Question: The National Ignition fusion recently [announced](https://www.llnl.gov/news/national-ignition-facility-achieves-fusion-ignition) the achievement of nuclear fusion "ignition", i.e. more energy released from a sample undergoing nuclear fusion reactions than was directly input into the sample by lasers (*not* the total energy required to power all the lasers, which was still much higher than the produced energy).

My understanding is that one of the [main design decisions](https://inference-review.com/article/the-quest-for-fusion-energy) for a hypothetical practical nuclear fusion reactor is whether to use deuterium or tritium isotopes (or both) as the hydrogen fuel source. Tritium has the advantage that it undergoes nuclear fusion at a lower temperature than deuterium, so it is a "better" fuel source in terms of intrinsic physical properties, but it has the disadvantage of being much, much more expensive and challenging than deuterium to produce in large quantities.

How much deuterium and tritium did the NIF's hohlraum contain in its ignition demonstration? [This](https://physicstoday.scitation.org/do/10.1063/PT.6.2.20221213a/full/) is the only source I could find that discussed the hohlraum's fuel contents, and it just said that the hohlraum contained a "deuterium–tritium fuel", but didn't give the relative proportions. Was it a roughly even mix of both isotopes?

(I'm also curious what pressure the "peppercorn" of hydrogen fuel was kept at before the lasers were activated, and what phase of matter it was in at that pressure at room temperature. Presumably it was a plasma when the lasers caused ignition.)

Answer: 
You have a loop composed entirely of wire from the positive to negative battery terminal. A "short circuit". A large current will flow. Neither the bulb nor the Earth connection are involved.
The earth connection is meaningless. You are putting a direct short across the battery so the potential will be essentially zero.

Now if you had connections to earth at two locations, as in the diagram below, then there would be a potential difference across the bulb, the amount depending on the relative resistance of the path through earth at the location, resistance of the bulb, and internal battery resistance.

Hope this helps.

[![enter image description here](https://i.stack.imgur.com/Al1z6.jpg)](https://i.stack.imgur.com/Al1z6.jpg)