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45

Redrawing schematics is a great way to analyze circuits, but also an exercise in why schematics are drawn in particular ways — to more clearly communicate to other engineers. simulate this circuit – Schematic created using CircuitLab The rearrangement above should be a little more clear. If you trace a path from one terminal of the battery to ...


11

I see a few answers already, but none provides the definition of "parallel": Two two-pins components are in parallel when the voltage across them is the same Conversely, with "series", you have: Two two-pins components are in series when the current through them is the same Bear in mind that "the same" literally means "the same" in this context. If you ...


8

No, you don't need a relay. There is never actually any reverse voltage across the "Ready to Arm" LED. It goes off because there is +9V on both ends of it, for a net voltage difference of 0V.


5

Assuming a buck converter that's 100% efficient, yes, you will be able to draw 1A from the output. In practice, buck converters can be very efficient, often over 90%, so you'll still be able to get around 900mA output, which is rather better than the 500mA you'd get with a linear regulator. An important part of the buck converter system is its input filter ...


4

It is hard to answer your exact question, since you want information on a scenario that has no commercial relevance, so companies are not going to test almost drained batteries just to know how much energy can be harvested from them. Those discharge curves they publish are used to design battery-powered products: the curves allow to estimate how long a ...


3

As others have noted, there's no problem as such with having both sides of an LED at +9V. It'll just be off, just as if both sides were at 0V. I do see a couple of other potential issues with your circuit, however: Your voltage notation is confusing. Assuming that the triangle symbols in your circuit represent ground, which is at 0V by definition, all ...


3

Yes, because there are two different flows: V1 -> R2 -> R1 -> V1 V1 -> R2 -> R3 -> V1 R2 and the combination (R1, R3) are in series, but R1 and R3 are in parallel.


3

Generally using a nearly-flat AA/AAA battery with a boost converter is not going to give you much, especially if the thing you're powering requires substantial power. When the battery goes down, not only its voltage decreases, it's internal resistance increases as well, sometimes dramatically. So the battery may show, say, 1.1V when measured with a ...


3

It's been a while and no one has decided to close your question, so I'l write something short. I think you are talking about this schematic: simulate this circuit – Schematic created using CircuitLab With KVL, starting at the lower left corner (at the ground symbol) and working clockwise, I get: $$0\:\text{V} + 5\:\text{V}-I\cdot R_1-V_{Z_1}=0\:\...


2

The classic 4-resistor difference amplifier will do what you want: simulate this circuit – Schematic created using CircuitLab This circuit keeps the difference between Vout and Vref equal to the difference between V+ and V-, effectively adding Vref to V+ if V- is grounded. The opamp tries to keep its two input terminals at the same potential using ...


2

Based strictly on the graph you provided: The curve intersects 1.2V at about 16 hours, by my reading, and hits zero at about 25.5 hours. That's about 63% of the lifetime, or 63% of the coulombs. Without trying to calculate area under the curve exactly, we can observe that it's monotone decreasing, and therefore the joules available before 1.2V are more than ...


2

For it to work in real life a few requirements need to be met: If the AC amplitude is high enough and the DC bias is low enough that the current flow through the DC source actually reverses, then the DC source needs to be able to both source and sink current. Many can only source current. Your schematic only shows the output terminals of both sources, but ...


2

According to the manual, overvoltage is limited at 1100 V, You will also see the voltage ranges for the AutoV (LoZ) mode are 600 and 1000 V. This indicates the meter will not be damaged by use testing 240 VAC circuits in AutoV(LoZ) mode. However, also take note of the warning, To summarize this warning: Do not use this mode for testing high voltage ...


1

When you state: if voltage is the amount of work needed to move a charge from A -> B not the actual the amount of work being spent across A -> B but the work needed [if] we were to move a charge from A -> B "Amount of work" is referring to energy. As you said on your post, Voltage is not an energy, energy would be described as Wh which is voltage * ...


1

Chip markings can be useful. So we have an ST LM317 A14A15 in the top, that one goes up to 40Vpp Vin-Vout. Then we have 3* JY21L, which, after a bit of searching reveals the high-side supply goes up to +150V. There also a TI LM339, which is rated up to 36V. And lastly a JY01 (can't even read it but after searching for JY21L you'll know it's there), but ...


1

Yep, they're in parallel - you can't complete a single circuit that goes via R1, R3 and the power source without retracing some previous part of the circuit R1 and R2 are in series R2 and R3 are in series R1 and R3 are in parallel Hope that helps.


1

historically the meter uses the force of electrons inside a multi-turn coil, that coil inside a strong permanent magnetic field, to cause a deflection of the coil to which the needle is attached. Lord Kelvin developed a version where the needle was replaced by a beam of light thus making the "needle" many feet long; this was used to detect electrons arriving ...


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