be used for power supplies- your suggested resistors would have an effective source resistance of about 56K\$\Omega\$ so a modest load of 1K would reduce the voltage to under 0.2V.
Efficiency of a linear regulator cannot be any better than Vout/Vin, and is often worse because the regulator needs to use some current to operate.
Also, the LM7810 is not comfortable with only a 2V headroom.
There are better regulators such as the LM2991 (negative) and the LD1117A (positive) that can be set with resistors and can supply significant current with little loss.
Be sure to read the datasheets carefully as to acceptable output and input capacitance and ESR, as well as thermal consideration and minimum output load current.
