# Tag Info

48

It's the voltage rating on the resistors that is important here. They are powered from rectified 230 V AC and they need to have the correct voltage rating to suit their application. Two resistors in series having an individual rating of 200 V gives a total voltage rating of 400 volts (near enough if you ignore tolerances on values). Take a look at the good ...

44

Congratulations for having the wit to know something was wrong! simulate this circuit – Schematic created using CircuitLab Figure 1. Parallel and series arrangements of batteries will have the same VAh rating. I will effectively have a battery of 20 V with a capacity of 2 Ah. That's the error. In parallel they can supply 1 A each for one hour. ...

38

Because UL 1741 does all the heavy lifting for you. That's what makes it "so easy". UL 1741 is a complicated spec for "grid-tie solar inverters". Aside from doing their usual inverter thing (itself no small matter), a 1741 inverter also senses the presence of the grid, and obviously syncs its output to the grid. A UL 1741 inverter is intentionally ...

32

These bulbs contain a shunt wire which is normally insulated from the bulb but closes when a high voltage is applied. From How Stuff Works: You can also see the wire in your image, below the burned out filament. The shunt wire is coated in an insulation with a low breakdown voltage. When a bulb burns out, the other bulbs look like wires (especially after ...

24

The problem with these theoretical examples lies in the fact the current is assumed infinite for 0 seconds. Crudely substituting this in the conservation law: $$\frac {\partial \rho }{\partial t} +\nabla \cdot \mathbf {J} = 0$$ $$\frac { \rho }{ 0 }+ \infty \neq 0$$ Since charge is conserved, the assumption of infinite current in zero time is wrong. ...

19

Reasons someone might put two resistors in series in a volume design: A bit higher power was needed than what the commonly stocked parts can handle. Let's say a company standardizes on using 0805 resistors unless there is a good reason not to. They therefore end up with many 0805 values in stock, with only a few values of other packages. Now you need a ...

17

In a series circuit, the current in each component of the series will be the same. This is true no matter what the component may be: resistor, capacitor, inductor, diode, battery, etc.

14

When you double the batteries, you double the current, and consequently quadruple the power in the resistor. You got all that correct. However you seem to have missed that by doubling the current, you also halve the runtime of your batteries. So it won't run for 10 hours, but 5. So the total energy is only doubled, not quadrupled. When you connect ...

14

It is really not that easy. It's easy for you because standards and regulations have been put in place that affect the power company and anyone selling you grid-tie inverters. The system is designed and regulated by law so that you can just buy a shiny box, pay someone to hook it up, and not worry. The grid-tie inverter is pretty complicated. It has to ...

13

It does! The capacity of a battery is correctly measured in watt hours (or equivalently, joules), not amp hours. A rough approximation of a battery's capacity in watt hours is its rating in amp hours multiplied by its nominal voltage. Putting two 1V 1AH batteries in series results in a 2V 1AH battery - which has twice the nominal capacity. If you were to ...

13

In addition to improving the gain-bandwidth product of the circuit, splitting the amplifier into multiple stages allows you to use choose different op amps which are designed to excel at particular characteristics. For example, you can choose an op amp with good input characteristics (i.e. low offset, low noise, etc.) for the first stage and a (possibly ...

13

When masses collide in an inelastic manner, momentum is conserved but energy has to be lost. It's the same with the two-capacitor paradox; charge is always conserved but, energy is lost in heat and EM waves. Our schematic model of the simple circuit isn't sufficient to show the subtler mechanisms at play such as interconnection resistance. An elastic ...

13

First off, I want to warn you just a little bit about putting battery systems in parallel. It's usually not a good idea because often the two batteries (or battery systems) don't have exactly the same voltage. If they are different, then the one with the larger voltage will supply some current into the battery with the lower voltage and this often isn't a ...

13

This is what your circuit looks like: simulate this circuit – Schematic created using CircuitLab When you close the DIPSW the LED will be shorted thus the current will flow through the DIPSW contacts instead of the LED. So it will turn off. If you want to turn the LED on when the DIPSW is closed then you should hook up the circuit according to this: ...

12

Your power banks provide 5V 2A output or 10W, and cost about $10. To power a 250W motor, you'll need 25 power banks which will cost you$250. Why not just buy a \$200 battery then? Also, think how bulky those 25 power banks will be on your bike, and what a pain they will be to charge. Hopefully these simple arguments are enough to persuade you, before I'll ...

11

The electrolytic cap in your circuit is reverse biased. The reverse biasing of the capacitor removes the isolating oxide layer, so it allows current to pass. If you connect the capacitor the right way after mistreating it this way, the electrochemical process that dissolved the oxide layer is reversed and the capacitor recovers. You have to be careful to ...

11

Your LEDs may or may not work exactly as you expect, depending on what you expect. The change in brightness when operating the switch will vary depending on whether you have 1 or 9 LEDs lit. With 1 LED, the change will be small, with 9 LEDs, the change will be large. With the switch closed, any number of LEDs will have the same brightness. With the switch ...

11

Noise: Say your opamp has GBW of 10MHz and noise of 1µV (to keep things simple). The source has 1µV RMS noise also. Each opamp will amplify its own noise by the circuit noise gain, plus the noise of everything upstream, of course, by the circuit's gain. So you want the gain of the first stage to be high enough (say, at least 10) so that the noise of the ...

10

The 300 mA rating is in the Absolute Maximum Ratings table - you don't normally want to go near those ratings. The 1N4148 is intended as a small signal diode. If you look at the Electrical Characteristics table, you will see that most specifications are given with a 10 mA test current, so you should only use a 1N4148 with currents in that range. As others ...

10

Question isn't clear. What do you mean with "sum"? Differences are pretty complicate but, I attempt the silly thing to describe it in 2 sentences. If you place DC sources in series, you get the voltage equal to the sum of all the sources' voltages. If you place them in parallel - given a diode between each positive lead, to avoid short circuits (author's ...

9

Would I be able to get what I need by soldering together 16 AA batteries in series? Or would internal resistance foil my plans? Probably it would at least make your goal difficult, if we assume alkaline batteries. Let's take some numbers from an Energizer application note, which says that the series resistance for a AA might be around 200mΩ. This value will ...

9

Your answer is right. To help gain some intuition ... Assume that the X is an open circuit, ie. that the series resistance is 22+ 180+ 75 = 277. that's just a little bit more than the desired result. Assume that X is a short, then that parallel network is 16//180 = 14.7 (// here means in parallel). the total resistance is 22+14.7+75 = 111.7. These ...

9

The cars analogy doesn't model what's going on. A better analogy is water in a pipe. Let's say you have a pump (like voltage source) that keeps a constant pressure from one end of a pipe to the other. With just a section of open pipe, the flow rate (like electric current) will be quite high. Now imagine installing a constriction (like electric resistor) ...

9

As the comment mentions, paralleling and short circuiting lithium batteries is potentially very dangerous if you don't know exactly what you are doing. Most Li-Ion batteries have a protection circuit which includes overvoltage and overcurrent protection, but it's still not a good idea at all. Much better would be to test this using a couple of small alkaline ...

8

simulate this circuit – Schematic created using CircuitLab Figure 1. Your three options with your kit. simulate this circuit Figure 2. Single or series connection using switched jacks. The inserted plug makes contact with the free arrow while it opens the contact with the touching arrow. Figure 3. A switched jack socket with NC switch contacts ...

8

I think the key part is the BMS system you mentioned. I believe it refers to charging control circuits that have circuitry to monitor individual cells or sets of cells, and circuitry capable of shunting charging current 'around' cells that are reaching full charge before others. As such, this arrangement is NOT strictly in series, as the additional circuitry ...

7

In theory, yes. providing you get the phase correct. It would effectively be a 6kV - 0 - 6kV transformer. Would I do it - NO. There is a good chance that the winding insulation will break down and the magic blue smoke will escape - even assuming a 50% safety margin they are not designed for that amount of voltage.

7

A cell or a battery is essentially a charge "pump". Now, to help form an intuition for the answer to your question, fall back to the hydraulic analogy. Two water pumps in parallel can produce twice the water flow of one (ideally). Two water pumps in series can produce twice the pressure (or head) of one (ideally).

7

I think the way to think about this is to think about load lines. (Public domain image from Wikimedia) What the load line graph shows is two equations that need to be solved to get the operating point of the circuit. I is the current going around the circuit in the clockwise direction. VD is the potential as indicated in the schematic. The diode response ...

7

I connected the negative on panel A with positive on panel B This is a good start and positive on panel A with negative on panel B But what is with this? This is how it should be done. And what is with the diodes? You should only use a diode when having a dual supply (panel and other supply) and you want to prevent current from the supply going into ...

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