What is a battery?

[4X4]

Batteries are the heart of any vehicle's electrical system. To understand why use of battery enhancement technology aids overall vehicle performance, it is important to understand the 3 functions of a battery
1. Supply power to the starter and ignition system so the engine can be started.
2. Supply extra power when the vehicle's load requirements exceed supply from the charging system.
3. Act as a voltage stabilizer in the electrical system, reducing temporary high voltages, which occur in the vehicle electrical system. These high voltages would damage solid-state components in the electrical system if it were not for the protection provided by the battery.

Battery plate sulfation occurs and increases every time your battery is used. It is part of the chemical reaction, which takes place in the battery. When a battery is sulfated, its voltage is depressed. The battery no longer meets the demands of the vehicle electrical system, and from an electrical perspective it "disappears". Electronic and electrical components then receive current directly from the alternator, and are subject to over-voltages. This results in premature failure of electronic components.

A lead acid battery is an electrochemical device, which stores chemical energy. This chemical energy is converted to electrical energy when the battery is connected to an external load such as a vehicle starter. The chemical energy is created by the chemical action between the materials which form the positive and negative plates of the battery, and the electrolyte:

Lead Dioxide (PbO2) Positive Plate
Sponge Lead (Pb) Negative Plate
sulfuric Acid (H2SO4) Electrolyte.

A battery relies upon clean plates and strong electrolyte to receive charging current and offer discharge current. When the battery is connected to a load, the sulfate (SO4) in the electrolyte combines with the active materials of the plates to form lead sulfate (PbSO4) and release electrical energy. Electrons flow from the negative terminal to the load and back to the positive terminal of the battery.

The Battery specific gravity (ie. the unit of measurement of the sulfuric acid content of the electrolyte) of a fully charged 12-volt battery is 1.300 at 26.7 deg C. This means that the sulfuric acid of a fully charged battery is 1.3 times heavier than pure water. As a battery becomes discharged, the strength of the specific gravity decreases because sulfur is leaving the electrolyte as it forms lead sulfate which adheres to the battery plates

State of Charge Specific Gravity Voltage (12V battery)
State of Charge Specific Gravity Voltage (12V battery)
100% 1.300 12.84
75% 1.250 12.50
50% 1.200 12.20
25% 1.155 11.90
Discharged 1.120 11.00

Thus by the time the battery is discharged, the acid becomes dilute as the sulfur has adhered to the plates of the battery as lead sulfate crystals. When a discharged battery is recharged, the chemical processes within the battery operate in reverse. The majority of the sulfate leaves the plates of the battery and returns to the electrolyte. However, a residue of sulfate remains on the plates of the battery. The quantity of this residue increases with each charge/discharge cycle of the battery. Over time, the battery plates become coated with an insulating layer of sulfate and the electrolyte is weakened because of the loss of Lead sulfur molecules from the solution. Both these factors serve to inhibit the electron transfers and thus the energy producing function of the battery.

Over time the sulfate deposits on the plates become hard and crystalline. When in this condition, plates will not accept a charge under normal conditions, and the accumulation of lead sulfate may cause short circuits during recharging or other mechanical damage to the battery. Often, hairline cracks appear in the plates causing open circuit conditions.

When a lead acid battery discharges or remains inactive, lead sulfate forms on the battery plates. Over a short period of time, sulfate gradually accumulates and crystallizes clogging the plates to the point where the battery will not accept or hold a charge. This process, known as sulfation happens to all lead acid batteries in all application. It is the leading cause of battery failure. Megapulse technology reverses sulfate accumulation in all lead batteries and it prevents sulfation from ever developing in new batteries. By pulsing a carefully controlled DC current into the battery, it re-energizes crystallized sulfates deposited on the plates and returns them to the electrolyte as active sulfur molecules. With the plates kept clean, batteries will provide more power, faster recharge and longer battery life.

Batteries commonly fail because of sulfation. sulfation occurs when a battery is discharged. The deeper the discharge, the more serious the sulfation. A battery relies on clean plates and strong electrolyte to both receive charging current and offer discharge current. A sulfated battery can do neither. sulfation also occurs when batteries are in an undercharged state. Battery theory states that cell voltage should read 2.45 volts per cell (i.e. 14.7 volts in the case of a 12 volt battery) from time to time to allow the negative plate to "form". If this does not occur, the negative plate remains mushy and subject to erosion from motion, vibration, etc. In automotive systems, alternators seldom exceed 14.2 volts. Battery theory states that 12-volt batteries must receive a minimum of 14.1 volts to maintain a charged state.

 

[millsy]

Great stuff 4x4. Thanks, learnt a lot!

And a bit of basic theory; A battery is a group of cells joined together.

Most of us usually use the word battery when we should be saying cell. e.g. one D cell, or one C cell, is correct, rather than a battery.

A car battery has six cells joined in series to increase the voltage to 6 x 2 = 12 volts ( approximately - actually 6 x 2.15V = 12.9V ).

The chemicals in D cells are different to those in car batteries, so each 'dry' D cell only produces 1.5 V.

If you connect 2 D cells in series to make a 3V battery, and then connect the battery to a small bulb that needs only 1.5 V, then the 3V battery will produce a large current and the bulb will be very bright. Great, but the cells will go flat quickly due to the large current flowing through the cells.

The other choice is to connect the cells in parallel ( side by side, with positive connected to positive and negative to negaive ). Now the voltage remains at 1.5 V. But the advantage will be that when the 1.5V bulb is connected, although it will only be 'normal' brightness the circuit will last twice as long as the series connection.

 

[grit]

If you connect 2 D cells in series to make a 3V battery, and then connect the battery to a small bulb that needs only 1.5 V, then the 3V battery will produce a large current and the bulb will be very bright. Great, but the cells will go flat quickly due to the large current flowing through the cells.

.

This is not actually correct... the globe is rated in watts which equates to volts x amps... by doubling the voltage you are effectively halving the amperage as the bulb will still only consume its rated wattage. Since E/I = R and R is a constant.

 

[Grumpy]

Well got me EI and R, I'm totally confused

 

[millsy]

Hi Grit, and Grumpy. Bulb's will use more or less power depending on the voltage. I will try to explain as best I can! Hopefully clearer than mud!
A light bulb has a fairly constant resistance. ( But not perfectly Ohmic - i.e. its resistance actually increases a little as the current increases, and hence the brightness of the filament also increases. The more current flowing the more heat and this increases the resistance. A perfectly Ohmic conductor is able to maintain a constant resistance whatever the current. ) But all this is not important for this particular discussion!

The main point is that when the voltage across any fixed resistance, like a 'perfect' light bulb, is increased, then the current will also increase in proportion to the increase in the voltage. This is Ohm's Law.

Ohm's Law can be used in three different ways depending on what quantity you are calculating; I = V/R ( Current = Voltage/Resistance ), or R = V/I , or V = I x R
So since I = V / R, if the V increases and R remains the same, then I will increase.
If you increase the voltage across a bulb the bulb gets brighter. The brightness of the bulb gives an indication of the size of the current - the more current, the brighter, and as is the case with a flat battery, less voltage ( less electrical pressure pushing the electrons through the wire ) means less current flowing, means dimmer head lights. I think we have all seen this either in cars, or in torches with flat batteries.

The power rating of a bulb is just the ideal working conditions for that bulb. The manufacturer is telling you that for this bulb to work as it was designed to operate, then you should use this particular voltage as a power supply, and if you do, then the power consumed from your power supply will be so many Watts. If you decide to use more voltage, then the current will increase ( I = V/R ) and so you will be dragging more energy per second ( power ) out of your power supply ( P = V x I , Power = Voltage x Current ). If the power supply happens to be a battery, then that battery will run flat more quickly.

The kids in my year 9 science class do this 'experiment' every year. To start the Electric Circuits topic I give them each a few cells, bulbs and wires, and some simple circuit diagrams to construct. Of course the first thing they discover is that the more cells on one bulb the brighter the bulb. It becomes a 'game' - who can get their bulb shining the brightest. Before you know it bulbs start blowing all over the place and I have to tell them that these bulbs are rated at 2.5V .3A and that they are allowed to go up to four cells for their thrills ( 4 x 1.5 =6V ) but no more!

The area of electric power consumption where some people, especially newspaper journalists, get there words all mixed up, is when they try to write about watts, kilowatts and kilowatt hours!

 

[grit]

Hi Grit, and Grumpy. Bulb's will use more or less power depending on the voltage. I will try to explain as best I can! Hopefully clearer than mud!
A light bulb has a fairly constant resistance. ( But not perfectly Ohmic - i.e. its resistance actually increases a little as the current increases, and hence the brightness of the filament also increases. The more current flowing the more heat and this increases the resistance. A perfectly Ohmic conductor is able to maintain a constant resistance whatever the current. ) But all this is not important for this particular discussion!

The main point is that when the voltage across any fixed resistance, like a 'perfect' light bulb, is increased, then the current will also increase in proportion to the increase in the voltage. This is Ohm's Law.

Ohm's Law can be used in three different ways depending on what quantity you are calculating; I = V/R ( Current = Voltage/Resistance ), or R = V/I , or V = I x R
So since I = V / R, if the V increases and R remains the same, then I will increase.
If you increase the voltage across a bulb the bulb gets brighter. The brightness of the bulb gives an indication of the size of the current - the more current, the brighter, and as is the case with a flat battery, less voltage ( less electrical pressure pushing the electrons through the wire ) means less current flowing, means dimmer head lights. I think we have all seen this either in cars, or in torches with flat batteries.

The power rating of a bulb is just the ideal working conditions for that bulb. The manufacturer is telling you that for this bulb to work as it was designed to operate, then you should use this particular voltage as a power supply, and if you do, then the power consumed from your power supply will be so many Watts. If you decide to use more voltage, then the current will increase ( I = V/R ) and so you will be dragging more energy per second ( power ) out of your power supply ( P = V x I , Power = Voltage x Current ). If the power supply happens to be a battery, then that battery will run flat more quickly.

The kids in my year 9 science class do this 'experiment' every year. To start the Electric Circuits topic I give them each a few cells, bulbs and wires, and some simple circuit diagrams to construct. Of course the first thing they discover is that the more cells on one bulb the brighter the bulb. It becomes a 'game' - who can get their bulb shining the brightest. Before you know it bulbs start blowing all over the place and I have to tell them that these bulbs are rated at 2.5V .3A and that they are allowed to go up to four cells for their thrills ( 4 x 1.5 =6V ) but no more!

The area of electric power consumption where some people, especially newspaper journalists, get there words all mixed up, is when they try to write about watts, kilowatts and kilowatt hours!

Sorry. I don't see Ohm's Law supporting this statement.

Auto manufacturers have been seriously planning increasing the voltage output of car batteries up to 4x - to considerably reduce the thickness (and weight) of wiring to run the same wattage components & in particular lighting. This is only possible due to the reduced current needed to supply the required wattage to the bulb.

Hope this analogy helps clarify (in more practical terms) my inability to simply agree with your statement. Might I also add that a light bulb (incandescent) makes a great in-line fuse.

 

[millsy]

Yeah, sorry Grit. I should not have got off the track talking about Ohmic conductors.

But to try to try to clarify; Some conductors have a constant resistance whatever current or voltage is being used. But the greater the voltage ( electrical pressure ) the greater will be the current, and in direct proportion - double the voltage means double the current. This is by Ohm's Law - I = V/R so if R stays constant then if V doubles I must double according to the formula.

Moving on to the power issue. In the example above, if the power usage is calculated after the voltage and current are both doubled then using the formula for power P = V x I, since V and I are both doubled then the power has in fact been increased by 2 x 2 = four times.

You said that the automotive circuits that manufacturers are planning will use maybe 4x the voltage ( 48V ) to save energy overall. Without actually finding the particular info on a quick google search that matches your descriptions, I can see that they are proabably looking at this in two ways; Firstly, the energy wasted in wires that carry large currents. produce a lot of wasted heat.

So if components, such as headlights are re-designed to have a (much) higher resistance, then Ohm's Law (I = V/R) ....gives a smaller current if R is larger.

This smaller current flowing through the harness wires means less energy wasted and so less power being dragged from the car battery. Also cooler wiring temperatures means insulation around the harness does not become brittle and breakdown.

Secondly, the power being consumed by the actual headlight will be a lot less;

P = V x I

but V = I x R, so P = I x R x I i.e P = I x I x R

So if V is increased 4 times, as you said, and R is increased by say 32 times,
then by Ohm's Law ( I = V/R ), I will be decreased to 4/32 = 1/8 the initial value.

Then by the formula P = I x I x R the power used will change by 1/8 x 1/8 x 32 = 1/2

i.e. only half the power being used.

Let's try increasing R by 64 times!

The new current would be I = V/R 4/64 = 1/16 of initial current.

And the new power usage would be P = I x I x R = 1/16 x 1/16 x 32 = 1/8. Wow!!

Only problem, how to increase the resistance of any particular component by 64 times and still have it operating as required. A complete re-design needed.

I tried looking up the resistance of the new HID ( discharge ) Xenon lights, but could not find it as yet. I think they must be fairly high compared to the old fashioned filament bulbs, and so this is why they have such a low power rating ( about 35 watts from what I could find on the Net).

Hope you got through all those formulae Grit! Cheers.

 

[grit]

Yeah, sorry Grit. I should not have got off the track talking about Ohmic conductors.

But to try to try to clarify; Some conductors have a constant resistance whatever current or voltage is being used. But the greater the voltage ( electrical pressure ) the greater will be the current, and in direct proportion - double the voltage means double the current. This is by Ohm's Law - I = V/R so if R stays constant then if V doubles I must double according to the formula.

Moving on to the power issue. In the example above, if the power usage is calculated after the voltage and current are both doubled then using the formula for power P = V x I, since V and I are both doubled then the power has in fact been increased by 2 x 2 = four times.

You said that the automotive circuits that manufacturers are planning will use maybe 4x the voltage ( 48V ) to save energy overall. Without actually finding the particular info on a quick google search that matches your descriptions, I can see that they are proabably looking at this in two ways; Firstly, the energy wasted in wires that carry large currents. produce a lot of wasted heat.

So if components, such as headlights are re-designed to have a (much) higher resistance, then Ohm's Law (I = V/R) ....gives a smaller current if R is larger.

This smaller current flowing through the harness wires means less energy wasted and so less power being dragged from the car battery. Also cooler wiring temperatures means insulation around the harness does not become brittle and breakdown.

Secondly, the power being consumed by the actual headlight will be a lot less;

P = V x I

but V = I x R, so P = I x R x I i.e P = I x I x R

So if V is increased 4 times, as you said, and R is increased by say 32 times,
then by Ohm's Law ( I = V/R ), I will be decreased to 4/32 = 1/8 the initial value.

Then by the formula P = I x I x R the power used will change by 1/8 x 1/8 x 32 = 1/2

i.e. only half the power being used.

Let's try increasing R by 64 times!

The new current would be I = V/R 4/64 = 1/16 of initial current.

And the new power usage would be P = I x I x R = 1/16 x 1/16 x 32 = 1/8. Wow!!

Only problem, how to increase the resistance of any particular component by 64 times and still have it operating as required. A complete re-design needed.

I tried looking up the resistance of the new HID ( discharge ) Xenon lights, but could not find it as yet. I think they must be fairly high compared to the old fashioned filament bulbs, and so this is why they have such a low power rating ( about 35 watts from what I could find on the Net).

Hope you got through all those formulae Grit! Cheers.

I'm actually enjoying flexing the grey matter that resides under my grey matted hair.

Its good to see that we now both agree on this previous inconsistency

The practicality of increasing the circuit voltage sometime in the future will not see a change in the wattage of the headlights used (although more efficient headlamps are available, for the purpose of this discussion, this should not be considered a factor). This provides us with 2 constants. E.M.F. (expressed in voltage) and Load (in this case expressed simply as Wattage). I believe that we now both agree that the increased Voltage (48 volts) when applied to a constant Load (lets say 65 watts) will indeed reduce the current (in this case 1.35 Amps). With a lower voltage of 12 Volts applied to this same load (65 watts) the amperage is 5.4 Amps. You can see that the Amperage has effectively been reduced to 1/4 of that currently needed to produce the same light. I have expressed this in practical terms so that we don't become consumed by theoretical calculations and so that it is expressed in a way that may be more readily understood in common practice. The result would be smaller & lighter batteries (new generation Li-Ion will further improve size, weight and output), wiring will be lighter & more likely used in a 'bus-bar' style configuration. Headlamps will become more efficient although there may be a rethink on Xenon discharge lamps (HID) as they have been found to effect the driver's peripheral vision.

You have made many valid points and your calculations and formulae are sound. Had I considered for one second that you would have put up such a convincing argument, I would have attempted to add a more convincing rebuttal to my initial comment. When adding a second battery in series to the circuit the available power is effectively doubled when the current remains constant. This produces the brighter glow. If the current increased in proportion with the E.M.F the wattage would effectively be 4 times that of one battery. A bulb will glow brighter when additional voltage is applied. A bulb makes a great fuse as it is not tolerant of additional current.

Best workout my brain has had all week!

Thanks again.

 

[millsy]

Hello again Grit! Yeah, we are having a real tussle here. Likewise, am enjoying it!

Now for round three! I still think we are a fair way apart in many fundamental aspects - particularly your idea that a component such as a headlight has a fixed power consumption regardless of the applied voltage.

There are many phrases in the first paragraph that I would disagree with, but I think that I will start with a phrase in the last paragraph - " . . . the available power is effectively doubled when the current remains constant.". This exemplifies. I believe a misunderstanding about the nature of an electrical conductor that has a certain FIXED RESISTANCE to the flow of CURRENT. Also a misunderstanding of the concepts of VOLTAGE ( and how it causes the current through a resistor based on the size of the resistance ( Ohm’s Law ) , and the concept of POWER and how it can change depending on values of voltage and current changing.
Add to this list the use of the units wattage when talking about load ( Wattage, or Watts measures power, Ohms measures resistive load ) and the use of Electromotive Force, EMF, instead of Voltage. These are two different quantities, but both related to the function of a battery or any other electrical power source.

Here I go again - rambling on off track! So, " . . . the available power is effectively doubled when the current remains constant.".

A battery is a source of stored energy. ( Stored energy = potential energy, in this case electrical potential energy.

POWER is the rate at which this stored energy leaves the battery, if focusing on the battery and how quickly it loses its store of energy, OR, the rate at which the LOAD converts this electrical energy to heat and light energy. The two quantities are different because a lot of the energy being put out by the battery is wasted in the wires connecting the battery to the bulb. So you could also talk about the power being wasted in the connecting wires.

If we assume a near perfect wiring loom ( i.e. no resistance to electric current at all - very thick wiring made of pure gold! ) then, and I make this point as strongly as possible, the POWER being used by the load IS NOT A FIXED VALUE BUT CHANGES AS THE SUPPLY VOLTAGE CHANGES, AND SO AS THE CURRENT CHANGES (this current able to be calculated using Ohm's Law using the applied voltage and the value of the FIXED resistance).

If the voltage is doubled, that does not just mean that the “available power is effectively doubled”. The power being used depends also on the current being drawn from the battery. ( P = V x I )
In the case of a simple incandescent bulb, the resistance is fairly fixed. So by Ohm’s Law if the voltage is doubled, then the current will also double. By using the formula for power we see that the power output from the battery, and the power consumption by the bulb increases FOUR TIMES. ( Resistance is constant, voltage doubles, so current doubles and since P = V x I, then power consumption is increased by a factor of 4. )

In your various explanations you seem to think that the power consumed by a ‘passive’ component like a light bulb always remains constant. This is not so. The only way the ‘wattage’, as you say, could remain constant is if the component mysteriously changes its resistance (upwards). As I have said, a light bulb is a passive, FIXED RESISTANCE. It cannot decide to increase its own resistance to the flow of electric current just because a larger battery provides more electrical pressure. This would be like a car deciding to let air out of its own tires every time the driver stepped on the accelerator, or maybe applying the brakes without the driver doing so!

Finally, the resistance of a simple resistive load is measured in OHMS, not Watts, or Wattage. Some loads are more complicated, in particular the components used in circuits that are called reactive circuits, involving capacitors and coils (inductors). These loads are called reactive loads and the formulae to calculate these are more complicated. These are the circuits used in radios to tune oscillating currents to be fed into, or out of, antennae. I must admit I have not used them for over 30 years and have largely lost touch with them!

As far as E.M.F. is concerned , my understanding is that it is the electrical voltage that the chemical reactions provide before this voltage is reduced by any build up of chemicals produced by the ongoing (discharge) reaction. These chemical products provide what is called an internal resistance in the battery, so that the actual voltage provided at the battery terminals, to be connected to the outside circuitry, is less than the E.M.F. .

A fully charged battery has few of these ‘waste product’ chemicals fouling up the cells, so has less internal resistance, and this means less electrical energy is used just getting the current out of the battery – i.e. the terminal voltage is near its maximum possible value. As the battery is used, more reactant chemicals are being used up inside to provide the current, more residue (product) chemicals build up around the electrodes (internally), and although the E.M.F. remains the same ( because this is the voltage produced right inside the battery’s cells) by the time this current flows through this increased internal chemical conductor, this ‘internal resistance’, the available voltage at the battery terminals has been reduced.

Phew! Are we still playing?

 

[grit]

Wouldn't miss it for the world

Again I see this discussion rushing out on an almost disconnected tangent. As for finding a direct association between the known values and the effects of increasing the voltage over a simple load consisting of one bulb we have; - The wattage of the known load (the bulb) and the known voltage applied. Unknown factors are resistance and current. We can however use the known values to establish the current required by the load this being P=E x I (I only express E as E.M.F to remain true to the original expression of Ohm's Law.

As you continue to hold fast to referring to the bulb in terms of resistance load, I offer my own assertions of Ohm's Law to your following statement...

"In the case of a simple incandescent bulb, the resistance is fairly fixed. So by Ohm’s Law if the voltage is doubled, then the current will also double."

Does this statement assert that you believe that a bulb now glowing far brighter (and hotter) would be of exactly the same resistance?

& why would the same bulb mysteriously suddenly require twice the current?

I love the fact that we are effectively arguing the most basic laws of electricity from such an enormous point of difference.

To remain focused on the most basic difference in our points of view I will not wander into the logistics of the internal workings of the voltaic cell.

I find myself at a point now, where I believe we might never see eye to eye on this particular point, yet want you to know that I have a new found respect for both your ability to convincingly express your point of view and your knowledge of the theory that you have expressed so soundly throughout this thread.

Cheers.

 

[millsy]

Well I will either convince you about Ohm's Law and fixed resistors, or you will convince me otherwise. Question is, how long will it take! Its a classic arm wrestle!

Quote; "Does this statement assert that you believe that a bulb now glowing far brighter (and therefore hotter) would not be of higher resistance?"

No, I do not believe this, but the increase in resistance due to the heating of the filamant is not as great as the increase in the voltage and current. I guess I will need to look up some data on the actual increase in resistivity, but that would require knowing what type of metal is being used for the element. A typical material could be looked up also I suppose.
I remind you of my simplifying statement from a few posts ago;
" A light bulb has a fairly constant resistance. ( But not perfectly Ohmic - i.e. its resistance actually increases a little as the current increases, and hence the brightness of the filament also increases. The more current flowing the more heat and this increases the resistance. A perfectly Ohmic conductor is able to maintain a constant resistance whatever the current. ) But all this is not important for this particular discussion! "

" The main point is that when the voltage across any fixed resistance, like a 'perfect' light bulb, is increased, then the current will also increase in proportion to the increase in the voltage. This is Ohm's Law. "

An Ohmic conductor is one that maintains a constant resistance despite increases in temperature, and so obeys Ohms Law at all possible voltages being applied to it. Not being an electronics engineer, just a physics teacher, I could not say how well the resistors used in electronic devices maintain their nominated resistance values as they heat up. Again I guess this info would not be too hard to find on the Net. As you say, a light bulb filament does increase its resistance, but as far as I can remember from some of our classroom experiments over the years I do not recall the increase being too large.

Can I just clarify your understanding of what happens when you increase the voltage across a resistance, above the manufacturers specification. I get the impression that you believe that the nominated power value, the wattage, remains constant despite the increase in voltage?

This impression is reinforced again from . . . quote;" why would the same bulb mysteriously suddenly require more current? "

Assuming that your answer to my question about the current remaining constant is a yes, I will try to convince you otherwise.

Voltage is the push that causes the current. The current can be linked to the speed of the electrons flowing through a particular piece of wire ( of a certain metal, and of a certain diameter).
The harder you push, the faster they go - i.e.the greater the voltage, the greater the current.

There are lots of analogies I could use. Heres one. My age and fatty heart means that when I go swimming it is best that I roll the arms over at a 'specified' stroke rate, and kick at a 'specified' beat rate. By sticking to what feels good I slowly make my way to the other end. But still being a little competitive I sometimes try to ignore specifications, usually when trying to keep up with some-one in the adjacent lane. So I push harder. Guess what - I go faster. More voltage - more current!

Again, a quote from an earlier post; " The power rating of a bulb is just the ideal working conditions for that bulb. The manufacturer is telling you that for this bulb to work as it was designed to operate, then you should use this particular voltage as a power supply, and if you do, then the power consumed from your power supply will be so many Watts. If you decide to use more voltage, then the current will increase ( I = V/R ) and so you will be dragging more energy per second ( power ) out of your power supply ( P = V x I , Power = Voltage x Current ). "

So I assert that when you increase the voltage applied to the bulb, its not a case of what current the bulb "requires" (your words), but a case of what current is being forced throught it by the bigger push.

Have I convinced you yet? Look forward to your answer to the question - Does the current stay fixed even when you increase the voltage?

And yes, I liked your intro to the last post. We can certainly use the "known" voltage and power specs of a certain bulb to calculate the current. Although I see that word 'required' again. Yes, that is the best current for that bulb if it is to work well for many years without blowing, and if it is to give adequate lighting for the job. But, if it gets more or less voltage supplied to it, then bad luck - it will have to cope with more or less current - again, this current calculated using Ohm's Law and the known resistance ( assuming no great increase, or decrease due to higher or lower temperatures as explained at the start.)

Lets do those calculations with a 12V 24W brake bulb;

P = V x I so I = P/V = 24/12 = 2A . . . . . R = V/I ( Ohm's Law) = 12/2 = 6 ohms

So I would say that if this brake bulb was given a voltage of 2 x 12V = 24V, then R would remain pretty close to 6 ohms ( just a little more due to heating, but not a lot more, so for simplicity, R = 6 ohms )

New current will be I = V/R = 24/6 = 4A and new power will be P = V x I = 24 x 4 = 96W

Like I have been saying - double the voltage - double the current, and four times the power consumption. The bulb likes to run at 12V 24W but bad luck, it takes what it gets!

If you can get some more wood for the camp fire, I'll go have a look for those resistance temperature figures. Looks like the Sun is coming up!

 

[millsy]

P.S. I think you forgot to " offer my own assertions of Ohm's Law ". Or will this be covered by your answer to my question; "Does the current stay fixed even when you increase the voltage?"

Looking forward to your reply, Millsy.

 

[grit]

P.S. I think you forgot to " offer my own assertions of Ohm's Law ". Or will this be covered by your answer to my question; "Does the current stay fixed even when you increase the voltage?"

Looking forward to your reply, Millsy.

I prefer to stand on safer ground and simply state that if this circuit were run through an amp metre the comparative amperage would not be double.

When a light bulb is purchased you will request a particular operating voltage and wattage, this is what we are measuring. The wattage divided by the applied voltage will determine the amperage.

I doubt it can be expressed any simpler than that and see no reasoning in any attempt to do so.

 

[millsy]

Ok Grit. I don't know about you, but I reckon we would both get a good price if they sold us off at a mule market! I'm doing a quick pack of the truck to head off bush somewhere for the evening. Catch you next time.

 

[4X4]

I just got back. Did I miss anything?

 

[millsy]

Me too. Drove 150km out east to find a little conservation park to camp in, Google called it Lowan CP, just to find that it was actually on private property and you had to contact the owner for permission. Was not going to do that at 10.00 pm! So turned around and and had to come home too soon!

As far as your 'What is a battery?' thread is concerned, Grit and I had a bit of fun, but I think we have done it to death now. Looks like we we are agreeing to disagree as they say. Thanks again for the great info that you started off with. It was just what I wanted. Whacked the new battery on my little charger for a couple of hours the other day, after three weeks in the standing VH. Did not hit 14.7 volts though, like the big truck battery did when I first started taking measurments.

Now heading off to plan a trip with a young fellow that I borded with when I first started work in the 70's. We are heading back to the Big Desert area, and hopefully Sunset Country, in a couple of weeks.

 

[oldrtack123]

Hi Millsy
"light bulb has a fairly constant resistance. ( But not perfectly Ohmic - i.e. its resistance actually increases a little as the current increases"

Sorry mate to have to correct you on this because the rest of your post is very good basic theory.
BUT electric light bulbs have far from a fairly constant resistance, whilst it may not be so easy to measure on low voltage bulbs ,a cold resistance check of say a 240v 60w bulb will show its resistance is only 75 ohms app . 75 ohms connected to240v draws 3.2 amps
3.2x3.2 x 75 = 368 watts!!!
This is the very reason many bulbs blow on switching on, they simply can't take the initial surge current before they heat up to an operating resistance of 960 ohms & .25 amps for a 60w bulb on240v. Before the days of solid state this feature had many useful applications.
I T WAS OFTEN RECOGNISED WHEN USING EXPENSIVE OR HARD TO GET BULBS THAT YU O COULD ENCREASE THEIR LIFE BY LIMITNG THIS INITIAL SURGE CURRENT WITH A SERIES RESISTANCE & BY PASS SWITCH.
The feature was also used as a series ballast to maintain reasonably constant current in some circuits [ it did require careful selection of the correct bulbs .the very best for this purpose was the ORIGINAL iron filament bulbs]
A light bulb was the worst choice to use to illustrate ohms law because it has far from constant resistance with temp rise. Any reasonable quality fixed value resistance USED WITHIN ITS WATTAGE RATING would be the right choice IE wire wound power resistances or even simple carbon resistors.WATTAGE RATING IS ,IN THE CASE OF RESISTORS, THE ABILITY TO ABSORB OR USE THAT POWER WITHOUT DAMAGE FROM OVER HEATING. GO BEYOUND THAT & IT CAN BE DAMAGED OR BE NO LONGER STABLE.
In the case of light bulbs its wattage is the power in the form of both heat & light it has been designed to handle . applying higher voltages will push it beyond its capacity causing premature failure . So the wattage rating of any electric devise is the power it is capable of safely using without being destroyed. The wattage rating is a constant, it does not change,you can force the device to use more power, in the case of a resistor by increasing the applied voltage, DOUBLING THE VOLTS WILL DOUBLE THE CURRENT & THE WATTS x4 IF THIS ENCREASED WATTAGE IS IN EXCESS OF THE RATED WATTAGE THE RESISTOR WILL BE DAMAGED.
THE SAME APPLIES TO ALL ELECTRIC DEVICES THEY SHOULD BE USED WITHIN THEIR RATINGS

 

[ian059]

This thread has been an illuminating read! Now, can some explain to me - "What is a battery?"

 

[grit]

This thread has been an illuminating read! Now, can some explain to me - "What is a battery?"

A battery is a collection of one or more cells contained within a single unit.

 

[grit]

This discussion should have stood as an example of how two people can express a very different point of view on something so basic.

Firstly, I'd like to set the record straight... Millsy your definition was right on the money and had you used the tried & tested analogy of 'water' my argument would have been blown out of the water 'quite literally'. Voltage is analogous to the water's pressure, Amperage is analogous to the rate of the flow & Resistance is analogous to obstacles within the water pipe. The obstacles are the only constant (or should be for the purpose of calculating electrical properties within a simple DC circuit), the only way to remove this constant is to remove the bulb from the circuit. Bulbs consume their 'rated' Wattage ONLY when the 'rated' Voltage is applied.

This thread could have become quite heated but did not and to this end you are deserved great credit. You strongly expressed your point of view but never to the point of personal discredit.

... & I did like the 'mule' analogy.

Hope you had a great weekend and agree that 4 wheeling is one constant we can all agree on.

GRiT