# Tag Info

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What is potential difference? It's pretty simple, actually. a charge repels other like charges (with same sign). That means, an electronic repels other electrons. if you at one point have, say, 10 electrons then they will try to move as far away from each other as they can. this point of many electrons (that is, this point that electrons are strongly ...

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if we take a battery and put in beaker and pour a hydrochloric acid in beaker it will definitely react because strong acid soon react with batteries when batteries are cut down

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some times it will explodes mandatory

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The battery terminal appears "small" when you hold it in your hand. But from the perspective within the wire, close to the terminal, the battery terminal appears large. The electric field close to a large object like this is roughly constant (pitcure the constant field between capacitor plates). Further away from the terminal, inside the wire, the charges ...

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No, it will rotate at a speed determined by the load. Witness that the current in, and thus the magnetic field produced by the stator coils is either in-phase with or anti-phase with the rotor current, with the $\pi$-phase change triggered by the split ring commutator. So the torque in each half of the rotor's rotation will throb at twice the AC line ...

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You have two sources of error: error in the source voltage, and error in the current measurement. Since your graph is log-log, at the bottom left corner (small currents and voltages) any errors are greatly magnified; even a 50mA or 100mv error looks big. At the top right corner, however, the (probably of a similar magnitude) errors are swamped by the signal. ...

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At very low voltages (and currents) the power produced is small, so the filament doesn't produce much heat. At these low temperatures the bulb tends to have significant interactions with the surroundings, especially drafts which will cool the bulb in an unpredictable fashion. Another source of error may be the bulb contacts, which may be showing significant ...

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Theoretically, these requirements arise from the way you connect the measurement devices to the rest of the circuit. A voltmeter is connected in parallel, as you said. Say that you are trying to measure the voltage drop across a resistor $R$ through which passes a current $i$. If the internal resistance of the voltmeter is comparable to $R$, then the ...

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I like your question because it puzzled me a lot when I started to learn electronics. Here is the answer. Just imagine a wire. There is a voltage V applied to its ends. What is the voltage at every section of the wire? If it is uniform wire, every meter of wire causes the same amount of voltage drop. Say, you divide the wire into n sections so that there is ...

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The current flows through both resistors. What works for me is to use the analogy of fluid through pipes under pressure. Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The ...

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Consider resistors $R_1$,$R_2$,$R_3$ connected in parallel (voltage $V$ is across them, $I$ is the total current from $V$. and $I_1$,$I_2$ and $I_3$ are currents in each resistor) The voltage across each resistor is same, which is $V$. So current would be same through it, even hundreds of resistances are connected in parallel. So if we add more resistances ...

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The solution is very simple. [EDIT: My diagram is almost similar to yours, just without the internal resistance r. I have talked about r in the last few lines. Anyway my diagram totally logical and valid for your problem.] See, your circuit sums up as follows: Now since the switch is open, you can consider that the lower branch of the circuit has been ...

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In the following situation you have the voltage source ensuring a potential difference V = 60 Volts between its terminals. The source's upper terminal is connected to the switch's upper terminal, so they have the same electric potential. The switch's lower terminal is connected to the resistor's upper terminal, so they also have the same electrical ...

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The problem here lies with the assumption that the resistance of the battery, wire and switch are negligible. There are two solutions that I can think of: 1) Because the two branches are connected in parallel, the branch with the resistor will still have $I_R=\frac RV$ current through it while the switch $S_1$ will have "infinite" current going through it ...

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A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have Q(t) = CV ...

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An ideal capacitor never "dissipates" energy, it merely stores it. The amount of energy stored in a capacitor is given by the formula you mentioned: $U = \frac{1}{2}CV^2$. In the case of the LC circuit, the energy stored in the capacitor moves into the inductor in form of magnetic field energy and then goes back and forth from them. In the case of an ideal ...

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Your question could use a context to be more clear. But I think you mean that the battery supplies a certain $\mathrm{emf}$ to the circuit, and the circuit elements require a certain voltage $V$ for the current to run. Now, if the voltage $V$ over all circuit elements (summed up) is less than the $\mathrm{emf}$ supplied, then some is lost. Meaning, some ...

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Your question makes no sense. Volts is merely a common unit EMF is measured in. To say that volts means a different kind of energy per charge than EMF is just plain wrong, making your question non-sensical.

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To be clear, current is charge passing through a certain area per unit time. This does not imply a second parameter in the denominator of the formula for current dq/dt; just a guideline for how to measure dq. The larger the cross sectional area, the larger the perceived current will be. This is why resistance is seen to decrease for larger cross sectional ...

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Yes, inductance exist in a DC circuit. The problem here is the similarity between the words inductance and induction. Inductance is not about change. In fact, inductance is measured in Henrys, which is a Weber per Ampere. Hence, there is no change. In contrast, induction is about change and does not exist is a DC circuit. You would be amazed how many ...

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What is the difference between the live and neutral wires? LIVE WIRE The live wire is connected directly to the generators of the electricity supply company.It carries current at high voltages (about $220-230\,\mathrm{V}$). NEUTRAL WIRE The neutral wire returns the electricity to the generator after it has passed through the appliance.The neutral wire ...

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The neutral is always at zero potential and it is taken as a reference. It is from this reference we mark the voltage at live wire. In short keeping the neutral always at 0, the potential at live wire varies. Now, the earth wire is another neutral at zero potential. It is nothing but a wire connected to earth on safety lines.

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Both holds true. If you use Ohm's law, you can easily see that $$i_1 R_1 = i_2 R_2$$. So, $$10 \times 1 = i_2 \times 0.2$$ gives $$i_2 = 50\,\mathrm{ampere}$$ Again by power conservation, $$V_{left} i_{left} = V_{right} i_{right}$$ And current in left loop will increase to be $i_{left} = 500\,\mathrm{ampere}$. As you can see, here both Ohm's Law as ...

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If your power supply is sourcing a positive current toward the ground, that corresponds to a flow of positive charge from the supply to ground. This is equivalent to a flow of negative charge from the ground to the power supply. In a real wire, only negative charges can flow, so the second thing will happen: electrons (which have a negative charge) will ...

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Given a particle of charge $q$ placed into an electric field, the equation of motion is $m \mathbf{\ddot{r}} = q\,\mathbf{E}(\mathbf{r},\mathbf{\dot{r}})$. The direction of the electric field comes into play in the difference of potential $V_A - V_B$, according to whether your ground is at a lower or higher voltage (usually grounds are at lower voltages). ...

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First of all, if you are dealing with a network of finite number of resistors, try redrawing it in some form in which you'll be able to recognize the parallel or series connections. Secondly, take a look at Delta-Y Transform which might be really helpful in some cases. If these fail, turn to Kirchoff's laws i.e. put a test generator between the points ...

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Let's say we hook up a capacitor to a battery (and maybe toss a resistor in there). The battery will pump a charge difference between the plates, which creates a potential difference between the plates. When the potential difference reaches the potential difference of the battery, current that takes and adds charge to the plates stops flowing, as current ...

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The overall resistance of the circuit is 8 ohms. $$I = \frac{V}{R} \\ I = \frac{16}{8} \\ I = 2 \textrm{ A}$$ 2 A will flow through the first four ohm resistor, 1.33 A through the 12 ohm and 0.67 A through the 6 ohm resistor. You can now calculate the power dissipated in each resistor by using $P=I^2R$.

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The charge is the same on either side of the two points, but the energy that the charges have is different.

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How do the $\vec{E}$ and $\vec{B}$ look like in dependence of the time? I think Hadrian Evan's animation is good, +1. But there's an issue: the field is the electromagnetic field. The electric field and the magnetic field are "are better thought of as two parts of a greater whole". You mentioned Jefimenko, see this quote: "...neither Maxwell's ...

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QUESTION Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load. SOLUTION Current throughout the circuit with a resistor is constant , no doubt about that. But to be fundamental, current in a circuit is set up by the electric field, not by electrons. For ...

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To maintain a current you need to "push" the charge through any obstacles on the path. If there is resistance against the current, then the "push" must be large enough to overcome this. The potential difference is this "push". Of course, as soon as the resistor has been passed, then a large "push" is no longer needed to make the current keep moving. Now ...

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In the body of the question, you mentioned the limit of R approaching 0. Let's begin from there. In this case we have what is called an L-C Oscillating Circuit. For convenience I will assume that the left plate of the capacitor has charge $q(t)$ such that $q(0) = Q$. Similar to the simple discharge of a capacitor, the upper plate of the capacitor begins ...

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The potential difference or voltage is actually the difference in the level/numbers of electrons or positive charges on the two terminals of battery or across the terminals of passive elements. Law of conservation of energy states that the total energy of an isolated system remains constant, that is why the sum of the potential differences across the passive ...

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To understand the answer, you need to be aware of the concept of electric potential. Electric potential is a scalar quantity. In any circuit, there is a potential at any given point on the wire. The difference in potential between any two points in this circuit is sometimes called potential difference or voltage. You can understand the difference between ...

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You have essentially answered your own question. You are correct that the potential as we call it is a measure of the amount of work that would be done per unit of charge as it is moved through that potential. Electric fields are conservative in that the amount of work that must be put into moving a particle from one potential to another is exactly equal ...

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Yes. there is a very simple method to determine that. If the two components are connected in such a way so that they have exactly one common terminal and no other component is connected to that common terminal then the two components are connected in series. The two components are said to be connected in parallel when they have two common terminals Any ...

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I don't know if this answers your question, but whenever I talk about in-series and in-parallel resistors I like the water analogy. In this analogy the battery is a pump that lifts the water from low (potential) to high (potential). The electric current is the water current, and the resistor is a wheel or a constriction in the pipe. So I have In-Series ...

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When thinking about inductors on a conceptual level, the thing to remember is that they oppose change in current. In other words, if the current, $i$, is dropping, they provide voltage in the direction of that current; if $i$ is increasing, they provide voltage in the other direction (this can be very loosely thought of as resembling "inertia" in the ...

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If your source term is the charged capacitor, I am imagining a switch in the system that you close at t=0, then you can find the current in the system. Of course you care about the current in the system: $$\oint \vec{B} \cdot \vec{dl} = \mu_0 (I_{Enclosed} + I_{Displaced}).$$ $$I_{Displaced}= \epsilon_{0} \frac{d}{dt}\Phi_E$$ And while the concerns above ...

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"Short" answer: You can't really have a short circuit as such in the first place. Superconductors aside, everything has resistance. Even free electrons at rest have a little bit of mass and will resist applied voltage just a wee bit. Put those electrons inside something and resistance can only go up. (Again, superconductivity aside. It doesn't just happen ...

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Every form of energy has inertia. Have a mass $m$ with speed $v$. It won't simply stop unless there is a force that makes it stop. Same way, have a current $I$. It won't stop unless there is a "force" applied. In our nice Earth, friction stops the mass $m$. And, resistance stops the current $I$. But in space, there is no friction, and mass $m$ can't stop. ...

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An ideal short circuit is that circuit element for which the voltage across is zero for any current through. One can think of the ideal short circuit in two ways: (1) as an ideal resistor in the limit of $R \rightarrow 0$ (2) as an ideal voltage source where $V_S = 0$ In both cases, there is current without voltage. This isn't necessarily a ...

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In a circuit that is shorted the current is governed by the internal resistance of the power supply . let's say the supply resistance is 1 ohm and the supply voltage is 35 volts the supply will deliver 35 amps to the shorted circuit. You will not measure a voltage drop across the wire shorting the power supply but there will be a voltage drop inside the ...

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But how can there be current without electron potential (voltage)? In the case where there is no resistance, current (once flowing) does not require any voltage to continue flowing. If you start a current flowing in a superconductor, then even with no applied voltage, it continues to flow. It doesn't take any force to keep a ball rolling if there is ...

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I can theoretically be anything EDIT: Oh, you mean the current,$I$, not the pronoun I.... Ohm's law applies only to components which could obey Ohm's law. It is an observational, phenomenological law. If R truly equals zero, then something else is governing the relationship between the current in the circuit and the potential drop. A filament in ...

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Since lurscher thinks he is so smart but obviously doesn't care to help you with the physics of this problem, I will. Get a square tub and fill in a couple of inches of saltwater. This will simulate a dense conducting mesh. Now make yourself two flat electrodes from copper or steel that are approx. 5-10% of the width/length of the tub. Connect them to an ...

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In class we learned about point charges, electrostatic force, voltage, current etc. and discussed circuits along the way. And then you hopefully learned that voltage isn't a general concept, and that the scalar field is an entirely gauge dependent concept in electrodynamics. I first thought about an electrical circuit as a 1-dimensional "restricted ...

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