New answers tagged

4

This type of wire, used for making coils, is commonly called "magnet wire". https://en.wikipedia.org/wiki/Magnet_wire It looks like it's bare copper, but it's actually coated with a very thin layer of transparent insulation. Otherwise, you're absolutely right -- if the wire were really bare, the coil wouldn't work because the current could cut straight ...


3

You are seeing the Electrolysis. Namely, separation of water into hydrogen and oxygen. Salty water is conducting better than a non-salty one, so you see a stronger reaction in it. But the non-salty one is still impure (it still has some minerals dissolved), so some conductivity exist in it too, but the reaction is weaker.


0

If you connect an ideal inductor across an ideal DC source, the current is initially zero, and rises linearly with time. The voltage, of course, is the applied DC voltage, so the current rises as di/dt = V/L amperes/second. With an inductor that has some DC resistance, the current does not increase indefinitely but exponentially approaches the applied ...


0

The current is zero but only instantaneously. Inductors give graphs of voltage that upward slope as DC voltage surges and then gradually reaches its set level (which it technically never reaches but gets continually closer). The graph starts at 0V and then ascends. It doesn't immediately leap up to max voltage like a square wave


4

Because the N-channel MOSFET can carry higher current and it is cheaper then its similar size counterpart P-channel MOSFET. N-ch MOSFET can be used also as high side switch, but it requires an additional charge pump. The shunt is placed on lower side, because the common mode voltage would be too high for the operational amplifier. Again, it is possible to ...


0

When you connect \$C_1\$ and \$C_2\$ in parallel they will share the same voltage \$V'\$. A portion of the initial charge from \$C_1\$ (\$Q_{1_0}\$) will flow to \$C_2\$, so that in the end we'll have: $$ Q_1'+Q_2' = Q_{1_0} \text{ (conservation of charge)} $$ $$ \frac{Q_1'}{C_1}=\frac{Q_2'}{C_2} = V' \text{ (same voltage for two components in ...


2

At the current state of our universe, charge is conserved. (This wasn't necessarily always the case. See this article on dark matter, for example, discussing the possibility that charged particles created shortly after the Big Bang lost their electric charge during the inflationary period.) This means that the sum of two relative charges held by the two ...


1

The scenario you describe is nonsensical and cannot be analyzed using normal circuit analysis techniques. Suppose you have two ideal capacitors with two different voltages across them. The voltage across a capacitor cannot change instantaneously because an infinite current would be required. So if you connect the two capacitors together with ideal wires ...


1

if each individual current enters via Phases R, Y and B, where does the sum of the three currents (that is Ia+Ib+Ic) flow after meeting at the common point of the three phases? The "sum" current has to be zero because it can't flow anywhere. if I were to keep an ammeter, I can individually measure the value of the three currents right? So how ...


0

Figure 1. Three-phase currents sum to zero. Here we can see that the black and blue phases are at 0.5 of max current and positive. Meanwhile, red is at -1 of max current. They sum to zero. Here we see that black is at +1 while red and blue are at -0.5. They sum to zero. Here we see that blue is at zero. Red is at \$ +\frac{\sqrt 3}{2} \$ max current while ...


1

The relays are designed and manufactured to be operated with the stated nominal voltage across the coil when energized. If you want to make a relay to be operated from a 5V power source, you will use fewer turns of thicker wire than if it is to operate from 12V or 24V, all other things being equal. For the relay to work mechanically, the ampere-turns of ...


0

Some are 5V some are 12V and so on.. . What I want to know is, why there's difference is in this value? Relays are made with different coil voltages to match the voltage used in the circuit that is controlling the relay. Do it have something to do with the force with which switching contact are touched and released? Indirectly. The coil has to exert ...


0

Yes. From "BEST of the TRIMMER PRIMERS" by Bourns: DRY CIRCUIT CONDITIONS AS FAR AS POTENTIOMETERS GO, DRY CIRCUIT CONDITIONS RESULT FROM EXTREMELY LOW VALUES OF WIPER CURRENT. PAST STUDIES SHOW THAT UNDER EXTENDED TIME AND TEMPERATURE CONDITIONS, OXIDE FILMS CAN FORM AT THE JUNCTION OF METALLIC ELECTRICAL CONTACTS, SUCH AS THE POINT OF CONTACT ...


0

Oxidation happens with both heat and moisture. The heating likely comes from the diode itself. But really, it’s just age and time taking their toll on a vulnerable component (exposed wiper and all that.) For example, old audio gear will usually need its scratchy pots cleaned or replaced. And, yes, you might be tweaking the diode bias current just a smidge ...


2

To remove further confusion. There is only one Vcc pin. AVcc is a different power domain only for the ADC Clock system and PORTC. Read the notes below the table. This is the right answer: 300mA is total current into all VCC and out of all GND pins. See also note 3.1 The sum of all IOL, for all ports, should not exceed 300mA. and The sum of ...


1

I think you're missing a VERY important text in how you're reading the datasheet. I think you found this on page 235: The "electrical Characteristics" apply to the whole chapter so all tables in this chapter. Now what you missed: Absolute Maximum Ratings This section can be found in almost any datasheet and they relate to values that should never be ...


1

Edit: I read the link you posted, https://forum.arduino.cc/index.php?topic=161354.0, and it does contradict what I say here. It states that the limit is per pin. It appears to be a quote from an official support person, but the answer is so different from what I would have assumed that I would personally verify with their support again if I was going to be ...


0

Overview I gather the existing circuit is something like this: simulate this circuit – Schematic created using CircuitLab Incandescent lamps have a widely variable resistance based upon their temperature. But in your case, the lamp is usually lit because the doorbell switch isn't pressed. So I'm assuming you've somehow measured the current through ...


2

Sure. If you plug I = V/R into P = IV, you just get P = V^2 / R. You could calculate the value of I and then plug it in, or you can solve it symbolically and possibly save a few steps.


2

You really really should do a basic web search before asking questions like that, here or anywhere. The matters involved are so basic and fundamental to the subject that you cannot properly understand the answers you'll get if you don't have a look at what is readily available first. [[Get out of jail free card: In a facility where you can post to SE but ...


3

There is one, I=V/R is only valid for resistors, so as a general definition of power it is better to go with P=IV which applies regardless of component. a schematic or an explanation of the actual situation would get you better answers. That said, defining power with P=IV is more general so it is better in most situations.


1

\$P = IV\$ is the general equation relating current and voltage to power. Specifically it relates the current through a device and the voltage across it to the power generated or dissipated by the device. In a resistor, the voltage and current are related with Ohm's law, \$V = IR\$. This relates the voltage across a resistor to the current through the ...


0

Please, stick to the NFPA 79 Standard, Table 13.5.1, which specify the allowable ampacities for conductors with smaller sections (30 AWG to 10 AWG) than the NEC ampacities (14 AWG to 2000 kcmil). This standard rates 2A for 24 AWG, 90 deg., single conductor. And considering 8 carrying conductors inside a same cable, we should apply a 70% derating factor, ...


3

Yes, both of those V represent the voltage drop. The formula P=(I^2)R is for the case of constant current and (V^2)/R is used for the case of constant voltage. Considering a constant voltage source, we can see that when the resistance increases, the current decreases. Ultimately it comes down to P=IV. So at constant voltage, power is inversely proportional ...


0

The worst shock of my life was 700 VDC for a moment. It was only a moment because the involuntary jerk quickly broke the connection. But I had a nifty little burn blister punched through my skin and into my meat that took a long time to heal. I was in high school at the time, and my dad never found out (or my electrical engineering career would have been ...


0

If the energy inside the cap is decreasing linearly, how can the voltage be also decreasing linearly according to that equation? The initial conditions aren't specified which are important for any capacitor calculations. But the current source will try to force 100mA through the capacitor, the only way to do this is to continually increase (or decrease ...


1

You will see a linear decrease in the capacitor voltage, but that is because you are drawing constant current, not constant power. The power you're drawing from the capacitor decreases with decreasing voltage. Energy is the integral of power over time, so it makes little sense to refer to "constant energy" unless no power is being drawn at all.


3

If the energy inside the cap is decreasing linearly, how can the voltage be also decreasing linearly according to that equation? In your circuit, the current source is absorbing or supplying energy, so the resistor is not the only place the capacitor energy can be transferred to. When the capacitor is charged above 0.1 V, the current source will be ...


4

Start with the information that is given for the current gain \$h_{FE}\$, you should be able to extract a "ballpark figure" from the datasheet: Note how that information is quite "basic", all that the manufacturer guarantees is that \$h_{FE}\$ will be larger than 1000. So you will not be able to find an exact value because \$h_{FE}\$ will never have an ...


1

The largest part of the voltage drop will occur along the filament of the bulb. If you were to measure ON the filament you would find that one side is 12V, the other is 0V and if you measure in the middle you will find 6V. However, your assumption that points A, B, C and D are all 12V is incorrect. Wires have resistance (unless you are working with ...


7

In real life, the points are not exactly at the same potential, since wires have some very small resistance. However, for many situations, it is fine to consider the wire having zero resistance, which means that electrically, points A, B, C, and D are the same voltage because they are the same point. More precisely, they are all the same node. If you want ...


0

You're right to be confused; 5V at 900mA is only 4.5W. The specifications given are probably simply wrong. But it's hard to say for sure which parts of the specs are erroneous. This is part of the reason why on EE.SE we like to say "No datasheet? No sale!"; With a proper datasheet, you have much more information at a higher reliability. Even if the ...


0

Your two metal blocks (assuming they're insulated from each other) make up a capacitor. Yes, current will flow as the capacitor charges. Then when you remove the source, the charges will find some path, however high the resistance, and the capacitor will discharge at a rate determined by the capacitance and the resistance.


0

There will be a brief flow of current when you first connect the battery to the metal blocks. That current will charge the capacitance between the blocks. Once that capacitance (and other stray capacitance) is charged, there will be no further flow. If there is no leakage current between the blocks, the charge on the capacitor formed by the blocks will ...


2

All switching circuits will require some current flowing through the switch. It can, however, be kept small so that a miniature switch can control a larger switch rated to connect and disconnect the load. simulate this circuit – Schematic created using CircuitLab Figure 1. Simple relay circuits. Figure 1a shows a simple 12 V relay switching the load....


1

7 milliWatts spread over 60 centimeters will not be detected. The heat will be dumped into the chest, and cooled by the blood. Sunlight is 1,000 watts per square meter, or 1,000 watts per 10,000 square cm. We easily sense sunlight, which is 0.1 watts per square cm. Your heat density is 0.1 milliWatts per cm length or about 0.1mW per 1cm*0.2cm (assuming ...


0

LEDs require a certain minimum voltage to turn on at all. For an infra-red LED this would be around 1.5V, maybe a little less. With the button open, almost all the voltage is across the lamp. The peak voltage across the LED is only about 0.33V, so it doesn't light at all. I have used the peak voltage here, not the RMS voltage. The peak is about 1.4 ...


0

Radiation is a negligible component of heat loss for a wire close to room temperature in air. It becomes a factor for long wires in a vacuum, and high temperature differences, but that's not the case here. Convection is the primary source of heat loss for a long wire in still (or moving) air. If it's in contact with skin, conduction will be a big factor. ...


0

What do you mean by a wire? Is it an electrical wire, with an insulation? As you have already said, the temperature will depend on the environment (including air flow), surface area of wire and insulation. Additionally, it will also depend on biological factors like blood flow, electrolyte status, perspiration etc. For biological experiments, keep in mind ...


1

If you put two identical resistors in parallel, the resulting resistance is 1/2 of each resistor. Therefore, twice as much current will be drawn from the power source. So, yes, if you put four motors in parallel, they will try to draw quadruple the current as one. (Actually, not quite 4x as the internal resistance of the battery will drop the voltage to ...


0

Pass 300ma through a transistor with its collector base connected. Use this collectors base voltage to bias 4 similar transistors in parallel as in current mirror configuration to obtain 4 x 300 = 1200ma.


0

You want to measure ESR and also Voltage for State of Charge. There's many ways to do this. A quick short circuit on the 10A scale might be hard to get a steady reading. It should peak then decay slowly but only be done for < 1 second to not waste power. The ESR=Voc/Isc for open circuit voltage and short circuit current. Meanwhile it is heating up ...


0

A 4 ohm 10W resistor is good place to start. Read Measuring Internal Resistance of Batteries Note: Battery Internal resistance has significant influence in battery current. You also could find Questions regarding CR-P2 battery specification. There are few links including the one referenced in the comments that you might find interesting


5

When you select a motor, controller and power supply, you must start with the motor. Then select a controller that is adequate for the motor. Finally select a power supply that is adequate for the controller. For the motor, you first determine the speed and torque that the load requires. The motor that you selected appears to be capable of supplying 3000 ...


2

The motors in these fans are single phase, with a single winding that is reversed each time a magnet passes the Hall sensor (positioned between the poles). Current through the motor is then reversed, either by reversing the connections, or having two windings that are alternately energized. To start up, the poles are asymmetric, and the magnets align ...


3

You put a motorcontroller for 3A on a motor of 5.28A. So better change the motorcontroller


1

However, I've always thought about a wire's capacity by total power transmitted, not just the current. You should think the other way around: A wire should not be selected by total power transmitted, but by the total power NOT transmitted. The total power NOT transmitted is defined by the (undesired) power dissipation in the wire ( \$I^2R\$).


0

Assuming arcs and corona discharges aren't a problem, could I transmit say 10 kW via a #40 AWG wire at 10mA, 1MV DC? The ampacity isn't violated, but that doesn't look right for wire that's around the width of a hair. That's correct, you could transfer a high power through a thin wire and not violate the current rating of that wire. However, the caveat you ...


1

The amount of heat dissipated when you pass a current through a wire is given by I²R, where R is the resistance of the wire. Notice that the heating goes up with the square of the current, and is completely unrelated to the supply voltage. Suppose I have a piece of wire that the manufacturer has rated to 20A, with insulation good for 600V. 600x20 = 12kW, ...


-1

Back ground: So effect(Watt) is only dissipated in ohmic resistance (pure resistance). as the cable heats up / cools down, this resistance changes proportionally - as with "ohms" law, the current is the voltage / resistance. Thought experiment time!: Long cable Think an infinitely long wire - huge resistance, huge voltage drop across said resistance. so ...


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