# Tag Info

## New answers tagged current

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The charging current warms the battery as it charges, and because of mass-energy equivalence, an unmeasurably small amount of mass is indeed added to the phone in the process. However, this is not what is measured by your electric utility in order to bill you for energy. The insignificant mass added to your phone when charging is lost partially when the ...

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But what happens with the current when the resistance equals 0? I mean... if Current = Voltage / Resistence ( 0 ) Couldn't be divided by Zero ._.

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Do you mean a coil in which conductors are made of ferromagnetic material instead? If that's the case, you would probably not be better off: the field which is created is concentrated around the conductor (mostly at core of the solenoid) and in another direction, so it would not benefit from the increased magnetic permeability (which is what ferromagnetic ...

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I think you have some misconceptions about voltage. You mention "net voltage" but voltage is always a difference in electric potential. In a circuit, that means you never talk about a "net voltage" or a voltage at a certain point. Voltage is always meant to be read as "the voltage between two points" or "the voltage at A with respect to B." It is never ...

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The net voltage around the entire circuit is zero. However, the voltage across the resistor is equal in magnitude and opposite in sign to the voltage inside the battery. You could view this conceptually as electrons flowing from the minus to the plus terminal (through the resistor) while the internal mechanism of the battery produces more electrons at one ...

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Draw the graph of $\frac 1 x$ . You can see that it is a decreasing function for positive $x$. Hence conductance decreases as resistance increases. We could have defined conductance as any other decreasing function also but $\frac 1 R$ appears in many equations so we defined it that way. You might want to look at derivation of $J=\sigma E$ to get better ...

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The electrical resistance of an electrical conductor is the opposition to the passage of an electric current through that conductor; the inverse quantity is electrical conductance, the ease at which an electric current passes. Consider resistance of $0.0001$ ohms, what is $\frac{1}{0.0001}ohms$? It is equal to $10000$ siemens. I hope this helped you in ...

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Conductance is indeed defined as the inverse of resistance. $$G \equiv \frac{1}{R}$$ To see what this means physically, consider that when the resistance is large (i.e. it is "difficult" for current to get through), then conductance is low, and when resistance is small (i.e. it is "easy" for current to get through), then conductance is high. There's an ...

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Faraday's law could not be explained from basic principles at the time of discovery. In that sense, it doesnt have an explanation. It was incorporated as a new law of nature, and included in what today are maxwell's equations.

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First, your force equation is wrong, as you're missing the electric field. Wait what electric field? That's the point! A changing magnetic field induces an electric field $\nabla\times E=-\frac{\partial B}{\partial t}$, and this "pushes" the current. Note that the applied magnetic field is perpendicular to the circuit/wire, so that at least part of the ...

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It is showing the direction of current flow reversed through the meter with the reversal of the leads.

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A Short answer is here, Firstly the capacitor gets charged.Though it takes infinite time to reach the battery potential the current is reduced to considerably low values. So obviously bulb won't glow. Note:As capacitor gains more and more charge its potential increases finally reaching the value same as the battery.So potential across bulb tends to zero. ...

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You can't talk about "correct" without a definition of what you are trying to achieve. However, you can rule out some obviously bad arrangements if the placement of the meters keeps the circuit from working, even though we have to guess at what the intended "working" is. This is a poorly specified problem, or there is context surrounding it that you ...

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The ammeter actually creates a short across the battery in the second figure (so that one is certainly wrong). Very likely, this would cause damage to the ammeter, or at least result in a blown input protection fuse. I think the other diagrams are OK, depending on what it is you are trying to measure (which is open to interpretation). This answer assumes ...

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It's quite opposite. All are correct except the second one. Ampere-meter should be positioned in series configuration and voltmeter should be positioned in parallel configuration to the measured element. Notice that voltmeters should have very high resistance, so that most of the current will flow through the measured element. The bigger resistance of the ...

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Actually is the other way round, the wrong one is the second. Ampere-meter has to be connected in series and Voltmeter in parallel. Think that Ampere-meter measures current so it has to be "a part of the circuit" as opposed to the voltmeter which measures differences of potential, so it has to be connected to two point of the circuit. Ideally Ampere-meters ...

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in welding a plasma is created, it's a mix of ions, electrons and atoms. alltogether they are a neutral mix. once you get plasma you get a ton of UV coming out of it, very dangerous to eyes not only on the direct contact, but also via reflection from other objects. in your case, I still don't know what exactly is the device you are creating. It sounds like ...

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All materials emit thermal radiation (such as light). The hotter the material, the more the radiation is shifted to high frequencies (shorter wavelengths). The radiation comes from oscillating electrons (regardless of whether there is an electric current). Welding reaches temperatures high enough to cause significant emission of UV light. Oxyacetylene and ...

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From a microscopic point of view you can image metal (conductors) in a lot of different ways. The easiest model is the Drude model in which atoms are fixed in the space and everyone have one or two (in a metal) free electrons. When you apply an external electric field this particles move as a consequence of Coulomb force. It's important to say that electric ...

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Yes, until energy is conserved. Kirchhoff's law is amongst great basic laws which help in constructing electronic physics. Kirchhoff's law is based on the Law of Conservation and we also know that energy is always conserved according to our latest experiments and analysis. If in future it gets false then we must once have to think about the applicability ...

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Imagine identical charges moving through a conductor. Said conductor has a cross-sectional area $A$. The volume of an element of length $\Delta x$ of the conductor is simply $A \Delta x$. If $n$ represents the number of mobile charge carriers per unit volume, the mobile charge $\Delta Q$ is given $\Delta Q = (nA \Delta x) q$, or the number of carriers times ...

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Ohm's law is not quite precise in our daily life. It varies according to external factors such as temperature, pressure, material of wire used. There are also errors while calculating it. You have maintain standard temperature and pressure (i.e. 273 K and 1 atm). You should also be aware that the conductor whose resistance you are calculating must not get ...

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Well, the thing with Ohm's law is that the ordinary expression $I=\frac{V}{R}$ (sorry, I'm on mobile and writing "nice math" here is a pain), works only for what is usually referred to as "static resistance", that is $R=\frac{V}{I}$. However, this implies a material whose resistance does not vary. There are some materials whose resistance varies inversely to ...

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Ohm's law isn't actually a law, in the sense that everything has to obey it. It's more like "This simple pattern that Ohm noticed". Devices exist that don't even come close to obeying Ohm's law, and they're used all the time. Ohm's law basically states that how well something resists current doesn't depend on how big that current is: the device isn't any ...

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The first equation is the definition of power which is, in words, the rate of energy conversion: $$P \equiv \frac{dE}{dt}$$ Where it is understood that $E$ is the amount of energy converted. For example, in a mechanical system where gravitational potential energy is converted to mechanical kinetic energy and vica versa. In electrical circuits, the power ...

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Picking up on jinawee's answer, current is charge per unit time: $$I = \frac{Q}{t}$$ So substituting for $I$ in your second equation gives: $$P = V \frac{Q}{t}$$ But $VQ$ is just the work done, i.e. the energy, in moving a charge $Q$ through a voltage difference of $V$. So substituting $E$ for $VQ$ gives us: $$P = \frac{E}{t}$$ which shows that ...

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Power is defined as: $$P(t)=\frac{dE(t)}{dt}$$ This is valid for any system. If energy is constant, then: $$P(t)=\frac{E(t)}{t}$$ If you're dealing with a resistance in a circuit, the dissipated power is given by Joule's law: $$P=VI$$ So the last one is a particular case of the first one.

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Ampere's law says: $$\oint_C\vec B\cdot d\vec\ell=\mu_0I_{\rm{enc}}$$ To get the result you mention we apply cylindrical symmetry to the problem and see that $\vec B\parallel\vec\ell$. Then we get the simple case where the LHS becomes $$B\int_0^{2\pi}rd\theta=2\pi Br$$ which gives the expression you mentioned above. If the symmetry condition is relaxed, ...

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$20V$ is not normally dangerous. You may not even feel it. It is not the voltage that causes danger, but rather the current it generates through your body. Anything over about $10mA$ will be unpleasant, above $50mA$ it gets dangerous. The current is determined by both the voltage and the resistance of the body. That resistance depends on how you touch the ...

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This was a comment but it got too long.. With the level of information you can provide this is not a physics question at all. 10 kWh over 48hr may be very high or very low consumption, unless you have traced the circuits you don't know what that meter supplies. The fact you run out of hot water tells us nothing useful at all about power consumption, it ...

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