I couldn't find anything on mikes.com and I couldn't find part# 17-1603 on mikesxs.net but if you are talking about 17-6903 Ultimate High Output Coil on mikesxs.net it say it is: Possible Output as high as 82 Kv. That just means it is capable of reaching that voltage under ideal conditions based on the primary side voltage, dwell time and current (and who knows what voltage,current or dwell time they used for their basis for those claims) not that it is the normal operating range for the coil.
Notes:
Increases in cylinder pressure and spark plug gap, as well as less than ideal mixture ratios (either too rich or too lean) will increase the voltage required to create a spark. In more simple terms, if you step on it, requirement goes up. If you increase your engine's load while towing i.e. more fuel volume enters the cylinder (upping pressure), requirement goes up. If your plugs wear and the gap burns larger, requirement also goes up. As well, any changes that you voluntarily make such as exhaust modifications, bigger carburetors or fuel injectors, camshaft changes, or upping your plug gap to get a bigger spark, all increase the firing voltage requirement.
Some Math
Having more available voltage results in more stored energy in the coil, and since physics dictates energy must be conserved in an equation, we can do some very simple math. Please bear in mind that this is an example to express the basics of the math with simple numbers and leaves out the pages of complex calculations.
Lets compare two hypothetical coils. The math was purposely kept simple for demonstration purposes.
Coil #1 produces a maximum of 50,000 volts with a current of 0.01A (10 milliamps). Total available power would work out to 500 watts (for the spark duration).
Coil #2 produces a maximum of 40,000 volts with a current of 0.01A (10 milliamps). Total available power would work out to 400 watts (for the spark duration).
If we have a condition that requires 20,000 volts to initiate a spark, with a spark current of 0.0075A (dictated by the circuit resistance - plug wires, distributor, spark plug gap and cylinder pressure) :
Coil #1 would work as follows: 500W divided by 20,000 V would equal 0.025A which is 2.5 times the current of the secondary spark since maximum output voltage is not required.
Coil #2 would work as follows: 400W divided by 20,000 V would equal 0.020A which is 2 times the current of the secondary spark since maximum output voltage is not required.
Now for spark duration (we are almost done), our conditions that required 20,000 volts to initialize a spark also dictate that the spark current would be 0.0075A for the particular situation. So Coil #1 would provide 0.025A divided by 0.0075A = 3.33 times the spark duration. Coil #2 would provide 0.020A divided by 0.0075A = 2.67 times the spark duration.
Both coils satisfy the requirements of the test, however Coil #1 produces more spark duration. Since a mixture burns in a cylinder, a longer duration spark does two things. It ensures ignition of the fuel mixture and, as the mixture swirls in the cylinder, the spark can be igniting a larger portion of the mixture helping ensure more complete combustion. With high energy ignitions, it is not uncommon to be able to increase your spark plug gap a little to get a larger spark for even better combustion. Aside from ensuring combustion, a good ignition system can improve efficiency by releasing the most energy from the fuel charge and leaving very little to go out in the exhaust.
Of course just picking a coil or ignition system that advertises high voltage is half the battle. If the system has less output current than your present system it may not perform any better. A good parameter to look for is maximum available output power/spark.
Notes taken from
http://www.auroraelectronics.com/ignition_systems_-_basics_to_high performance.htm