22
\$\begingroup\$

I am a student and have a very basic knowledge of volts, ampere, watts, charges, etc.

My question is fairly simple: Why can't we use 9 V DC batteries to run everything in the world?

  1. Because it's DC, and we need AC to run appliances.

Answer: Use an inverter to make it AC.

  1. Because they have 9 V to provide only.

Answer: We can use a step-up transformer to boost its voltage and amperes.

So my question and some counter questions are there. Please explain to me what I am missing here.

My goal with this question is to learn the science behind it and not to produce an efficient power supply. Also, in my question, 'a 9 V' battery means just one single battery, not more than one.

\$\endgroup\$
25
  • 55
    \$\begingroup\$ The same reason you can or can't use a tea cup to empty a swimming pool. \$\endgroup\$
    – DKNguyen
    Sep 13, 2022 at 14:14
  • 5
    \$\begingroup\$ For the same reason we don't run all ICE vehicles with 50cc engines. \$\endgroup\$
    – Puffafish
    Sep 13, 2022 at 14:38
  • 30
    \$\begingroup\$ This is an excellent question. Just because it asks about something so fundamental that most of cannot remember learning it, doesn't mean we shouldn't bother trying to write a good answer. \$\endgroup\$ Sep 14, 2022 at 2:13
  • 7
    \$\begingroup\$ A conspiracy of the power companies, that is. \$\endgroup\$
    – fraxinus
    Sep 14, 2022 at 11:49
  • 17
    \$\begingroup\$ @Christian - StackExchange is "created and run by experts and enthusiasts". It's FOR everyone. Even beginners. Especially beginners, really. If you're an expert, you're probably here to answer questions (from beginners) moreso than to ask them. \$\endgroup\$
    – JamieB
    Sep 14, 2022 at 14:13

14 Answers 14

37
\$\begingroup\$
  • Batteries are chemical energy stores. The size of the battery determines the amount of electrolyte that can be accomodated and that determines the amount of energy that can be stored.
  • The battery can provide power at a rate limited by the size of the electrodes and its internal series resistance. In general, the smaller the battery the higher its internal resistance. Thus small batteries are limited in the amount of instantaneous power they can produce. Try to draw too much and the voltage collapses.

We can use a step-up transformer to boost its voltage and amperes.

In general, when you use a step up transformer to boost the voltage then there is a decrease in current. This comes from the transformer's characteristic "power in = power out" or $$ P_{IN} = V_{IN}I_{IN} = V_{OUT}I_{OUT} = P_{OUT} $$ You can't magic the power out of nothing.

\$\endgroup\$
3
  • 6
    \$\begingroup\$ This answers the question as asked. Most other answers do so only partially. \$\endgroup\$
    – ATCSVOL
    Sep 14, 2022 at 0:19
  • \$\begingroup\$ Also, Pin x time = E = Pout x time. \$\endgroup\$ Sep 15, 2022 at 0:35
  • 1
    \$\begingroup\$ Thank you, that last bit is the answer I was looking for. Power remains constant. So if voltage increases, current must decrease to balance the equation. In conclusion, I cannot use transformer to power whole world :) \$\endgroup\$
    – Waleed
    Sep 16, 2022 at 16:48
53
\$\begingroup\$

Archimedes said, “Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.”

This is the nature of the thought exercise you’re asking. And the answer is, yes, given a suitable conversion system (our fulcrum and lever, so to speak), you could ‘run everything in the world’, at least briefly, with the familiar 9V (PP3) battery.

Without getting hung up on the practicalities, energy is energy, power is power, and a 9V battery certainly has some.

Now, consider that a 9V battery is a pretty small (and non-renewable) packet of chemical energy compared to, say, an off-grid battery bank, a generating plant, or for that matter, the entire world’s power grid.

How small? A typical alkaline 9V battery is good for about 400mAh, or about 3.6 Wh (0.4Ah * 9V) of electrochemical energy from that little guy.

Compare that to the total global energy production, which in 2021 was about 28,000 TWh (link). This works out to about 3.2 TW of power being used worldwide at any given moment (28,000 TWh/365d/24h.)

In round figures then, it would take about one trillion 3.6 Wh 9V batteries per hour to supply the global energy demand. Or, one 9V battery could supply one trillionth of that demand. Or it could supply it for one trillionth of an hour, that is, 3.6ns (as noted by @Neinstein).

You could do similar analyses for, say, a whole household or a even single appliance, and come away similarly disappointed with the 9V battery’s ability to power anything but the smallest of devices.

And that’s okay. The 9V battery isn’t designed for bigger stuff. Internally, 9V batteries are actually a stack of six tiny 1.5V cells connected in series. The internal resistance of that stack is pretty large, severely limiting the current that the battery can deliver.

Rather than powering your house, or the world, the 9V battery’s mission is to power small electronics like ‘transistor’ radios: that is, devices with a small current drain but in need of a higher voltage to operate. And for that it’s pretty darn good, if pricey.

The 9V battery: great for your smoke detector, not so much your toaster oven making the smoke.

You might do slightly better if you Mr. Fusion’d the battery, converting its entire 45g mass to energy. Then to ‘power the world’ you’d need to torch just three per hour: using e=mc^2, each battery yields 4.05 x10^15 joules (4.05 petta J, PJ) or 1.125 TWh.

(1 J = 1 watt-second; 1 watt-hour is 3600 watt-second or 3600 J.)

Fun fact: the largest US nuclear test, Castle Bravo (15 Mt), released about 62 PJ of energy, or roughly sixteen 9V batteries’ worth of mass into energy. Castle Bravo's energy output would power today’s world for about 3 hours. I suppose you'd have to upgrade to Tsar Bomba (50 Mt, 209 PJ) to get longer play time.

Realistically though, both science and economics limit the 9V battery use cases. A 9V battery is heinously expensive per unit of energy delivered, very resource-intensive to make, non-rechargeable in its common form, and limited in both energy content and power output.

Sadly, as things stand today most alkalines end up in landfills. Only about 4% are recycled at present, although there are efforts underway to improve this (link).

Li-ion and related chemistries come closer to meeting the need for scalable energy storage, which is why they're the go-to choice right now. But even these have problems when viewed as a life-cycle. Li-ion recycling is only now being looked at seriously (link).

So what else can a 9V battery do? How about rewiring your brain? Yes, it’s a thing. (Disclaimer: you can hurt yourself doing this. Not recommended as a DIY.)

\$\endgroup\$
8
  • 3
    \$\begingroup\$ Actually, converting a battery's mass entirely to energy would require annihilating it with another battery made of antimatter. To, still very much hypothetically, fuse all the lighter elements in the battery down to iron would still generate a huge amount of energy, but would convert only a small fraction of the mass to energy. \$\endgroup\$ Sep 14, 2022 at 8:02
  • 2
    \$\begingroup\$ A very good answer, however I'd like to add that while the energy inside one battery is certainly a limiting factor, so is the power which it can provide. There is a limit to how much amperage you can get ouf of the chemistry, and increasing the voltage outside of the battery doesn't increase the power limits. \$\endgroup\$
    – Mookuh
    Sep 14, 2022 at 9:07
  • 9
    \$\begingroup\$ Had to do some more pointless math. A battery would power the world for 3.6 nanoseconds. During this time, light propagates for about a meter (in vacuum); so the light from the streetlights you're powering wouldn't even reach the streets before your battery dies. \$\endgroup\$
    – Neinstein
    Sep 14, 2022 at 10:01
  • 1
    \$\begingroup\$ As far as getting the energy out? You could theoretically draw down the battery at its 1C rate and store it for conversion. Rinse and repeat, one trillion times. That’s the lever and fulcrum. \$\endgroup\$ Sep 14, 2022 at 15:13
  • 1
    \$\begingroup\$ The Q is a way to do some truly silly (but hopefully correct) dimensional analysis. Another use of 9V batteries is as weights in kinetic energy storage. Or for throwing at Flyers games. \$\endgroup\$ Sep 14, 2022 at 18:17
12
\$\begingroup\$

It's not just the voltage. You can step it up to 120 V, but it will deliver microamps. 9V batteries have a high internal resistance so they can't supply much current. You would also need a really large step-up transformer and an inverter as you can't power a transformer with just DC.

\$\endgroup\$
6
  • 2
    \$\begingroup\$ You assume that "9V battery" means a single 6LR61 battery. But there is no reason, in principle, why I could not power, say an electric hair dryer, from a huge bank of 6LR61 batteries. By my calculation (using data from data.energizer.com) a bank of 4000 of them should be sufficient to run a typical hand-held hair dryer for about ten hours. [P.S., Would not "step up" the voltage: I'd connect 30 or so in a series "string," and many strings in parallel.] \$\endgroup\$ Sep 13, 2022 at 14:50
  • \$\begingroup\$ @SolomonSlow, sure but there's what would realistically would be built in volume and then there's just spinning one's imagination, which doesn't really produce useful results. This answer addresses the former. Incidentally, the OP pretty clearly meant PP3 batteries rather than anything else. \$\endgroup\$
    – TonyM
    Sep 13, 2022 at 18:29
  • 1
    \$\begingroup\$ @TonyM, Looks to me as if, "PP3" was an EverReady product number. "NEDA 1604," "IEC 6F22", "MN1604," and "6LR61" all are different names for batteries that have the same form factor and same nominal voltage as the original EverReady model. I don't know which of those names designate a particular chemistry and/or internal configuration, and which (if any) only refer to the form factor and voltage. en.wikipedia.org/wiki/Nine-volt_battery \$\endgroup\$ Sep 13, 2022 at 19:29
  • \$\begingroup\$ @SolomonSlow : Note OP actually says "Also, in my question, 'a 9 V' battery means just one single battery, not more than one." \$\endgroup\$
    – MPW
    Sep 14, 2022 at 17:34
  • 1
    \$\begingroup\$ @MPW, that sentence was added to the question some time after I wrote my first comment, above. \$\endgroup\$ Sep 14, 2022 at 17:50
8
\$\begingroup\$

Because not everything can be powered by 9V batteries.

You can't get enough power out of it.

You can boost voltage, but then you lose current, because it's the power you need.

\$\endgroup\$
8
\$\begingroup\$

Well, a single PP3 9V battery can't supply much power and contains very little energy. It's also very expensive as a source of energy. More than other primary batteries.

If the power from the wall socket costs 10 cents a kWh (that's not cheap) consider that an alkaline 9V battery contains about 5Wh. So if that battery costs $1 that's 2,000 times as expensive. So instead of it costing $0.02 to heat your mac'n cheese in a microwave it will be $40.00. In fact you'd never likely get there because the food would be cooling faster than you could swap 40 batteries out and in as they were depleted.

To make just one pound (454gm) of aluminum takes 7kWh, so that's $1400 worth of $1 batteries.

Basically, almost nothing we enjoy with our modern high standard of living would be economically possible using 9V batteries as a power source.

\$\endgroup\$
5
\$\begingroup\$

There are two key limiting factors here. One is the power capacity of a 9V battery. Each battery only has so much energy stored in it, canonically measured in Joules (1W of power for 1 second) but often written in Wh, watt-hours, because that tends to make the numbers mean something for humans. A typical 9V battery has about 5Wh of energy. That means it can power a 1 watt device for 5 hours, a 5 watt device for 1 hour, or any other combination that multiplies to 5. Starting a car consumes about 2.4 Wh, so starting a car twice would deplete the 9V battery. You'd have to get a new one for basically every commute you make!

The other issue is more subtle. A 9V battery doesn't always output exactly 9V. The chemical reaction which pushes electrons from the cathode to the anode can only progress so fast. If you deplete the electrons faster than it can replenish them, the electrostatic forces which form the basis of voltage get weaker. This gets weaker and weaker until the battery and the load reach some "agreement", a voltage level where the battery can just barely replenish the charges as fast as the load consumes them. I call this an agreement, but it's nothing fancy -- its just a stable point in the equations.

In most practical batteries, this is modeled as an "internal resistance." For most batteries, this effect is linear enough that we can model it as if there was an extra resistor in series with a perfect 9V source and the load. In this model, it is easy to see that a power hungry load will have a very low resistance, and the current will soon be limited by this "internal resistance." This can be remedied by putting more batteries in parallel, but your question explicitly focuses on a single battery.

In the end, if I want to run my 90 amp arc welder, it will require about 3000W. 3000W at 9V is about 333 amps. The poor chemical reagents inside the battery will never be able to keep up with that.

If you want to see what happens when the generator doesn't keep up with the load, look up what happens during a brown-out. That's the power-grid wide version of this scenario.

\$\endgroup\$
4
\$\begingroup\$

Since you've restricted it to a single 9v battery, the answer is because it's too low Power (physics term) for many applications. As an analogy, a tiny motor could, given enough time, do as much work as a giant engine, but the giant engine can do the work quicker, so we say it has more Power. This is going to be a problem:

A real 9v battery can likely only output about a max of 350mA at 9v which is only about 3 watts. Consider a helicopter trying to take off. It can't just spin the blades 1000 rotations over the course of an hour and expect to take off. It has to achieve a rotation rate. Or consider toasting bread.

If you're allowed to use capacitors, you can store that energy and expend it in a much shorter amount of time to achieve the power needed, but at that point, you're just charging a different battery, which surely violates the principle of the question.

If your hypothetical 9v battery isn't restricted to a power output, then this limit wouldn't apply, but that's not a 9v battery, it's a magical device, and you could in fact power everything with this.

\$\endgroup\$
3
\$\begingroup\$

Why can't we use 9V DC batteries to run everything in the world?

The simplest answer is that to make a battery and instil it with energy is a very inefficient way of delivering power to a load. Then there is the disposal of the mega-multitude of batteries into landfills or maybe some reclamation plant. It's just not a cost effective way of doing things and, neither is it a green way of doing things.

\$\endgroup\$
7
  • \$\begingroup\$ No sir, my question is to understand the science behind it. Consider that I am the richest man and I can just buy all batteries in the world. But why can't I run everything off just 9V single battery? \$\endgroup\$
    – Waleed
    Sep 13, 2022 at 13:52
  • \$\begingroup\$ Your comment comes down to "why can't I waste a lot of money". There is no science involved; it's purely economics and ecology. \$\endgroup\$
    – Andy aka
    Sep 13, 2022 at 13:56
  • 4
    \$\begingroup\$ @Waleed, Re, "Why can't I run everything off just 9V single battery?" Because a single 9V battery cannot supply even one Watt of power. The richest man in the world could, in fact, power everything in his life with 9V batteries, but he would need to connect hundreds of them together to power his kitchen refrigerator, thousands to power his hair dryer, tens of thousands for a clothes dryer, and hundreds of thousands to power an electric car. \$\endgroup\$ Sep 13, 2022 at 15:00
  • 4
    \$\begingroup\$ @Andyaka, Yes, I considered that "9V battery" could be interpreted broadly, but I decided to pretend that OP meant what most laymen mean when they walk into a shop looking for a "9V battery." At least one other person answering here appears to have made the same assumption. \$\endgroup\$ Sep 13, 2022 at 15:15
  • 3
    \$\begingroup\$ @SolomonSlow I hope his 9V battery powered factories can make the batteries faster than they are depleted. \$\endgroup\$
    – DKNguyen
    Sep 13, 2022 at 18:17
2
\$\begingroup\$

The actual electrode material that provides energy is only a part of the cost, weight and bulk of a battery. There are many other components, for example sealed packaging to keep the electrolytes in, contacts, etc.

In a lower capacity battery format, you save on electrode material, but not much on everything else. For example an alkaline AA contains about 2500mAh and an AAA about 1000 mAh, but the price of both isn't that different. The amount of materials in the case, contacts, etc, isn't in proportion with the capacity.

The 9V battery contains 6 tiny cells, so it has even more packaging materials that don't participate in the battery capacity, and very little actual battery inside. Capacity is low, and since the cells are small, they can't deliver much current.

The opposite philosophy would be a bigger cell like AA or 18650 which contains a lot more battery material and less "skin", which translates to a lower cost per stored unit of energy, and much higher power and current capability.

So the point of the 9V battery used to be convenience and small size, it was well suited to low current analog circuits of past decades. Now, more modern analog circuits can run on much lower voltages, and DC-DC converters have become ubiquitous and cheap, which means more devices tend to use AAs as a cheaper option. Two alkaline AAs offer about 3 Wh while a 9V battery offers about 4 Wh, so stored energy is similar, but two AAs are much cheaper and more convenient.

\$\endgroup\$
2
  • \$\begingroup\$ You speak of the sealed packaging that keeps electrolytes inside the battery. Obviously, you are not talking about Duracell AA and AAA cells, which are guaranteed.... to leak. I do not know if this problem extends to their 9 volt batteries. I have had many flashlights, radios, etc. ruined by corrosive leakage before I switched to Energizer cells. \$\endgroup\$ Sep 15, 2022 at 19:50
  • \$\begingroup\$ Most alkalines leak when completely drained. \$\endgroup\$
    – bobflux
    Sep 15, 2022 at 21:25
2
\$\begingroup\$

You could, theoretically, but so much energy would be lost to heat, and AC/DC conversion, that it would be horribly cost-ineffective, and wasteful. The amount of batteries you would need to get the right amount of amps is ridiculous (each one does about 500milliaps for an hour until they die.). You would also need to quickly replace them. Even a simple phone charger running off of 9V batteries would need (about) three batteries to get to 14.5W. (iPhones can only do 15W). Scale that up to larger, more power-hungry devices, and it gets out of hand very quickly.

\$\endgroup\$
1
\$\begingroup\$

We don't really have a device that keeps 9 volts over its terminals no matter what.

That's a useful mathematical model because it approximately describes the behavior of a real-world device as long as certain constraints apply. Your proposed usage would severely break the constraints, and the model would no longer apply. The voltage of a 9-volt battery quickly drops if you try taking too much power from it.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Related: a piece of wire doesn't really "electrically connect" whatever is at its terminals. Or rather it does, approximately and as long as certain constraints apply. \$\endgroup\$ Sep 14, 2022 at 13:18
  • 1
    \$\begingroup\$ A more accurate model than the constant voltage source is a constant voltage source in series with a resistor. And a more accurate model than that is a variable voltage source in series with a variable resistor where the voltage decreases and the resistance increases according to the charge remaining in the battery. \$\endgroup\$ Sep 15, 2022 at 19:59
1
\$\begingroup\$

Yes it is theoretically possible to power everything using 9V batteries but you need very large converters and transformers and super many batteries to get enough power.

To use a normal 1000W microwave you need an at least 1000W DC-DC converter and an at least 1000W inverter and like 1000 batteries to just start a single microwave.

That would use so much resources and cost a lot of money.

It would also create a lot of heat due to power loss in the circuits.

So in theory it is possible but in practice it’s just a waste of resources

\$\endgroup\$
1
\$\begingroup\$

Where are you going to get your 9v batteries from? Unfortunately, making 9v batteries requires energy; and less energy is recouped from a 9v battery than is put into making it (actually, much less is recouped, because 9v batteries are optimised for powering small light loads whilst being cheap to make, rather than being efficient energy stores, but this only speeds the process up).

At this point it's a simple matter of diminishing returns. Even if you could neglect conversion losses, you could never power the world on 9v batteries for very long, since you would run out of energy to make 9v batteries with. (Cheating with the question, such as defining 'everything' to mean 'all essentially electrical equipment' is fun, but it won't actually change this by much, because the process is so very inefficient.)

However this does raise an interesting philosophical question: in that case, how come we can get any net energy out of any system? And indeed, in the long run, we can't: the second law of thermodynamics is the bane of all those glorious self-powering motor-generators 'the power companies don't what you to hear about', and also of everything else. In the long run energy is lost from any given system.* Fortunately for us, there are lots of good energy sources which don't require horrible amounts of energy to get at (we can recoup energy put in in the past to make them, albeit we never get 100% out). Unfortunately these energy sources are mostly rather polluting, and finite. ('Renewable' energy isn't infinitely renewable either, but we don't need to plan for infinity, only for a timescale much shorter than the heat death of the sun.)

*Energy is not lost, it's only lost from the system: in total mass-energy is neither created nor destroyed. It's usually lost as waste heat, although it can be lost as any kind of radiation, or as kinetic energy in e.g. vibration.

\$\endgroup\$
1
\$\begingroup\$

You could make the following analogy:

You can compare your circuit to a water pipes system.

You can view the voltage as the water pressure in the pipes (high pressure is the equivalent of high voltage).

You can view the current as the amount of water which goes through the pipes (high current is the equivalent of a big amount of water).

Since we are talking about a battery, this means we are talking about a voltage source. A voltage source will do everything it can to maintain their voltage (in your case 9V). If we go back to our analogy this mean you would have a water pipes system which always try to maintain the water pressure but with a limited water supply.

Assume the water pipes system equivalent to your 9V battery is able to spray 1 garden for a day (can be viewed as your load). What happens if you connect your water pipes system to additional identical gardens (splitting you main pipe in two, can be viewed as a a parallel circuit)?

Since your system tries to maintain the pressure (the voltage) then you need twice the amount of water. But your system only had enough water to fill the needs of 1 garden for the day. So now after 0.5 day your water tank (current) will be depleted.

In theory you could add as many garden as you want but for each garden you add your water supply gets depleted faster

(Assuming batteries don't have a current depletion limitation, which they do, for example 9V is 450mAH, you must also consider the impact of the internal resistance of the battery.)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.