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See: https://www.smh.com.au/business/companies/huge-tesla-battery-in-south-australia-primed-for-big-upgrade-20191119-p53byo.htmlWhat's the environmental impact of that much lithium? You'd need 17 of those to match the power of Dinorwic, and 70 of them to match the energy.
Yes I can.In which case you'd have to apply the same logic to the voltage and current and say they're not changing either. My comment was in response to the OP who was talking about double voltage with half current, but no change in resistance.
It is 0.4 Ohms in each cell whether that cell is in series or parallel.
My point- I thought I had laboured it hard enough but, here we go again- is that the cell doesn't know what's happening outside. So each cell has a constant internal resistance.Yes, I know, but that’s just repeating what I’d already said here:
My original point to the OP was that you can't double the voltage in series without doubling the resistance, and you can't double the current in parallel without halving the resistance as he seemed to be arguing.Snip.JPG (181.46 kB . 1528x617 - viewed 1160 times)
Snow is very anechoic, too.If you drive close to a wall you will hear your own tyre noise reflected, and the lack of tyre noise on fresh snow is remarkable.Yes. Actually, I think a significant amount of the noise "from" the other car is noise generated by one's own car which is reflected off of the other car. I remember noticing this effect as a child when driving with the windows open: I could tell when we were passing by parked cars, even with my eyes closed.
No I didn't miss it.You start by arguing that there isn't any resistance,Did you miss this bit?if the load resistor is zero.
Conversely, if a large surge in demand happens- the cliche reason is the end of a popular sport event leading to everyone going and making the - the system has to compensate.There's also pumped storage. Dinorwic empties the turbines with compressed air whilst on standby, and spins the generators at synchronous speed so that when the sluice valves are opened they can go from zero to full power in 16 seconds. It's quite an impressive beastie if you go for the guided tour.
As Evan has pointed out, there are systems that will regulate the voltage and frequency, but that takes some time..
There's another set of automatic controls. Some large scale users of electricity have an agreement with the supplier where, they get their power at a reduced cost in exchange for agreeing to let the grid controllers disconnect them when there's a spike in demand.
An automatic system disconnects them from the grid if the frequency of the mains drops by some set amount.
These users are things like huge refrigerated warehouses. Cutting the power to the cooling plant for a few minutes won't make any real difference to the temperature, but it gives the power suppliers some leeway.
The clever bit is that the mains frequency is a very accurate indicator of the load on the grid, and it's automatically distribute to all users.
No it isn't. With a series loss only the efficiency is Rl/(Rl+Rs), which is an asymptote to 100% as Rl tends to infinity. With a source that has series and parallel losses (eg a transformer), the load for optimum efficiency is the geometric mean of Rs and Rp (provided that Rs<<Rp).You would never normally do that with a battery though, the efficiency's only 50%.Which is the best you can get...
This is self contradictory. You start by arguing that there isn't any resistance, and then include the internal resistance in your calculation, at which point there is resistance.Actually, I think you (sort of) can; if the load resistor is zero.I am not considering different resistors or anything like that.You've told us that you're drawing 10A from 2V, and 5A from 4V. You can't do that and keep the resistance the same as well. V=IRIf we have one Cu/Zn battery that has a voltage of 2V and current of 5A, it will produce power of 10W.A battery with an open circuit voltage of 2 V and short circuit current of 5A is plausible- it would have an internal resistance of 2/5 =0.4 ohms
And, if you short circuited it, it would dissipate 10W
All as heat in the battery- no electrical power would be delivered.
Now, if you put 2 in series you get an O/C voltage of 4V and an internal resistance of 0.8 ohms.
The current through the short circuit is still 5A
And the power "wasted" is 20 W of heat (and no electrical energy in the external circuit)
OK, Now connect them in parallel and, again, short circuit them
The total current flowing in the wire is 10A i.e. 5A through each cell
And the total power - still all as heat- is 20 W.
Again, it's clear that batteries are unable to see.
One more bit of additional info: You will get the maximum power to an external load when that load's resistance is equal to the battery's. (impedance matching)You would never normally do that with a battery though, the efficiency's only 50%.
I am not considering different resistors or anything like that.You've told us that you're drawing 10A from 2V, and 5A from 4V. You can't do that and keep the resistance the same as well. V=IR
In case of fully submerged resistive heater, almost all of the energy is transfered to heated media.The thread is about hobs, not kettles
Problem is that an induction cooker won't heat an aluminum block directly.That's why they're using stainless steel.
One watt-hour is 3,600 joules. A year is about 31,557,600 seconds long. A watt is 1 joule per second. So 3,600 joules divided by 31,557,600 seconds equals about 1.141 x 10-4 watts.Wow, that's a long winded way of going about it.
The figure for resistive elements refers to resistive elements, and the figure for induction hobs refers to induction hobs.Wikipedia are quoting the US DoE efficiency tests as an average of 72.2% for induction hobs, and 71.2% for resistive ones.I think the number refers to resistive heater where the heating element is located below the pot.
https://en.wikipedia.org/wiki/Induction_cooking