# Tag Info

72

CPUs are not 'simple' by any stretch of the imagination. Because they have a few billion transistors, each one of which will have some small leakage at idle and has to charge and discharge gate and interconnect capacitance in other transistors when switching. Yes, each one draws a small current, but when you multiply that by the number of transistors, you ...

54

To be accurate, a current shunt must have a specific resistance value. The method used here to get that resistance just right is to build the shunt with a too-low resistance, then trim it by cutting into the bar until the measured resistance (or rather, measured voltage output for a known current input) increases to the wanted value. It would also be ...

49

It's the watts dissipated in the fuse itself not the watts in the system. Therefore since the fuse has resistance (R) it's the current, which provides that power I^2*R. The voltage has nothing to do with it : at 6V, 12V or 240V, the fuse still blows at 20A. However you cannot use a low voltage fuse in high voltage applications : it will still blow at (...

45

You placed an ammeter in parallel to a voltage supply, which created a short circuit through the meter and most likely blew the 10A fuse in the ammeter socket. What you want to do is put the meter between the charger and the battery, and then connect the other end of the battery to the charger. Diagram will help: simulate this circuit – Schematic ...

40

A wrench does not have a resistance of 0.5 ohms, it's way lower. Your basic multimeter cannot measure resistances to better than an ohm or so, the resistance of leads, and the unreliability of contact resistance make it impossible. The way resistances as low as a wrench are measured is to use a 4-terminal Kelvin method. What you do here is to pass a ...

30

The components that make up DC adaptors (inductors, transistors, capacitors, diodes, ect) are all rated for a certain current and/or power dissipation. Components that can handle 1000A vs. components that can handle 5A are orders of magnitude apart in cost, size, and availability. For an example let's look at an inductor that could be used in a 1000A ...

27

Imagine the situation where you have a new card you have designed, never been powered before... You plug it into a 28V supply and set the current limit to 5A because "well it will only draw what it needs to work". Now ideal case it should only draw 0.1A but you now have hooked it upto a 140W supply. There is an unknown problem in your design be it Circuit ...

27

There is no real inherent distinction between breaking one side of a loop or the other side- it's all in series so breaking the negative or the positive side of the supply keeps electrons from flowing. When you have an electronic switch and are turning off part of the circuitry with other circuitry it's easier to break the negative - called a low side ...

25

understand that voltage is relative to ground, I prefer to disagree. A voltage is against a reference point. Often that reference point is ground but not always. Taken the above into account your current is defined the same way. Take a pin/port of a component or circuit. You can now define the current going into that port/pin as positive from which it ...

22

This is one way of doing it. Ref A normal transformer can't deal with DC currents. Therefore the operating principle of DC current probes is rather different from AC probes. Here also is the current carrying conductor the primary winding and is inserted through the core opening. There is also a secondary winding, but now it functions as a compensation coil. ...

21

To dissipate $1\mathrm{kW}$ at $14\mathrm{V}$ you need a resistor with a resistance of $R = \frac{\mathrm{V}^2}{\mathrm{W}} = \frac{\left(14\mathrm{V}\right)^2}{1000\mathrm{W}} = 0.196\mathrm{\Omega}$. You can buy a $0.25\mathrm{\Omega}$ $1 \mathrm{kW}$ resistor on Digikey for \$54.95 (Part no. FSE100022ER250KE). Using two or 3 of them in ...

21

2.9 ohms (measured with power disconnected through circuit) As the lamp heats, its resistance rises sharply. Ohms law is never wrong except when misapplied.

20

I'll answer your first scenario only, because I think it's more interesting. Let's dissect: the battery I'm using can only provide 20 amps But what that really means? What would happen if you try to pull 21 amps? If it is a smart battery, some sort of overcurrent protection might kick in. If yes, then you have the answer - the motor will just not run. OK, ...

18

WARNING: If "trying this at home" be aware that there is a small potential for significant hazard occurring - see below. What you propose is a viable and useful and potentially dangerous method. I consider it is extremely unlikely that you will harm yourself doing this but need to note that the possibility exists. I would not do this with other than ...

18

So if a fuse is rated for 12V DC and 20 Amps, this would be equal to 240 Watts. If a different voltage is supplied, will this change the Amps at which the fuse will break? Does the fuse technically 'blow' at 240 Watts? All the fuse knows (before it blows) is the current passing through it. This might be: - 20 amps from a 1 volt supply feeding a 0.05 ohm ...

17

A 240 volt motor will only be connected to the two live wires, not to neutral, so your example draws 6.6 amps at 240 volts. If it is connected to operate at 120 volts, it will be connected between one live wire and neutral, and will draw 13.2 amps at 120 volts. I think it is misleading to say that, in the 240 volt case, the motor draws 6.6 amps from each ...

17

According to Wikipedia, top CPUs released in 2011 had some 0.5 to 2.5 billions of transistors. Assuming a CPU with 1 billion of transistors consumes 64A of current, the average current is only 64nA per transistor. Considering operation frequencies of several GHz, it's actually surprisingly little.

17

Meters are often left connected to voltage sources for purposes of monitoring them. Further, they often use the same knob for both mode selection and power on/off. If a meter uses the same knob to turn the meter on and off and control the mode, it may be very easy for someone turning the knob to accidentally turn it to a current-measurement mode while it ...

16

First, it highly depends on the battery. Some cars have much beefier batteries, measured in Amp Hours. We arn't even talking about Electric Vehicle battery banks which are massive. Then it depends on the type of battery. Some chemistries are different. Some are 24V instead of 12V. Some cars have more than one. Etc. That said, the normal peak current is the ...

15

I think people are distracted by the PMOS, but the real reason is a voltage spike when the 24V is connected, due to inductance of the wiring that is even higher when the multimeter is connected, vastly exceeding the max voltage of the LDO. I assume you are using a ceramic capacitor at the LDO input, which is a reason someone may be experiencing this ...

15

It's a moving iron meter hence this symbol: - Taken from here Moving iron meters are used for DC AND AC so, the meter needle will only move in a clockwise direction irrespective of current polarity. If it used a magnet it would be polarized and with AC, the needle would barely move (because the needle would try and move clockwise and counter-clockwise at ...

14

Your hunch that batteries have a current limitation is correct. In general, it's hard to tell the current rating [A] from capacity [A·h]. You have to look it up in the datasheet. A lot depends on the design of the battery. For example: coin cells with 500mAh capacity may have only 3mA max current. Another (opposite) example: automotive starter ...

14

Switch damage due to excess current goes one of 3 ways: 1) If the damage occurs as the switch is opening (which is the common failure condition when dealing with motors and solenoids) the switch contacts burn out, and the switch will no longer close the circuit. 2) If the damage occurs as the switch is closing (which is likely when switching resistive or ...

14

This all brings me to wonder, how much of what we use is calibrated to someone else's meter. And how many of those meters, are calibrated to yet someone else's.. and on and on. Seems like one big house of cards. and... Is there some sort of certification sticker one should look for when buying a meter or other equipment that indicates it is indeed ...

14

That is how the shunt is calibrated for accuracy as it reduces the area to pass the current.

14

In general the current rating of a wire is limited by the wire's ability to dissipate heat to its surroundings without getting too hot. That in turn depends on the maximum safe wire temperature and the thermal resistance to ambient. Unfortunately figuring out exactly what the thermal resistance to ambient is for a given installation method is nontrivial. ...

14

There are three clues: The scale isn't linear so that implies that it is a moving iron meter. The ≂ symbol indicates that it can handle DC and AC. (The '-' indicates DC and the '~' AC. The terminals are not labelled. If polarity mattered they would be. The meter can be wired either way. The moving iron works on the principle that the iron place near ...

13

If you want to avoid cutting or disconnecting any wires, then you would use a clamp meter. The clamp is put around one of the wires of the circuit to measure the current through that wire. However, this requires being able to access a single wire rather than both wires in the circuit; if the clamp meter were put around both, the opposite currents would ...

13

I agree with VillageTech in incriminating the PMOS. It has a capacitive load in the source and an inductive load in the drain (your mutlimeter wires), add a bit of parasitic inductive/capacitive coupling and you get a colpitts oscillator. However, since your supply is 24V and you're using a LDO to make 5V at 80mA, I wonder why you need to use a PMOS to make ...

13

A wire that will carry 15A without overheating, will carry 15A whether it is 1 meter long, or 1 km long. What will change is the resistance, and resistance is directly related to voltage drop. If you are powering a 120V load that draws, say 10A., if your wire has a resistance of 0.100$\Omega$ (half that in each direction), then there will be a voltage drop ...

Only top voted, non community-wiki answers of a minimum length are eligible