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Voltage And Current

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What Is The Nominal Voltage And Current Of Acer Ac Adapter Mean?

What is the nominal Voltage And Current of acer ac adapter mean?

The nominal voltage of power adapter normally refers to the open circuit output voltage, that is a voltage value not take any load and without current output. Therefore, this can be considered as the output voltage limit of ac adapter. For the internal use of the initiative power supply unit or voltage regulator components of the benchmark, in general the use of high resistance DC voltmeter can directly measured the nominal voltage, even if the mains voltage fluctuations in a certain place, the output is also a constant value.

The current of power adapter. Regardless of any acer laptop ac adapter has a certain amount of resistance, so when the acer adapter output current, the voltage drop in the internally , which is equal to the output current multiplied by the ac adapter resistance. Resulted in two things, one is the heat generated is equal to the square of the output current multiplied by resistance, so the power will be hot.

the other is nominal voltage becomes the output voltage minus the internal voltage drop, resulting in lower output voltage. When the usually design completes considering the heat issues, then generally restricts a current value, when the output current achieve this value, output voltage is reduced to 95% of nominal voltage, or other ratio, the manufacturers can set a higher or lower proportion according to the differern needs of products, the current value called nominal current.

Such as the nominal current of 72W, 16V ac adapter for acer is 4.5A. If the load resistance is too low, resulting in the output current exceeds the nominal current, there will be two things happens, one is individual components heating exceeds cooling capacity result in buring and cause the damage of power adapter, the other is thermal design left margin, merely a reflection of the output voltage is further reduced, if reduced too much may cause the load not work correctly.

Are the same nominal acer ac adapter voltage, output current differences, can be used in the same notebook?

The underlying principle is the nominal current of small power can be replaced by a large nominal current of the power. It might be misunderstood, someone thinks great nominal current of the power will burn Notebook, actually how large the current depends on the load in the case of the same voltage, that is, the work of the acer laptop, the current was bigger when the notebook in a high-load operation and smaller when the laptop turned into standby state, in short, the current equals to the voltage added to the notebook divide by the notebook's equivalent resistance. Large nominal current acer laptop ac adapter has sufficient current margin and will not overheating or low output voltage after replace a small nominal current power.

If you use a small current replaced a high-current acer adapter will ouccers such dangerous. But some friends instead of 72w with a 56w acer adapter is also no problem, that is because power adapter is usually designed to leave a margin, load power must be less than the power, therefore it is feasible, but the margin on the remaining acer ac adapter power are only a few.

Once you have received a lot of notebook peripherals, like two usb hard drive, then cpu running at full speed, the drive at full speed to read the disk and coupled to a battery charging at the same time, that would be dangerous, keep your hands touch the acer adapter to feel whether it could already boiled eggs. The 16w that the 72w adaper more than the 56w adapter is to deal with this situation. So it is best not to use a small high-current power instead of high-current acer ac adapter.

The same type of notebook, why mine always hot when someone else's power is just warm?

Do not doubt your acer ac adapter problems, first check that what are you doing with your laptop, whether like the above said two usb hard drives, cpu running at full speed, hard disk read and write crazy, drive at full speed to read the disk, while charging the laptop battery, stood loud music, the screen brightness of the largest wireless network card has been detected signal and so on, make good use of power management.

Is there will be any problems when the nominal supply voltage much higher than my laptop battery voltage?

acer adapter to the laptop is different from the battery supply, the battery's output is pure DC, very clean, directly access to DC transformer modules, the battery voltage is neither possible nor necessary to design very high, the voltage requirements of microelectronic circuits analog signals and digital signals is 5V for the sector, to get rid of PSA module efficiency and pressure requirements, 10.8V is enough, and Li-ion battery chemistry EMF determines the output voltage of a power-saving cores can only be around 3.6V.

Therefore, many cells are based on three series of the way, 10.8V have become a very popular battery voltage. Some batteries' nominal value is a little larger than integer multiple of 3.6V, for instance, 3.7V or 11.2V etc, in fact it's in order to protect the battery, or a manipulation in the calculation of the design capacity of the acer battery.

If you use the acer ac adapter, the situation is more complex, first you need to steady voltage and do further filter to ensure that stable wokes when the power performance is not in a good situation, after the regulator output voltage is divided into two parts, one part for the work of the acer laptop ac adapter, another part to the battery charge.

acer ac adapter to that part of what happened when the same with the battery-powered, and to that part of the battery through the battery charge control circuit can be added to batteries, the control circuit can be very complicated, so the supply voltage must be greater than the acer laptop battery, then it will have sufficient differential voltage is supplied to the charge control circuit margin each unit. Final truly voltage added on the batteries is by no means 16V, please rest assured.


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What electronic tools/components is this?
1.) An electronic tool specifically used to measure relative electrical parameters such as voltage,current and resistance. 2.) An electronic device that opposes, limits the flow of current. 3.) An electronic component used to charge and discharge electrical charges. 4.) Devices do not amplify but maintain the necessary operating voltages and current of certain devices. 5.) Used to adjust the proper meter setting to be used in checking. 6.) The measure of the opposition of current. 7.) The measure of energy storage capability. 8.) Devices that can oscillate, rectify and amplify sine waveform. 9.) An electronic testing tool used to measure AC/DC voltages. 10.) Used for touching the terminals of the component under test. 11.) Also known as capacitor or the leyden jar in early times.

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If the universal transistor DC bias is configured as a common-emitter amplifier...?
If the universal transistor DC bias circuit is configured as a common-emitter amplifier, calculate the resulting voltage and current gain. Assume a load resistor of 1 kohms. where Uni transistor DC bias Vcc = 15 V, R1 = 10 kohms, R2 = 2.2 kohms, Rc = 680 ohms, and Re = 100 ohms. Assume ? = 200 and Vbe = 0.72 V. I really suck at circuits please dumb down the explanation, don't however dumb don't the math. I'm awesome at that. Thanks for the help.

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watts vs voltage and current?
does watts have to do anything with a subwoofer/speaker or do you just need the correct voltage and current?

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Resistivity question help please? 10 points.?
Question: The resitivity of aluminium is twice that of copper. However, the density of aluminium is one-third that of copper. a) For equal length and resistance calculate the ratio: mass of aluminium over mass of copper. This topic isn't on density, it is on resitivity and ohm's and voltage and current etc. If anyone could point me in the right direction I would greatly appreciate it. Resitivity equation is: p = RA/L Resitivity = p Resitance = R Cross-sectional area = A Length = L

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How does not including the internal resistance of a circuit effect V and I calculations?
If you are measuring the voltage and current across two resistors in parallel.. The voltage is kept constant with a variable resistor.

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I dont understand this paragraph on electricity!! Please help!?
The tungsten filament does not obey Ohms law because the temperature of the filament rises as the current increases. We can infer from the graph that the resistance of the filament increases as its temperature rises. 1) by V=IR, if the temperature rises and resistance rises, then doesn't voltage rise? and so therefore by ohms law voltage and current rise together?? The resistance of the filament at its normal operating temperature ( 3000 K ) is ten times greater than when it is 'off' - that is, at room temperature. This gives rise to a momentary 'surge' of current ( and therefore power) when the lamp is first switched on. Is this part basically saying that when the lamp is of the resistance still remains but it is low. and when you turn the lamp on, the resistance takes a while to built up, so there is a large current suddenly? Resistance is caused by the vibrating positive ions in the crystal lattic of the metal impeding the flow of electrons. When the temperature of the metal is raised, the amplitude of the vibration of the lattice ions increase and as a result, there is an increased interaction between the lattice and the ' free ' electrons. In terms of drift velocity, I = navq, A and q are constant for a given wire. For a metallic conductor n does not depend on the temperature and so n is also constant. As the temperature rises, the increased vibrations of the lattice will reduce the drift velocity, v , of the electrons and so I will also decrease, that is, the resistance increases with temperature. BUT AT THE START OF THIS WHOLE PARAGRAPH IT SAID THAT CURRENT INCREASES WITH TEMPERATURE!! ahhhhhhhh!!!! This is talking about resistivity in metals. are the tungsten filament lamps not metal?? so confused.

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The voltage and current delivered to a load is V = 100?30° in and I = 10?60°.Calculate the average power assum?
Calculate the average power assuming that in V and I are already rms values.

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