# Tag Info

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DC component. Replace AC source with its equivalent impedance (0). Apply Kirchoff, with your knowledge of C's impedance at DC. That should give you the DC component of the voltage at Node 1. AC component. Replace DC source with its equivalent impedance. Apply Kirchoff, bearing in mind the two impedances are complex. That should give you the AC component. ...

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Assume the capacitor as a tiny generator. When the alternating source reaches its positive peak, the capacitor has a positive polarity towards the alternating source. So a negative polarity towards Node 1. And when the alternating source reaches a negative half cycle, the polarity is reversed. So you will get output within 2.5 +/- 0.2 volt.

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Norton Resistance You already did the essential steps, e.g. remove load, replace voltage and current sources with short and open circuits respectively. $$R_{norton}=5k\Omega||15k\Omega=3.75k\Omega$$ Norton Current Simplify the circuit by converting current to voltage sources: simulate this circuit – Schematic created using CircuitLab Calculate the ...

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I tried to fix same LED Christmas's light. It has resistor and diode inside. One string 50 LEDs divide to two parts. 25 LED bulbs connected in series and one diode and resistor. Measuring current gave half sine wave, amplitude 60 mA. So the average is 20 mA. Smart inventors just replaced incandescent bulb with LED and added diode and resistor.

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Voltage at non-inverting input is 3.75V and therefore voltage at inverting input is also 3.75V. Current through top 30k resistor is 3.75/30k = 125uA and this will be the current through R irrespective of its value. With R = 0 Ohms, current through 10k load is 3.75/10k = 375uA Therefore with R = 0 Ohms, Io = 375uA + 125uA = 500uA which is the minimum value of ...

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Half a series of 100 led Christmas was off. The problem was one of the two resistors. Changed for a 4.9k ohms, 1/2 watt, back to work!

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It depens on the component you are looking at. One LED and one resistor are in serial. All LED and resistor pairs are in parallel to each other. For distinction between parallel and series circuit its always a good measure to track where the current flows and look how many loops you can draw from Vcc to ground. In this case you can draw 5 loops(indicates ...

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I suspect the transformer is ideal, and it is where you may be stuck. Current into a polarity dot on one side means current out of the dot on other side. Their magnitude will be exactly per turns ratio. So, draw an arrow going into left side dot and assign it a current variable. Then draw an arrow coming out of the other dot (or in to non-polarity dot side ...

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Consider the situation where Vin > Vout, and both are modeled by ideal voltage sources. When S1 closes, the capacitor charges up to Vin and gets a charge of Qvin = C*Vin stored on its plates. Now when S1 opens and S2 closes, initially the capacitor is charged to Vin, which is greater than Vout, but is also now connected to Vout. In general, an ideal ...

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[edit 1] Note: as JRE points out in a comment, if the unit is still under warranty, try that route first. If so, see if you can reverse your potentiometer mod and bring it back. Even if there is no warranty, you may want to get an estimate on a proper repair. Then you can decide if you want to customize it. My rule with taking apart electronics is that I ...

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Photons are quanta of energy that can behave like particles. But they are not 'stored' in the system in the form of 'particles'. They are the result of the interaction of your system with other systems or the environment. So, you should ask yourself: if I have energy coming out of my system (one electron, or one atom, or an ensemble of atoms like a solid ...

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Assuming that a DC motor is running at 1200rpm and there is a power cut, the motor will stop in 40 seconds. Apart from inertia, unpowered stopping time depends on the load, internal friction, windage and iron losses. These factors are not directly related to the effect of a braking resistor. However if you know the operating power consumption (voltage x ...

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I do not believe the active users of Physics SE are patient enough to write or read explanations that can be understood by most of us - the practical electricians. I write an answer here. Elementary concepts such as "radiation is emitted when electron returns to lower energy orbit" unfortunately explain too loosely what happens in solid materials. ...

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Equivalent resistance is zero. It is only one wire, connected in close loop. All connections are to one wire and it is a short circuit.

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Here's a schematic of the voltage divider for the kilovoltmeter. R5 is a must to keep 2.5 kV from floating near the panel, should the voltmeter coil open for some reason.

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You're correct. Keep in mind that power, temperature and voltage ratings will be critical in your application, and if you intend to run the resistor hot it may need to be derated in voltage - read the datasheet carefully. For 2.5kV I'd be using a resistor or resistors that can handle twice that. Say, with 1kV resistors I'd use 5 of them in series, with their ...

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This question has got much attention in 2019 and later. Here's a little more. Pure negative resistance R=U/I, where U and I are 0Hz DC but have different signs can be simulated with an opamp circuit as already shown by others. Dynamic negative resistance where U and I only change to different directions in some operating voltage and current zone, but still ...

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For calculating the appropriate resistor, you need to find out the forward voltage drop of the LED, and its forward current rating. These numbers are in the datasheet of the LED. Different colours require different voltages, and Extra Bright LEDs may have different voltages and currents specified than their dimmer cousins. A calculation example: A typical ...

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Capacitors are commonly used to supply extra current under a high load transient. Depending on how long the high load will be present, it might not be feasible to find capacitor(s) large enough... 3 seconds is probably way too long. If you don't need a TON more current, then you could put a resistor in series with the cap to increase the discharge time (you'...

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Right now, it sounds like the battery voltage and the load voltage are one and the same. You can put a resistor between the battery and the load, and that will help keep the BATTERY voltage higher. But it will make the load voltage lower. Capacitor can't really help. I also want to point out that there is a fundamental contradiction in what you are trying to ...

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A 4_watt resistor likely needs force_air cooling, or mounting on a metal panel to remove the heat. Depending on the leads, soldered onto PCB boards, is not the path to reliable systems. How do we know this? The thermal resistance of PCB traces, of standard 1.4 mil thickness from the standard 1 ounce of copper foil per square foot, is 70 degree Centigrade per ...

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Trying to understand this circuit. What is the resistor R1 for? R1 biases the transistor so that it can be fed an AC signal via the capacitor C1. It also supplies some signal feedback to keep some semblance of slight quality about things but, as it stands it isn't HiFi (it's very LoFi) and impractical for most loudspeakers due to the DC current that passes ...

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Conductance is reciprocal of resistance. $$G=\frac{1}{R}$$

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If I wanted to know the voltage going thru the resistor, only using the Ohms law triangle, I would get: V=10A∗100kΩ=1,000,000v! While this is theoretically correct, this is even more voltage than the power source has to supply! This is because the a priori assertion that the current is 10 A was incorrect. In your given circuit, assuming an LED with a ...

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While most of the uses of the Ohms law triangle, in the way I have used them may be theoretically correct, they are not at all practical. Ohm's law only works when the ratio between voltage and current is a constant, which is called resistance. It does not work in other cases such as LEDs and loads whose relationship between voltage and current draw is not ...

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Your confusion seems to stem from the fact that you think the load will pull 10A under all circumstances. It will not. It is rated for 10A @ 12V which means the load itself has a resistance of 1.2 ohms. You connect the load straight to the 12V power supply. simulate this circuit – Schematic created using CircuitLab Only 10A will flow trough the load ...

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I'm assuming you have: A 12V Constant Voltage supply A load that will draw 10A at 12V And that you want to add an LED to this. I'm also going to assume that your 12V Supply is able to source the required >10A. The way to add the LED to it would be to add it in parallel to the load, not in series. This way, you don't have to "protect" the LED ...

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Ohm's law applies to any particular resistor, and the voltage across and the current through that resistor, not across other components. The LED will drop 3 V, for a huge variation of current though it. We can therefore regard it as a constant voltage drop, at least for the first round of calculations. With 12 V from the battery, and 3 V across the LED, this ...

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I wont give you the answer but I will tell you how you can get it. Consider the following circuit: simulate this circuit – Schematic created using CircuitLab Find out the power Pd dissipated in Rload Find out the condition where Pd is maximized

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The distinction is between film, and bulk, resistors. In a film resistor, most of the mass of the resistor is inert, serving only to support a film of resistive material. All the heat is generated in the film, and then has to diffuse into the support material, taking a relatively long time. For a very short pulse, only the mass of the film itself is ...

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You mention both voltage and power surges, and refer to several different types of resistors. Carbon film resistors are not great with handling power surges. Carbon composition resistors, as one might imagine are much better, but they are scarce these days. Metal oxide film resistors are often well specified for power surges and for mains use. Metal film ...

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Start with the number of color bands. Yours has four bands. A resistor with four bands has three for the value and one for the tolerance. The tolerance band is always the right most, and is usually spaced further from the other bands. That means you need to turn your resistor around. Now reading from left to right, you have blue, gray, orange, gold. From ...

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A google search turned up a dozen results: https://www.mouser.com/technical-resources/conversion-calculators/resistor-color-code-calculator?gclid=CjwKCAiAtK79BRAIEiwA4OskBoYzOw6cS6UT08Lgup-8PD1fVWpyjcs3gee_zKlh_g5VKpnUGWOyXxoCJIoQAvD_BwE If I see the colors correctly, 1st Blue, 2nd Green, 3rd Orange and 4th Gold (tolerance) = 65 kOhms 5% tolerance. If the ...

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There is current flow. Consider the picture on the RHS: the 5 branches are in parallel, driven by a given $v_i$ voltage (say, $v_1$). Then you can say the $n-th$ from the common node to ground will be: $$i_n = \frac{v_i-L_{M_n}}{M(x)}$$ and the total current will be: $i =\sum_{n=1}^4i_n+(v_i-L_R)/R$. Hence you can see each branch will have a current ...

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That's right. If the RC circuit has time constant of 100us, it has no chance to react to the 1us pulse, so it will pass on right through. If on the other hand you needed to pass a 100us pulse with a RC circuit having also 100us time constant, the 100us pulse would have enough time charge up the RC circuit, so the waveform would have droop. Same thing can be ...

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Normal LEDs are prone to go into thermal runaway when run from a fixed-voltage source without a series resistor. As the temperature increases, the amount of current they will pass at that voltage will also increase, until they've gotten so hot that they are at least partially destroyed. Blinking LEDs incorporate a temperature-sensing element, which will ...

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It doesn't match the 5 band color coding scheme for inductors, so that's out. The Digi-Key resistor decoder says 1.86 ohms and 1% tolerance. That's reading it as brown, gray, blue, silver, brown. You've interpreted the blue band as violet, but on my screen and to my eyes it is blue. At any rate, it should be 1.8 something ohms. If you are measuring 16, then ...

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Since you have an Arduino controlling things, why not have it also monitor the current in the string? Use a hall sensor, a low pass filter, and an ADC input to the processor. Leverage your programmed logic. This serves the goal of saving power and also gives more precise control over brightness. If you want very accurate control over brightness add a photo ...

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No. You would need to start with precision resistors and then measure them on a bridge to ensure you have a matched pair. Note that resistors are measured before the tolerance is assigned to them. A 20% resistor is guaranteed to be at least +/- 10% off. A 10% resistor is guaranteed to be at least +/- 5% off. Ditto for 5% and 1% likewise being off, and all ...

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Your LEDs in a given chain could all turn out to need 3.4V, so the max you could theoretically design for in a chain is 7 diodes. But they might all be at the 2.8V end of their spec. Allowing say 100 mV across the FET when on, that means you would have to drop the remaining 24 - (7 x 2.8) - 0.1 = 4.3V across a (4.3V / .35A = 12 ohm) resistor or whatever. ...

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Using the "electromechanical relay", or a mechanical switch or lever controlled with a solenoid or motor is probably the best way to do it. You could experiment gas discharge tubes. If you want the resistance to lower with a lower current you could use a PTC resistor. With high voltage the current raises, which heats the resistor, which causes the ...

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Having done this myself with one of the many pre-made LED PCBs you can get on aliexpress, I suggest you get some 1W 0.5ohm power resistors, temporarily put enough of them in series to give you a safe current, and then remove them until you're slightly below the rated power of the LEDs. If they're rated for 350ma then somewhere in the 300-325ma range sounds ...

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There is excess noise in a carbon resistor when current flows through it. It doesn't matter whether that current changed direction in the past and will do again in the future, or is steady, the noise at any given current is the same. The question is perhaps, why do people emphasise that noise is generated even when the current is DC? One possibility is that ...

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Also importantly, the forward voltage drop of a LED goes down as the junction temperature increases, which could lead to thermal runaway. The simplest approach might be to put a resistor of about 5R between the source of the FET and ground; as the LED current reaches 350mA the gate-source voltage will drop from 5v (I’m assuming you’re using a 5v Arduino) to ...

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It's a lottery. With two resistors in parallel, at worse they won't make any difference because their bias will be identical. At best they will compensate perfectly, their bias being opposite. Most of the time you will have something in between. To increase your chances of having complementary resistances, you should buy them form two different manufacturers....

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Indeed, the formula is irrelevant when the sum of the led voltages is equal or nearly equal to that of the supply voltage. In this case a resistor is not necessary but you must be careful to always use in your calculations the smallest voltage in the voltage range of the led. Each type of led has a voltage range (look at their datasheet). For example 2.8V to ...

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The answer is more complex than it seems because the PWM function of the arduino will interfere with the feedback of the power supply. If you use a constant current driver, The power supply will try to compensate for the relative drop in current by increasing voltage. If you use a constant voltage driver, it will not react fast enough to keep the voltage ...

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However this is AC and to control the 'amplitude' of a sin wave you must use an inductance and not a resistance ( that work only in DC based on Ohm's law). Calculations will differ accordingly so ideally one would use a coil-inductance and take the midpoint of it to obtain two equal amplitude sin waves. You can then be as accurate as you wish and build a ...

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Case: $t\to\infty$ Always redraw the schematic. Even if you don't entirely want or need to. It's just good practice to get into. See the Appendix below for details. I'll just redraw the entire schematic as you have it, at first: simulate this circuit – Schematic created using CircuitLab Now, for the case with $t\to\infty$, the switch has been ...

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Theoretically you could set the Vgs so that the current was right. The logic output needs a series resistor for it. PWM output with RC lowpass filter could set the right Vgs by having the right duty cycle (and high enough frequency in relative with the filter passband). But that would be like dancing on a tightrope. Fets are individuals, the right Vgs is not ...

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