# Tag Info

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 ...

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 ...

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

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 ...

0

I am still not sure of your problem physically. Mathematically, it seems to be about finding the solution to: $$\nabla^2 \Phi = \frac{J}{\sigma} \delta(\vec{r})$$ with $\Phi(\vec{r}=R\vec{e_r})=0$ for $\vec{e_r}$ being the boundary condition. And you are right, this can be solved using the method of a Green's function, the one obeying the auxiliary ...

2

Here is an answer I propose for the 2nd way. I'm for sorry for the bad paintings.

2

Your error is to assume that only your red charges generate the heat, ie the red charges go through area $A$ and they are not replaced by any other charges. If that were the case then the factor of $\frac 12$ would be correct. However as the red charges move through the resistor black charges to the left of the red charges would move into the resistor and ...

1

Since you equate $W$ with $NqU$, that means $W$ represents the amount of energy dissipated as heat during the time interval $N$ charges passed through the cross section. That time interval is $t / 2$, so the resulting power is $P = {{UIt / 2} \over {t / 2}} = UI$.

0

What happens during the charging process is this: During charging, an external electrical power source (the charging circuit) applies an over-voltage (a higher voltage than the battery produces, of the same polarity), forcing a charging current to flow within the battery from the positive to the negative electrode, i.e. in the reverse direction of a ...

0

Resistive systems are not conservative. This can be seen by the fact that the system experiences a potential drop whichever direction current passes through a resistor. The current is essentially experiencing a drag force which cannot be written as the gradient of a potential function; you can't 'draw' the potential landscape for a resistor, so it won't ...

1

In the series connection the current (charge passing a point per second) is the same every where because there are assumed to be no sources or sinks of charge and so the charge is conserved - the amount of charge entering is equal to the amount of charge leaving. This is Kirchhoff's current law and there is certainly no accumulation of charge rather the ...

2

It doesn't come only on one side. Suppose I define a new pair of constants $K_1$ and $K_2$, which obey: $$R = \frac{K_1}{K_2}$$ Then I can write: $$K_2V=K_1I$$ And I have a constant on both sides. For convenience, we usually just collect any constant values into a single constant, give it a name, ideally an intuitive one, and put it on one side. Which ...

1

So, lets step back a bit. First lets look at avalanche breakdown. Electrons are constantly scattering, off atoms and other electrons, with some average scattering rate under given conditions. In a semiconductor, avalanche breakdown occurs when the field is strong enough that a free conduction electron gains (through accelerating in the field) a threshold ...

1

Charges at rest move when a force is applied on them and this is due to Newton's laws. Now to apply a force, we need a field, like electric/gravitational field. Each field acts upon certain measurable properties of a system, like gravitational on mass, electric on charge etc. Now potential is just a fancy name of height in electromagnetism. I hope you're ...

0

The power will remain the same for a particular load as we are not changing the load. so if we increase the voltage, the current will decrease to make the net power consumed by the load same as before. If we increase the current, the voltage will decrease for making the power same. The power will only change when we changes the load.

2

Start by noting that the electrical potential is an energy per unit charge. In an electric field $E$ the field produces a force on a charge $Q$ of: $$F=EQ$$ so if we move the charge a distance $dr$ the work done is just force times distance or: $$W=EQ\,dr$$ The work done per unit charge is $E\,dr$, and this is what we mean by the change in the ...

Top 50 recent answers are included