# Tag Info

0

The statement "On the cylindrical surface $\mathbf{J}\cdot\hat{n}=0$..." refers to just inside the wire so $\sigma\neq0$ and $\mathbf{E}$ cannot be whatever it wants.

0

Based on your drawing, with the batteries all in parallel, the resistor will have the same voltage across it whether you have one battery in the circuit or more than one battery in the circuit. Assuming that each battery produces the same emf, this will give you one value of current through the resistor, regardless of how many batteries are in parallel. ...

2

According to Maxwell equations, steady currents and steady charge density won't produce EM waves. So if you have a steady current loop, you can calculate its magnetic field just by the Biot-Savart law, which gives an steady magnetic field in space. If a current loop would radiate energy, then it would be impossible to produce persistent currents in ...

9

If you consider that electric current is actually the flow of individual charged electrons, then as John Rennie pointed out, the radiation exists but is negligibly small. But if you were to imagine breaking the current into more and more point particles with less and less charge while holding the linear charge density $\lambda$ fixed, then the radiation ...

4

To back up John Rennie's answer, consider the Bremsstrahlung formula for velocity perpendicular to acceleration: $P= {{q^2a^2\gamma^4}\over{6\pi\epsilon_0c^3}}$. For all practical purposes $\gamma=1$, so we can simplify this to $P\approx ({q \over \mathrm{C}})^2 ({a\over \mathrm{m/s^2}})^2 {1\over{18.85\times 8.85\times 10^{-12}\times 2.7\times 10^{25}}}\... 27 Circular currents do produce EM, and indeed this is exactly how X-rays are produced by synchotrons such as the (sadly now defunct) synchotron radiation source at Daresbury. In this case the current is flowing in a vacuum not in a wire, but the principle is the same. Current flowing in loops of wire don't produce radiation in everyday life because the ... 0 The ampere is used, among other things, to derive the volt, which is the electric potential difference that a current of one ampere has to flow across in order to deliver a power of one watt. (The watt itself is defined mechanically, from second, meter and kilogram). One volt is around the same order of magnitude as the potential differences commonly ... 0 Electric current is due to the flow of free/conduction electrons in a conductor. By convention the direction of current is chosen as in the direction opposite to the flow of electrons. At any instant, the number of electrons leaving the wire is always equal to the number of electrons flowing from the battery into it. Hence, the net charge on the wire is ... 2 The range of a given voltmeter can both be increased and decreased. We know that for converting a galvanometer of resistance$G$into a voltmeter of range$V$, a high resistance$R$given by $$R = \frac{V}{I_g}-G$$ has to be connected in series to the coil of the galvanometer. In order to increase the range of the voltmeter, R has to be increased. It ... 2 It is the same thing.$V$being proportional to$I$means that$I$is proportional to$V$. But the proportionality constants are different, so the law might have been set up in the simplest possible way as $$V=RI$$ instead of $$I=\frac{1}{R}V$$ Direct proportionality is seen in both cases,$V\propto I$and$I\propto V$, just with different ... 0 Symmetry. At the left hand node the current can go through either the$2\Omega$or the$4\Omega$resistor and then the rest of the circuit,$\frac{2}{3}\Omega$,$2\Omega$and$4\Omega$. At the right hand node the situation is exactly similar in that the current comes from an identically arranged$\frac{2}{3}\Omega$,$2\Omega$and$4\Omega$, and then ... 0 The short answer would be the empirical Matthiessen's rule: the total resistivity of a crystalline metallic specimen is the sum of the resistivity due to thermal agitation of the metal ions of the lattice and the resistivity due to the presence of imperfections in the crystal (scattering). There are of course deviations from that rule: it assumes that ... -2 Your source voltage is a pressure of electrical charge, with a degree of resistance (obstruction) somewhere in-between zero and infinity, towards a lesser pressure of electrical charge. If the source pressure is not kept constant, by electrical input to the source, then this source pressure will diminish for as long as your current is kept constant. So any ... 1 No. The battery is already neutral, and remains neutral during operation. (-ve charge leaving one terminal has to be replaced at the other terminal.) If some -ve charge flowed to Earth the battery would become +ve, attracting electrons back to it. 1 Technically, you can neutralize the electrostatic potential of the entire battery this way. However, batteries do not primarily work by electrostatics. They work by creating a potential difference between the two terminals which encourages electrons to flow out of one (the negative side) and into the other (the positive side). This encouragement is ... 0 Regarding your question about a pole position error in either of the pictured electromagnets. No they are both correct. If you visualize wrapping the fingers of your right hand around those electromagnets in the direction of current flow, you will find your thumb points to N pole in both cases. You do have to look carefully at the left hand pictured ... 3 It depends on the sense of circulation of the current in the wire of the solenoid. If you take the solenoid in your right hand so that curved fingers copy direction of current in the wires, the thumb will show direction of magnetic field inside the solenoid. It thus points to the "north pole" of the electromagnet. This picture may help to understand this ... 0 Imagine that instead of a solenoid you have only one loop, apply Lenz law, that would give you the direction of the magnetic field lines generated by the electric current. Compare the field lines' pattern to the one which corresponds to the magnetic field lines of a natural magnet. Then change the direction of the current in the coil, and do the same ... 1 As far as I can see the steps you can do are the following: Find voltage over$R_3$with Ohm's law. Find voltage over the parallel portion with Kirchhoff's voltage law. Find current through$R_2$with Ohm's law. Find current through$R_1$with Kirchhoff's current law. Rind resistance$R_1$with Ohm's law. In short the laws... Ohm's law (for a component)... 1 The picture is correct. By the passive sign convention, the reference direction for current is into the positive labeled terminal of the circuit element and thus the circuit element is absorbs (not necessarily dissipates) power when the product of the voltage across and current through is positive. However, the reference direction for$I_S$is out of the ... 3 If you remove all resistors the voltage drop will be across the wire. (Because the wire probably has a very small resistance the current through the wire will be very big and the wire will get very hot). if there are resistors in series connected by wires, the resistance of the wires is usually neglected. You can easily see that this is reasonable because ... 0 We don't know the electirc potential of the individual plates, right? Yes, we do. Or, that depends on what we want to know. Electric potential is just the potential compared to some other point - some arbitrarily chosen reference point. If for example, you chose one of the capacitor plates as the reference, then this plate has an electric potential of$0\...

0

To know the electric potential of the plates you first need to define where and if there is a point with zero potential. This point is arbitrary. Once you defined that, you can calculate the potential of the plates. For instance, if one of the plates of the capacitor is defined to be at zero V, then the other plate will be at 2V. If you define it that way ...

0

Let's first look at cases where this is not the case. When the lamps are connected to a constant current source, current is indeed "divided" over the lamps in parallel. Assuming equal resistance R, the bulbs will both see a current of I/2, and the dissipated power in each is IÂ²R/4 or in total IÂ²R/2. In the case of the bulbs in series, the current I will ...

0

Lets start with two light bulbs, 120v AC, 60W each. If you connect them in parallel to the mains, each receives 120v, 0.5A (= 60W). If you now connect them in series, each light bulb now receives only 60v (due to voltage division), and assuming the same current (0.5A), it only receives 30W! Therefore, if the light bulbs light up at all, they each only has ...

0

The main question is unnecessarily complicated by alluding to a phone battery and its battery pack. Concentrating strictly on two "plain" batteries, one being charged to 5% of it capacity and the other charged to 35% of its capacity. The implication is that the one with the larger charge can charge the one with smaller charge. This is not necessarily true....

0

I may add a little bit of chemistry in the hope that it would be of some use to the physicists and engineers discussing the charging process of cell phone batteries using backup power source batteries. Betteries are devices that transform chemical energy into electrical energy and vice versa. The so-called secondary batteries operate in both directions ...

5

Voltage is not any part of this explanation. The answer is that each battery pack stores a certain amount of energy. This is measured in joules. At its most basic level your phone battery has a certain capacity in joules, you external battery bank also has a capacity in joules. When you charge the battery you are transferring a certain number of joules from ...

22

Connecting your phone to the battery pack doesn't directly connect the cells in parallel. I assume this is where your guess of an equilibrium with equal voltage -> equal charge percentage comes from. Shorting lithium-ion / lithium-polymer (LiPo) cells together like that would likely cause one or both to literally catch fire from the high currents, or from ...

0

In case of the battery packs, there is much lighter weight requirement, and also much smaller development / manufacturing costs. But it is important to be bigger (in the sense of Ah). If you fill a cup of tea from a large jug, the cup will be full (100%) while the tea level in the jug decreases only a little bit. The Ah capacities of the batteries are ...

2

I may be wrong, but I think Lenz's Law might provide an answer. The circuit with the straight wire takes in current i(suppose) once the switch is closed. The one with the looped wire, will having a changing flux through it once the switch is closed. Since any change is to be opposed, the current drawn this time will be less,(assuming the dimensions of the ...

3

I tend to agree with Sanya in that I am not sure about the universality of this. There might of course be instances where this is the case. A pure metal has a periodic lattice of ions. There is then a conduction band of electrons that fills the space between the ions. These electrons have wave vectors in the reciprocal space. In space the occurrence of ...

5

For an iPhone the battery voltage is a nominal 3.8 V and the battery pack would probably replicate the 5 V output voltage of a USB power supply. So the battery pack would be discharged as it was driving current into the positive terminal of the phone battery and thus recharge the phone battery. So only when the battery pack voltage was less than the ...

24

The key here is the voltage of both the batteries. The battery in the phone is generally at a voltage of 3.7V. The battery pack has a higher voltage or a circuit which gives a voltage of 5V to your phone. So, as long as the voltage with which you charge the phone is higher than that of the battery, the percentage of power in it doesn't matter and the phone ...

2

First of all, I want to see an (experimental) proof that any metal has a higher resistance than any alloy (at any pressure, temperature and volume). What I presume your teacher might have wanted to hear is something along the following lines: a perfect, perfectly static crystal would be, if I remember correctly, perfectly transparent to an electron, so there ...

29

Sometimes it is easier to understand circuitry in the context of water. What you're imagining is two tanks of water of equal size linked together by a pipe that has been sealed off. If one tank holds 5% water and the other holds 35% water, when you remove the seal, the tanks equalize and you end up with 20% in both tanks. What you're forgetting is that ...

1

Okay I'll have a go at answering this, although it may be a make belief scenario. Looking at where you got your inspiration from, he stated using a magnet as a core for an electromagnet, so your curiosity must have piqued from the idea that the wire carrying electric could itself be magnetic. Because of polarization and magnetization, all the magnetic ...

1

I'm don't know if you will find this answer satisfying, but suppose the EMF went the opposite way. Instead of opposing the current, it boosts the current. Then the higher current will produce a higher field and higher EMF which will boost the current, which will produce a higher field and higher EMF ... until the wire melts. Lenz's Law established ...

1

you are basically trying to undergo a transition from a law which is valid for static charges(or non relativistic speeds) to one which is valid for steady currents. That is why, simple differentiation is erroneous and does not include any magnetic field term in dE/dt.

2

It is like water in a hose. If the hose is full of water, water flows out the end immediately when you turn on the faucet. A drop of water at the faucet pushes a drop next to it, which pushes the next drop. Water doesn't flow that fast. If the hose is empty, it takes a while to reach the end.

0

Neutral is a circuit conductor that normally carries current back to the source, and is connected to ground (earth) at the main electrical panel. In the electrical trade, the conductor of a 2-wire circuit connected to the supply neutral point and earth ground is referred to as the "neutral". A difference can occur when either current is flowing down the ...

Top 50 recent answers are included