# Tag Info

0

To measure the response of your material you need to more clearly define what your input (excitation), and your output signals are going to be. If you decide your input is voltage then your choices for the output are probably limited to either current or displacement. Frequency is neither an input nor an output, but rather a property of the signals being ...

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

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

An ordinary voltmeter has finite input impedance which simply means that, to measure a voltage across, there must be some (tiny) current through the voltmeter. Thus, to measure the built-in potential of a diode with a voltmeter would require that the built-in potential 'drive' a (tiny) current through the voltmeter. But that would require that the diode, a ...

0

This is called the build in voltage (sometimes Vbi) and is used to find the different across the junction under 0 bias. K is the boltzman constant, T is temp in Kelvin, q is charge of an electron in coulombs i believe, Na is number of acceptor ions, Nd is donor ions, and Ni is intrinsic carrier concentration for Silicon usually. If you need the ...

4

1) The case of a moving conductor and a stationary conductor is fundamentally different. When the conductor is stationary, a changing magnetic field produces an electric field everywhere in space, whose curl along any loop enclosing the varying magnetic field is non-zero, given by Curl(E) = -dB/dt .Using stokes law, we easily find the emf to be the rate of ...

1

The mathematical definition that John Rennie gave explains it well, but I'd like to give an intuitive answer to your question. Imagine a rubber sheet horizontally stretched. The height of the rubber sheet at a point is equivalent to the electric potential at that point. Now, since as of now there is no disturbance at all, the height (potential) throughout ...

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

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 ...

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