I want to squeeze more power out of a 12V, 100W DC consumer. Can I achieve this by increasing the voltage above 12V?
(2 votes)
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For example, if I have the circuit here: How do I know which path the current takes? So does it just flow all the way around, or does it take the path I3 / I2 back to the voltage source? How do I know? Maybe a stupid question, but I don't understand it.
This depends entirely on the consumer: whether active or passive. You did not write anything by the consumer!
It is a heating cartridge with a cantal wire coil.
12V 70Watt, which only brings half the power… as promised, it only does 250 C° instead of the specified 500 C°.
That’s why I think you can squeeze more steam in here.
I’ll turn up to 14V and see if the temperature gets higher.
We measure the temperature in °C. We measure the power in Watt.
“…. it only does 250 C° instead of the specified 500 C°…”
You probably read something wrong. A Body can have a certain current temperature or can also survive without destruction, but not “strip”.
An electric Head (here heating wire) is brought to a certain power with a certain electrical voltage, but not to a certain temperature.
Of course, the temperature of a heating wire is decisively dependent on the supplied power, but not in a fixed relationship, because, depending on the experimental condition, the heat dissipation has a different size. And the temperature of a body results from the balance of supplied and dissipated heat.
Consumers have tolerance ranges and the data refer generally to the measurement values. In reality, the consumptions are always slightly different physical quantities, such as 11,8 V or 12,4 V, etc.
If a maximum voltage is specified, you should not exceed it. If it is not specified, and this is usually the case, it goes through youth research to ruin. Especially because not all consumers are linear. 1% more pure does not have to result in 1% more.
Otherwise, the power at the ideal, fairy-tale DC resistance applies
It is a heating cartridge with a cantal wire coil.
12V 70Watt, which only brings half the power… as promised, it only does 250 C° instead of the specified 500 C°.
That’s why I think you can squeeze more steam in here.
I’ll turn up to 14V and see if the temperature gets higher.
In this way you increase performance by 36%. The heating wire won’t last long. Then you can throw the device away. However, an additional heater would be helpful.
The device can do it. The heating cartridge also creates this – because it is standardized to 500 degrees.
Depends on the consumer:
12V/60W bulb: power increases with increase of voltage, as P=U*I.
Some electronics on the voltage regulator: As long as you remain in the green area on the input side, the power remains the same because the controller readjusts. However, the power loss increases.
If you squeeze out more power than provided by increasing the voltage and you succeed in this, the service life will be greatly reduced to a very high degree. This is also called “overvoltage”.
In the worst case, you immediately cut the (usually not so good smelling) soul out of the consumer and the consumer will no longer perform the desired performance at all.
It is a cantal wire designed for 500 degrees, but only 250 degrees. That’s why I think it can take more power. Kantal’s not a lump.
Electrical consumers are designed for a voltage. One can increase the power only in a certain frame by increasing the voltage without the consumer being instantly burned.
The higher the voltage, the faster the consumer burns through.
So what you’re doing won’t be
Please describe your plan. At least according to the current formulation it sounds very “unreasonable” in the direction of a greatly reduced service life and defect.
The rated power means that the corresponding equipment is designed and dimensioned under certain conditions.
If I just give you more load to wear than it corresponds to your powers, you break down below and the power is zero.
From the FS or from the consumer?
Power = voltage x current.
So yes, more tension is more power.
For example, the bulb would be brighter, but would also break faster.
What is it?
Depends on the device. In the case of motors, this can sometimes go quite well up to a certain point, but in all that has an electronic control, I would guess. Who knows how stress-proof the whole thing is? At the end, there’s something going on, and then power is zero.
The Rated power is the operational voltage predetermined in terms of construction. To change the rated power, you would have to reconstruct the device.
The increase in voltage increases the power of the consumer in square. With double voltage, we would have the four-fold power.
The voltage increase leads to more or less rapid destruction of the device because it has not been built for this load!
Of course.
More voltage also means more current with similar resistance.
More power and more voltage means more electric power.
And so the engine can do more.
However, usually only for (very) short time, since it is not designed for this extra power and its wires will quickly burn through with the greater current strength.
Wrong performance. The power of a motor is determined by the consumer (machine) in which the electric frame in which the motor also “represents”. Engines can be overloaded in terms of time.
Unfortunately incorrect: Because higher voltage = higher current = higher load.
We were even introduced a BASIC computer at school, with magnetic strips, must have been in 1973-74. Small one-line display with LED, like a much too big old calculator saw the thing.
Could you please get back in?
Please.
The first computer I had access to was a SER 2C. That was 1976.
Oh, you know, this forum is also used by lay people with “non-sensical” ideas and suggestions to inquire seriously. This variety of answers does not exactly help, unfortunately strengthen the wrong ideas and lack of understanding.
That’s why I’m a little tough. But still thank you, it would only be nice if you could put yourself on the side of the instructors or teachers.
Good luck!
— I don’t do that for good reasons.
The world does not consist of linear resistances.
But you like to flatten your world if it makes you happy. In the engine, the flow is increased – first – and then? What “physics” now ensures a higher load torque?
The transformer even goes into saturation, the condenser disintegrates spectacularly, as does the radio interference filter.
Oh, my God, you don’t learn that in your studies!
And at the ICE the record drive was no longer possible due to the dubious increase in voltage on record day, and the drive power was zero – too stupid if you don’t know what a 4-quadrant is!
As a genius, I recommend raising the supply voltage of your computer for more computing power. Small tip: Keep fire extinguishers ready!