A Veritasium puzzle has attracted plenty of interest via Youtube, Quora and several other technical milieux.Here is how to make sense of it.
A transmission line may be a long rectangular, square or circular section called a waveguide.It may be a circular section with a central conductor, called a coaxial cable.It may be two conductors at a given distance, with some means for holding the separation constant and some means of holding the pair clear of obstructions and the ground, called a ladder line.Ladder lines are usually rigged at least 4 interwire lengths above ground.
All transmission lines have three major parameters: characteristic impedance; signal loss per unit lengthand its variation with signal frequency; and power handling capability.For defining parameters in detail, a download from this site will provide TL details. https://ac6la.com/tldetails1.html
Here's the first point of interest: until an input signal is reflected back to source a transmission line may be fairly represented by a resistance equal to the characteristic impedance of the line.
Now onto the case in point. Veritasium postulates two transmission lines which are 3E5 km in length, terminated remotely with a short.As you can see, the ladder line he pictures is far from ideal: laid on the ground, its stray capacitance is high,and its frequency response will be poor, tailing off rapidly with frequency, and its transmission loss per unit distance is also high. Still it has some characteristic Impedence Zo which we will replace with a resistance value of R = Zo
Remember, this will only serve until reflections occur. If his lines were better supported, there would be some small signalreturned from the short-circuited ends after about two seconds.We will therefore confine our considerations to the first two seconds after the knife switch is closed.We draw a series circuit: battery, switch, R1 = Z0, lamp, R2 = Zo, back to battery.
There is a problem; the resistance of the lamp is not given, so we should consider three cases:Rlamp = 1* ZoRlamp = 10*ZoRlamp = 1/10*ZoCall the battery terminal voltage V
Case1 Volts across lamp = V*Zo/ (Zo+ 2*Zo )= V/3 Current = V/(3*Zo) Power into lamp = V^2 / 9*Zo
Case 2 Volts across lamp = V*Zo*10/(12*Zo) = 5*V/6 Current = V/12*Zo Power into lamp= 5*V^2/72*Zo
Case 3 Volts across lamp = V*0.1*Zo / 2.1*Zo = V/21 Current = V / 2.1*Zo power into lamp = V^2 / 44.1* Zo
We would like to compare these lamp powers to the lamp power when directly connected to battery and switch
Case 4 Volts across lamp = V volts current = V/R amps power = V^2 / R watts
Case 5 Volts across lamp = V volts current = V / 10*R amps power = V^2 /10* R watts
Case 6 Volts across lamp = V volts current = V /0.1*R amps power = V^2 /0.1* R watts
We conclude that if the lamp resistance matches the characteristic impedance of the ladder lines, the lamp power is maximized. With no ladder lines connected, the lower lamp resistance dissipates most energy.Hence if Veritasium wants to see an immediate glow from his lamp in the first two seconds, he should match the lamp resistance to the ladder line Zo.The physics estimates he received around 2% lamp power appear to be low compared with the 1/9 or 11% power available in the first two seconds.
AFTER two seconds, one expects some reflected wavefront - it is likely to be very very small.In other venues, some writers have asked: If the response is so delayed, you are supposing that the modest powerto the lamp would STILL be available - which as I hope you will now agree, it is appropriate to say "yes!"