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Turn it around and consider each resistor separately. If you push a current I through the first resistor, let's call it R1, you'll get some voltage difference V1. If you push the same current I through the second resistor, let's call it R2, you'll get another voltage difference V2. If R1 and R2 are identical, then V1 and V2 would also be identical. ...

3

There are a couple of ideas you are probably missing. One of them isn't often taught directly in an electronics textbook (but would be found in a physics textbook.) The other idea is simple enough and is where I'll start. Suppose you have the following schematic: simulate this circuit – Schematic created using CircuitLab With reference to the ground ...

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I understand Ohms Law and how the math works out but I’m asking intuitively why would voltage divide in this case rather than be equal across the “constrictions.” In the below diagram, something isn't right. 5V is listed for both constrictions, if you put two pipes in series, then the constrictions should read 5V and 2.5V for the first constriction, and ...

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Maybe analogy of a line with slope m helps. Consider y = -x + 10 line, start from the origin in horizontal axis. As you go 1 units (resistor) in +x direction; you started from 10 (voltage start) and ended at 9 (voltage end), the slope (current) is -1. Now go for another 1 units (another resistor) in +x again. You started from 9 and ended at 8. So, the ...

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The control (coil) of a mechanical relay requires more-or-less the nominal voltage specified and will draw approximately the specified current (varying a bit with temperature and actual voltage etc.). The relay contacts (assuming a mechanical relay) can switch a range of voltages and currents. If the maximum is specified as 28VDC at, say, 10A maximum, you ...

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It is the influence of rain, birds nests, kite strings and other contaminates to air that degrade the breakdown voltage ,BDV, of parallel line to line breakdown of approximately 1kV/mm ( and 3kV/mm for dry air plane to plane flat smooth rounded edges) Each country will have electrification safety codes for safety clearances that maybe >=1m in order to ...

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The speed of a brushed DC motor is roughly proportional to the voltage applied, given the same countertorque. You are running the motor at 1.5 times the rated voltage, thus you get 1.5 times the speed. There's a reason that motor is designed for 6V only. DC brushed motors are very forgiving, even the commutator and the bearing should stand 1.5 times the ...

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The equation is correct — the output voltage is indeed independent of the input voltage — but the equation only applies when the rest of the operating requirements of the device are met, specified elsewhere in the datasheet. Among these are the requirement that V_in > V_out + V_dropout. If this requirement is not met, then the first equation no ...

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ideally rail to rail Rail to rail won't happen because of the diode drop. Use a PNP bipolar transistor with base on op-amp output, collector to ground and emitter connected to VF3 via the diode. If your supply rail is genuinely only 2.5 volts then you probably won't require the series diode. Any supply rail greater than 5 or 6 volts requires a diode to ...

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According to your link the motor specs are: Rated Voltage: DC 12V Reduction Ratio: 1: 172 No-Load Speed: 30RPM Rated Torque: 14Kg.cm Rated Current: 1.1Amp With a permanent magnet brushed DC motor, speed is proportional to voltage and torque is proportional to current. Your motor is specified to produce 30rpm at 12V with no load, and 14kg.cm (presumably) ...

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If you reduce the voltage to 9 volts, the speed will be reduced proportionally. The torque capability will not be reduced. If the battery is too small, it will be discharged quickly and the internal resistance may reduce the voltage significantly below the nominal battery voltage. The rated motor current is 1.1 amps and the rated torque is 14 kg-cm. That ...

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Not for long! (And the speed would decrease as the battery wears out and the voltage falls) 9V batteries are pretty "weak" because the other half of the story is how much current they can supply at that voltage (Power = Voltage * Current). If you want to spin a big motor, the battery voltage will sag because of the battery's internal resistance. The lower ...

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In principle, it's probably possible to run a 12V motor on 9V -- it'll just run slower and with much less torque. But if by "a 9V battery" you mean a typical PP3 size battery (i.e, the rectangular ones with a snap connector on top), the answer is almost certainly "no". These batteries are designed for low-current applications like handheld radios, remote ...

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Everything has resistance. Period! That includes batteries (ESR), diodes(Rs), inductors(DCR) Caps.(ESR) and even Resistors (R) ;) “Kirchhoff voltage loop (KVL)” teaches how to use that “Leftover Voltage” this is after all the supply voltages are added or subtracted with the sum of all parts in series, which each have resistance thus KVL teaches you ...

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No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. There will always be the max resistance between power and ground pins on the potentiometer.

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First, a couple of CircuitLab tips. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no ...

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No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on. But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.

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The proper way would be to use... a buck converter. 10A at 12V is only 120W. Assuming 80% efficiency, you're looking at around 3A out of the 48V, which is pretty manageable. You will need to design it carefully, a bit of simulation and prototyping, as you're looking at dumping around 25W of heat. A good design could take that lower, but plan for worse and ...

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Use a 48v battery, and a buck down to 12v. Both are nice commonly used voltages, and there should be plenty of converters available to choose from in that power range. If you were only going to use a few mA at 12v, then it may be permissible to tap into an intermediate battery connection, and rely on your charge cycle to rebalance the batteries, but 10A is ...

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Pretty much every calculation involving batteries is an approximation. Battery life depends on a ton of factors (battery chemistry, temperature, load profile, battery age, self-discharge effects, other effects such as gas build-up). [...] the website lists the run time for all different models as 1.5 hr so clearly this information isn't correct! The ...

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You need a resistor that is small enough to stop the flickering which is presumably caused by leakage from your power supply, but large enough not to be destroyed. Power at 12V in a resistor is P = V^2/R = 144/R Or to dissipate a given wattage you can rearrange that equation so that Rminimum = V^2/P A resistor twice the Rminimum value is recommended for ...

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V = I*R R = V/I P = I*V = V^2/R = I^2*R You are trying to put 6W through a 1/4W resistor. Of course it burnt. Was there a reason you used 20 Ohms? Or was that just a random value? Use those equations I posted above, and you can figure out which resistor to use, and make sure it is at LEAST 6W

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Hi ForeverLearningJP I have two of these QDD boards with problems, the early QDD has the same problem as yours had, it was returned to me from a luthier who was making an instrument for me and he stated it does not work (it did when I posted it as I tested it) I now think he reversed the voltage by mistake? please can you tell me which diode you used to ...

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They are similar to symbols used for MOV's. The device is clearly a common-mode and differential mode surge arrestor. Is there a part number and can you get a data sheet? Can you get a device and physically inspect it?

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You can use a relay rated for 240Vac and 30Vdc if you use an RC snubber design to limit V=LdI/dt and Ic=CdV/dt within the linear operating range. Since we do not know your circuit or cable impedances , I cannot specify what will work. But I am sure a design can be made to work that is well damped and limits AC voltages with 170V transients absorbed in an ...

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It’s >2A and therefore not gold plated and needs a wetting current of 10% of rated current=10A. The low voltage does not matter. It is the current that matters. For 5V logic signals all you need is a low ESR e-Cap around >=22uF to discharge the current when the switch closes with any series that is suitable for the signal. With frequent use (1/day) this ...

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The voltage rating of a switch is precisely about what how much of an arc it can break while it opens, and how much voltage it can withstand when open. Your switch must be rated for the maximum voltage that may appear across it. This is true of any single pole of a switch, so it doesn't matter whether the switches you are using are SPST or DPDT. In ...

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Looking at the specs printed on top of your relay it seems like it has a DC rating on it. So the relay alone is probably fine for switching DC. Note that relays have different ratings for DC because there is no zero crossing in the current to extinguish any arcing. This can cause extra wear on the contacts or even cause them to weld shut if used outside ...

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Yes, basically make a sawtooth and feed that to a comparator. The venerable 555 can do this, although rather crudely. See here: https://www.falstad.com/circuit/e-555pulsemod.html The key to making a good sawtooth is to use a current source to charge the cap. This makes the linear ramp, improving the accuracy of your PWM. Another way to make a PWM modulator ...

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The motor is probably still good, but the motor drive electronics are probably toast. You're likely up for "repair or replace controller board" this will probably cost more than a new drive, but less than $1000. 0 reduce R1 by some amount, maybe try 10K or 22K there instead, a iM variable resistor between pin 7 top end of LDR would allow some control of the amount of hysterisis 1 This circuit may operate with the darkness. But your circuit is not suitable for your application. Because the light condition is rapidly change with the time when you riding in a street. I suggest you may need some delay between switching. The requirement is.. When you go into dark, you need your lights immediately. When you go into light, you don't need ... 0 Power matching is said to be used to match an output impedance to a physical system. ex, a precision led with a 27pF capacitance or a moving coil with a physical atenuation factor to it's free movement (A subwoofer). Also when you have a very large power source and very small loads in parallel which could have very different resistances each causing one to ... 1 There are basically three approaches to running a 3-phase motor from single phase power. None of them is as convenient as buying a single phase motor in the first place. Static phase converter. This involves a bespoke arrangment of capacitors to shift the phase. Here is a commercial supplier. It does not give you full torque and has relatively low ... -2 Due to the cost of matching PU plastic grid rated cap Impedances to motor impedances are greater than VFD, look only for the best match to your needs in current in a VFD solution. the costs in your range are ~<$10/A for 3 phase VFD’s. Up to \$25/A. e.g. https://www.ato.com/single-phase-to-three-phase-vfd. Which is not to say it is a preferred source , ...

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The old-fashioned alternative to a modern VFD, as mentioned in another answer, would be a "rotary inverter". A rotary inverter is really just an electric motor driving an alternator. Using a single phase motor and a three-phase alternator would give the required result.

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If your budget allows it you should get a VFD (Variable Frequency Drive). It can create three phase signals with different phase shifts and frequencies, giving you the ability to cintrol speed, direction, power etc. of your motor. These can be found on eBay for moderate prices, the more powerful they are the more expensive. There are probably other options, ...

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Those 3.7 V batteries need to have a voltage between 2.5 V and 4.5 V per cell, if they're outside that range then consider them to be damaged. Some cells might survive a short time below 2.5 V. But 0.1 V means that they're dead. 9 V might be on the edge of what a device that is supposed to work on 2 Lithium based cells in series can handle. Chances are that ...

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If you care about matching brightness (which I assume you do based on the nature of your question) then what you want to shoot for is equal current through each bulb. So there’s two ways of solving that problem. One is to wire the bulbs in series and use a boost LED DC-DC driver, and let Kirchhoff’s Law do the rest. This is arguably the cheapest and most ...

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Either way is fine. trust me. BLDC Fans with aerodynamic loading end up looking like a resistive Load so Voltage is divided equally guaranteed and fans run well. I used 24V x 2 from 48V in production of 10k units , no issues for voltage , but 1% failures unrelated due to dead start from Hall sensor errors from Korean supplier. I performed a ganged ...

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Use 12V and wire the fans in parallel. If you use a 36V power supply and wire the fans in series, the voltage might not be shared equally. This could even cause the fans to burn out. In parallel, they will all get 12V.

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What is the application? - it probably matters. Do you care about volts. If so why? You may. Or not. Brightness of about 2:1 can just be discerned by eye for most people. Tungsten bulb resistance rises with voltage so 18V:15V difference will be less than 1.2:1 brightness - but IF incandescent, colour change may be unacceptable. Use LEDs, assuming "bulb" &...

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You have come across the reason why high voltages are used in the home (230V, 110V), even higher are used to transmit power over longer distances (10kV and higher). This is also the reason many cars are moving away from 12V power to 48V, and lorries/trucks use 24V rather than 12V. As you mention, Power is the current times the voltage, P=IV. If you remember ...

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You will have 2W less losses at 5V, which means more current. 2W / 5V = 0.4A max assuming no losses...

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No. The simplest answer possible is that voltage is the density of electrons. That is, the "pressure" required to push them together against their repulsive force. Of course, this is complicated by other factors such as the medium in which they are moving.

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So you are using 5% resistors and are worried about stability over time. Your 1kΩ, 5% resistor can range from 950Ω to 1050Ω. There is no substitute for getting a good quality DMM and making appropriate measurements. From Resistor01: Shelf Stability/Drift. In addition to all the environmental and electrical factors listed above, resistors experience a ...

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The delay you require is only 25nS. I would consider simplifying your circuit to use two or all three of the other gates in the 74HC86 package to provide the delay, their nominal Tpd is 11nS at 5v into 15pF. Without extra capacitive loading, their delay might be a bit less. Their delay will be strongly affected by the rail voltage, so only use this method if ...

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However looking through the datasheets I found that MAX9010 outputs TTL levels, while 74VHC86 accepts CMOS levels (0.7 * Vcc). That's a good spot and I agree with you - maybe you should inform Maxim on their dodgy circuit. Shame on them. Should I pay special attention to this issue - what are the conditions when circuit may fail producing proper ...

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I thought you were way off with 1.5 A but a browse through Farnell's list of RJ22 connectors found this. Figure 1. Extract from MTP-44U - CAT5 RJ22 MODULAR PLUG, 4 POSITION, 1 PORT. The voltage isn't a problem as they have to take telephone ringing voltage which is about 100 V AC. I was surprised at the 1.5 A current rating but be aware that the you'll ...

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$10sin(\omega t)=0.8$ $sin(\omega t)=0.08$ $arcsin(0.08)=\omega t=0.080086\:rad$ $\therefore \:t=0.080086/200=0.40043\: ms\:\approx 0.4 \:ms$ Note: $sin(\theta)\approx\theta\:$ for small values of $\theta$, where $\theta$ is in radians. So we could get to the approximate answer without doing the $arcsin$.

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