Let’s pretend it’s the middle of the 1950’s. If you needed to get access to 1.21 gigawatts of power at once, and you didn’t have access to a bolt of lightning, how could you go about creating that power so that you could channel it into some sort of device?
They had power plants back then and capacitors.
Build enough power plants and enough capacitor banks you could do it. Once the capacitors are all charged let em rip.
Not sure how you could put 1.21 gigawatts in one go through a power line though but I suppose it is just another engineering problem.
Dilton, look, all we need is a little plutonium.
The Hoover Dam could supply it.
A very large capacitor bank like they use for artificial lightning experiments might be able to supply that kind of wattage.
And a DeLorean.
You could try a one-shot, self-destroying MHD generator using a solid rocket as the plasma source. Might take a bit of development work though.
Far more appropriate, IMO is the explosively pumped flux compression generator, no kidding!
If you went back to the 1940s, it’s possible that the power station at Oak Ridge might have been able to help. I don’t have exact figures, but the stuff it was doing for the Manhattan project was supposed to consume 14% of the US’s total power output.
1.21 Gigawatts is certainly not an impossible amount of power from a large power plant, even for the 1950’s. However, sending it through a single wire would be quite difficult.
1.21 gigawatts is around 5% of the electrical consumption of today’s province of Ontario. One large power station can easily produce energy at that rate, as Finagle’s link mentions.
The interesting question is: assuming I have something in my garage that needs that much power to function, how do I get it? Can I just call up Toronto Hydro and get them to put in 1.5-gigawatt service? Let’s take a look.
Electricity…
The largest transmission lines I’ve heard of run at 750 kilovolts (kV). Two amperes (A) at 750 kV gives us a transmission capacity of 2A x 750 kV = 1500 MW, or 1.5 GW, enough to handle our 1.21 GW load with a safety factor to spare.
But how much power can a high-voltage transmission line carry? There’s a high-voltage DC line running from Iguaçu Falls to São Paulo, that carries 6.5 gigawatts at + and - 600 kilovolts, for example, that consists of two transmission lines. So evidently transmission line can be built to carry around 3 gigawatts.
But look at the link to see the size of the equipment required to receive this level of power. It’s city blocks in size! Getting a line into your garage of that magnitude would be… interesting.
What about chemical energy?
I recall reading that the rate of energy transfer when you are filling your car is surprisingly-high. Here’s a cite:
20 US gallons in 4 minutes is 75.7 litres in 4 minutes. That’s 18.9 L/min, or 0.316 L/s. So in order to transfer energy at a rate of 1.21 GW by moving gasoline, we’d need to transfer 0.316 L/s * 1.21 GW / 10.8 MW = 35.3 L/s of gasoline.
Now, how big a pipe do we need for that?
Here’s a pipe-flow calcuulator. It says that to achieve 35.3 L/s with a flow speed of 1 m/s (pretty fast), you need a pipe 21.2 cm in diameter. That’s big, but nowhere near the size of the electrical transmission line. So lets oversize the pipe a bit and assume 25 cm. That’ll give us 49 L/s.
The real question is, how do we get the energy out of the gasoline and into the mysterious piece of equipment in the garage…
What about mechanical power?
According to Wikipedia, engine power output can be expressed as torque multiplied by rotational speed. Let’s say we have a shaft rotating at 1000 RPM. That’s 16.7 revolutions per second (rotational speed), or 105 radians per second (angular speed). A total power transfer of 1.21 GW at that rotational speed implies… lessee…
Power in kW = (torque in N·m * 2π * rotational speed in RPM)/60000, so…
1.21 GW = torque * 2π * 1000 / 60000
1 210 000 GW * 60000 / (2π *1000) = torque
11 550 000 N·m = torque.
Yep. 11.55 million newton-metres. Which is equivalent to a force of 11.55 million newtons–the weight of 1180 tonnes of mass–pressing at a distance of a metre from the axis to keep that shaft turning at 1000 RPM.
That’s a high rate of power, but we were never told how many joules total energy was needed. It could be quite modest if you only need 1.21 gigawatts for a tiny fraction of a second.
True, but since the only spec we were given was the power, the system has to be able to withstand the power for as long as it needs it. If it only needs it for a microsecond, it only needs 1210 joules of energy in total… which is around the amount stored in a charged carbon-zinc AAA battery. But can you get all that energy out of the battery in a millionth of a second?
Now I want to watch that movie again. 
The OP does not require the power be electrical in nature. Setting one’s house on fire is probably the easiest way of doing this.
Um. Doesn’t 2 A * 750 kV = 1500 kW, not MW?
Oops. You may be right. I guess it’s 2 kA.
Not from a battery, but perfectly plausible for a capacitor. For example, one of these pulse discharge capacitor can be charged to 15 kV, and max peak discharge current is 25 kA. That’s 0.375 Gigawatt peak power. Four of these would supply more than 1.21 Gigawatt, and it’ll all fit in a suitcase. That’s with today’s technology, but I think these are polypropylene film capacitors, and I believe polypropylene was available in the late 50s.
Although this peak power can be sustained for much less than a microsecond. (It takes 120 ns to fully discharge this capacitor at peak discharge current.) And this ignores the inductance of the load. If the load has a high inductance, it takes much more energy to ramp up to peak power.
Except you’ll need some way of harnessing that energy and transferring it to the DeLorean. Filming a video of your burning house and posting it to YouTube probably won’t cut it.
But the OP didn’t specify only the Delorean of the film. The Delorean’s high-power equipment was apparently fed by electricity, so the bolt of lightning was appropriate. But what if our mysterious piece of equipment needs the 1.21 GW as mechanical energy? Flowing water? Gasoline? That’s what I was wondering about–how many different ways could the power be supplied.
Not really.
The wire would have to be short
and fat
…and in all probability you could only do it …once 
side note for Una - I once had to troubleshoot a 25KV distribution recloser that had failed. Came to find out that the engineer who ordered it accidentally specified a 12KV recloser coil. It worked real good - once.
A wire doesn’t have to be fat to carry large currents for short times. In the steady state, the resistance of the wire has to be low enough that its surface can export whatever heat that resistance generates. But in the very short time regime, the heat does not need to be exported, it only has to be absorbed by the copper without raising its temperature too far.
I hear it is pretty common for 24 AWG telephone wire to carry a lightning strike of, oh, perhaps 10,000 amperes without melting.
IIRC, after the lightning strike we saw some long thin flaming puddles that were the cables connecting the clock tower to the Delorean.