The "speed", rupture capability, and voltage spec of a fuse are each separate traits that aren't necessarily related. They will be specified according to the nature of the application.
Many cheap multimeters (i.e. <50$us) will only have glass fuses, because they're cheaper. But 'high rupture' capability (the ability to break a circuit up to a rated voltage, when a high fault current is flowing) is important in a multimeter (and in a circuit breaker), especially in a DC scenario, because DC >40V can easily form an arc and continue the current flow, even though the fuse has blown.
For the "20A" fuse in your meter, a HRC type would be wise, and rated for the max V rating of your multimeter, and same speed as the original.
The question contains some errors but I will try to answer what I think you are really asking. You're talking about fuses (fuse wire) so clearly you are interested in trying to protect the circuit.
When batteries are connected in series, the same current flows through all batteries. So you need a single fuse to protect them all.
Actually fuses are often used to protect the cable, and so fuses should be set to a lower rating than the current the cable can carry.
If you really are talking about resistors and putting them in series with the batteries: then standard formulae apply: power W = I.I.R (I squared R). V = I.R , W = V.I and all the other permutations of these formulae.
So you have to ensure that the power dissipated by the resistor does not exceed the maximum power rating for that resistor. If it does, the resistor will become excessively hot and burn out.
The voltage you know, the maximum current you know or you can calculate, and so you can calculate the power dissipated by the resistor.
You don't need fuse wire in each position. Fuse wire - of the correct rating - in one position is enough.
Let's spend a moment talking about fuse wire. Fuse wire is specified in Amps, as to how much current it can handle before it breaks. Fuse wire is not specified in W - watts.
Watts = volts x current. And the voltage doesn't matter, because it's an electrical conductor (made of metal), the voltage at any point along the fuse wire is the same (that's not entirely true, but good enough for discussion here).
So you're only interested in current when talking about fuse wire, which is measured in amps.
Best Answer
Fuse ratings in a design should be set so they interrupt the current for an over-current condition. You determine how much current your design needs. If your load cannot tolerate large currents or there are extreme safety requirements, you may want many fuses. They should be rated so they won't trip during normal use, but trip if something shorts out. This really really depends on the load so follow these steps:
1) Find your loading current (If its an active load like a motor then experimentally find the nominal current with a power supply)
2) Give yourself some margin, if your load is drawing 2A normally, then multiply that by a saftey-factor so that if there is a voltage spike it won't blow the fuse out. A 50% saftey-factor would give you 3A (of not tripping). If the load shorts (lets say 1Ohm worst case) Then for a 50V pack you would get 50V/1Ohm = 50Amps). So you would need a fuse that blows with less than 50Amps of current, but give yourself some margin in that direction too, and maybe get a fuse that trips at 10Amps.
3) Find a fuse with the rated voltage and current, keep in mind every fuse takes time to blow or trip, will the load or source have a problem with that? If the fuse takes 1 second to blow, its probably not going to protect the batteries, if it takes 1us it probably will because not much heat will be dissipated anywhere.
Edit: 4) If you have a captivate load, you will want to watch the inrush current. (Li-pos are "kind of" capacative in the fact that they are a very low ohmic source and their chemsitry can deliver lots of current fast.) There are slow blow fuses that are more suited to this sort of thing. Double check the inrush rating and design for it.