# Tag Info

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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}}}\... 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 ... 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 ... 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 ... 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 ... 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. 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 ... 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 ... 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 ... 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 ...

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 ...

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

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

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.

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