# Tag Info

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Actually the shunt resistor has low resistance than the ammeter resistance. so most of the current will go through the shunt resistor(because of its low resistance) and only a small amount of current will flow through ammeter. then how the ammeter will damage?. According to me, the shunt resistor has low resistance so most of the current will flow through ...

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Because an ammeter has low 'resistance/impedance'. It is designed to measure the current. A low resistance increase the current passing through that wire. When an ammeter is placed in parallel almost all current flows through ammeter and not through the wire and the resulting current will 'blow a fuse'. Conversely, a voltmeter has a high impedance and is ...

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In series the current has to pass through the shunt resistor as well as the ammeter. If the current rises then the voltage to drive the current through the shunt resistor increases ($V=IR$) - this will reduce the voltage available to the rest of the circuit and the current will drop... In parallel the ammeter has a lower resistance than the shunt resistor ...

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All AC electric circuits radiate. Indeed, in the United States of America, all consumer electronic items must pass explicit FCC Electromagnetic Interference and FCC Radiation Safety assessment for their sale to be legal, so that one can make sure that the electromagnetic radiations emitted do not interfere with other electronic equipment and are safe for ...

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Displacement Current actually does not exist, it is a theoretical misnomer. When we consider a Capacitor as a low Characteristic Impedance Transmission Line we can think of energy flow between the conductors (Parallel Plates) We see a TEM wave (ExH) moving at the speed of light for the medium. What Maxwell thought of as Displacement current is the ...

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If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current. If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current. However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which ...

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The magnitude of the B-field is $a/r$ and circulates around the axis. By symmetry, you understand that the magnitude is zero on-axis. But if $a$ is anything but zero, your expression gives an infinite B-field magnitude. Therefore $a$ must be zero and therefore the B-field is also zero everywhere else inside the pipe. The result also follows from Ampere's ...

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So you have that $\vec{B} = \frac{a}{r^2}\begin{bmatrix}0 & z & -y\end{bmatrix}$ thus the magnitude is $B = \frac{a}{r^2}\sqrt{z^2+(-y)^2},$ where $a$ is unknown. Can you write that as a function of $r?$ Can you investigate what happens as $r$ goes to zero? Are magnetic fields continuous in empty space (a vacuum)? If so, try the next five: What ...

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NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.

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No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.

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I used the right hand rule to see where the forces are and for question 1 they are pointing out from the small loop in the xy-direction, which really would make the small loop stable due to the symmetry! It's true that the force on an inner current element due to the magnetic field of the outer loop is outward. But the magnetic field of the outer ...

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First of all, there is no need to worry over this. The reason we can get away with using conventional flow even though electron flow is what is really happening is that the two are indistinguishable except for in some experiments that are really hard to do. On to your questions. The GND is not an "electron reservoir". First of all, in most circuits there ...

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You are having some misconceptions Wire(more pricisely axle) is not rotated between magnetic field in a turbine generator.Wire(more pricisely axle) is used to rotate an coil (present between magnetic fields) present in generator. Magnetic field itself can not force electrons in a conductor to move untill it is at rest. We need an moving conductor so that a ...

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Both equations are valid. You just made an error in taking suddenly 20 $\Omega$ instead of 40 $\Omega$. $$P = I^2 \cdot R = 25 A^2 \cdot 40 \Omega = 1 kW$$ $$P = U^2 / R = 40000 V^2 / 40\Omega = 1 kW$$ The power is always determind by the current that runs through the part of interest times the voltage drop over it. If there are other parts in a ...

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If I understand your questions correctly: Yes, it can be somehow the other way around. But we do "know": There are two sort of particles in here, one of them has a certain charge and is light, the other has the opposite charge and is heavy. You can then claim that the heavy ones rather stay in place and the light ones sprint around and make up the current. ...

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Generally it is said that current is due to the flow of electrons; how can we make this claim? If the wire is in a magnetic field the moving charges will move in a circle based on the magnetic force. This happens until enough charge imbalance develops on the edges of the wire to produce an equal and opposite electric force. But measuring the voltage ...

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From classical models, the electron and a proton revolve around their mutual center of mass, which approximately lies on the proton itself, because the proton has a significantly higher mass than the electron. This is why electrons revolve "around" the proton, and hence form the outer layer of an atom. Quantum mechanically, electrons could never form a ...

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You are taking a shortcut when you say, "The voltage is zero." Voltage is always measured between two points. In electrical engineering, when we say the voltage at point X is V, we actually are measuring the voltage between point X and an implicit other point called "ground". In the electric power grid, "neutral" is ground, by definition. So the voltage ...

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The key thing to remember here is that we are talking about an ideal wire. This means that there is (effectively) no resistance in the wire, and therefore no voltage drop along the wire. Now if there is a voltage difference between the two ends of the wire, that induces a constant electric field along the wire, causing current to flow.

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Probably in the last case of slit direction of magnetic field is on the side walls upwards as shown and thus there is no interaction of this field with the magnet. Here it seems it is to be considered to neglect radial magnetic field (even though it do exist). In complete loop magnetic field is centralized so it interacts with the magnet making it go really ...

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Doping increases the free electrons (as well as holes) in the semiconductor. So it can increase the current through it (by decreasing the resistance). But it has a limit, over which the overlapped part between "n" and "p" will be damaged. When extra current will flow, the conductor will face Joule heating, so more thermally exited electrons will arise. ...

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The general derivation of Pouillet's Law is given in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance

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The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with ...

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You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$\frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...

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Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ... 1 Spin-resolved current in the context of scanning probe methods means that due to a finite magnetisation of probe and sample the current consists of electrons of one spin in a larger quantity than of the other. Spin current usually refers to current that consists exclusively of electrons of one spin direction. -2 if we consider a current of 6A flowing through circuit we will then get a equation E=6I*R 6IR = 5Ir = 5/6 ohm total current is 10/(5/6)= 12A 0 It is true that accelerating charges radiate energy in the form of electromagnetic radiation. However, the relationship is not a naive ratio between the acceleration and the amplitude of the electromagnetic wave. Rather, you need to invoke Maxwell's equations. Specifically, there is Ampere's law:$$\nabla\times{\bf B}={\bf J}+\frac{\partial{\bf ...

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The situation you posed is an ideal situation or somewhat fantasy in my view.Even though consider a loop with a particular magnetic flux and of course it is not possible for the magnetic field to not be there at the the time it is there so some change in time will be there but infinitely small.This would be the situation when rate of change of magnetic flux ...

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A metal conducts quite well because the there is an electron band that crosses the Fermi level. So, electrons can easily be excited to increase their momentum a bit and consequently move in one direction. Now if you add one electron to the wire, the Fermi level rises. However, you would not be able to see the increase caused by one single electron (or a ...

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The electromagnet is acting as an electromagnetic ring launcher. There is a nice explanation on YouTube here, or for more about the physics see the question Is the standard explanation for the ring launcher incomplete?. The pieces of metal are being repelled from the electromagnet because when the magnet is first turned on the changing magnetic field ...

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Connect the battery to this circuit. At this point, the charge doesn't know that there is a hole in the circuit. Negative charge therefor flows away from the negative battery pole, since it is repelled by this same charge, as far away as it can along the attached wire - this means that the charge will pass through the light bulb, and the light bulb will ...

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Adding extra electrons isn't going to make any difference, at least as a first approximation. Suppose we have a wire with some applied voltage $V$ and some current flowing $I$. What this means is that the power supply is injecting $I/e$ electrons per second into one end of the wire and extracting $I/e$ electrons from the other end of the wire. The ...

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