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1

The I-V characteristics of materials and devices should always be measured at the same thermodynamic conditions, i.e. at the same temperature. Mixing the actual isothermal I-V characteristic with the temperature dependence doesn't lead to any useful data for the purposes of physics (but it is occasionally done in electrical engineering and electronics design ...


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Even if the circuit has no resistance(which isn't possible as every wire has some finite non-negligible resistance $\rho L /A $), the source will have some internal resistance causing the circuit to have resistance, after which you can use Ohm's law to find the current. Though it isn't recommended to connect two ends of a battery together.(It reduces the ...


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if you were to have say a 5V source connected directly to ground, no components in the circuit at all, you would need to add resistance for the current to flow? No, for sure current will flow. If nothing resists the electrons, they will keep speeding up. Current $I$ will keep rising. Power will keep increasing, $P=VI$. For really no resistance, ...


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In a circuit, an electric field is created. This electric field forces electrons to move. As current must remain constant (since no charge build up is observed), the electric field must be strongest in the materials that are the least conductive (or most resistive). As this electric field moves the electrons, they gain kinetic energy. In order to conserve ...


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I will try to explain with some examples. There is never a potential difference when there is equilibrium. You can think of it as a height difference. Think of positive potential as a high point and the negative one as a low point or ground. So there is a height difference. A thing at the high point is bound to come down. Similarly, whenever there is ...


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Incandescent bulbs appear dim because the filament's temperature is below it's design temperature. It's running cool because the current flowing through it is lower than normal. Current + resistance = heat for most non-exotic conductive materials. The current is lower than normal because the applied voltage is lower than normal. As a light bulb's ...


1

In short it's low voltage. Current, voltage and power are all linked via the impedance (like resistance) of the thing in question. The voltage is a property of the grid, with something like a light bulb you can assume the grid voltage won't change. it might if you were running an industrial induction furnace maybe, but not with a light bulb. The current ...


0

Well... When the back emf is equal to the voltage supplied by the battery, it is not really hard or anything counter intuitive to realize how the current exists in such a case. See, all you need to realize is what is actually the back emf? When the charges in motion, tries to pass through an inductor - the inductor converts its kinetic energy into magnetic ...


3

If the emf due to the solenoid is assumed to oppose the applied voltage and have equal magnitude (in volts), there is zero electromotive intensity in the wire. Since current is assumed to be present, this means the current flows even while total electromotive force vanishes. This is possible for wire made of perfect conductor (superconductor). In practice, ...


1

What the voltmeter is reading in the top circuit would be, if you used a Kirchhoff's loop rule in the loop containing 2 resistors and the voltmeter, the difference in the potential difference across the top-right resistor and the bottom-right resistor. The loop rule calculation would look something like this ($A,B,C,\&D$ are the top-left, top-right, ...


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A negative voltage means that you have hooked your power supply across your device backwards. Purely resistive devices, like resistors and lamp filaments, don't care which way current flows through them, only how much current flows. Such devices will always have current-voltage curves which are "odd" functions, with $I(-V) = -I(V)$, as in your graph. It is ...


3

... but what happens if you connect it in series? Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-$\Omega$ and a 200-$\Omega$ resistor (series resistors). The voltage across the 100-$\Omega$ resistor will be 1 V, and across the 200-$\Omega$, 2 V. Next, connect a 1 M$\Omega$ resistor (a voltmeter) in parallel with the ...


3

Heat losses (aka $I^2R$ losses) occur because charge has a hard time getting through the conducting element. In a light bulb the filament is purposely made with a higher resistance, $R$, compared to the resistance of the metallic leads on which it is welded to. Heat is energy and power is the rate at which energy is transfered $$P=I^2R$$ and so with a ...


1

I think it's due to the high resistance of a voltmeter but I don't really see why this would be a problem: It just means that the current going through the circuit is shifted down a level, but all things remain the same right? Ideally, a voltmeter should have such a high resistance that it's equivalent to an open circuit. This means that when ...


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The "work" is the useful rotational pull you get from the motor, and excludes any wasted heat. If only 75% of the energy going in comes out as work, then you need to put in 90/0.75J of energy to get 90J of work out.


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If you have an isolated capacitor, so that there is no conducting path for charge to flow from one plate to the other, then the charge on the plates will be conserved as you change the geometry. Since $$Q = CV,$$ a drop in the capacitance $C$ is matched by an increase in the potential $V$. Note that the stored energy $U$ in the capacitor, $$ U = \frac12 ...


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Any physical quantity, if it can be expressed as a function that has all its' derivatives at a single point, can be expressed as a Taylor Series


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You don't need to use relativity to see what will happen in to a current in a gravitational field. We assume a wire of constant cross section with length in the vertical direction , and a constant current flowing through it. This must be done by applying an electric field $\mathbf{E}$ along the length of the wire, producing the current density (according to ...


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Current is a measure of charge flowing in some unit of time. The gravitational field will dilate time. The current will be reduced in the observer's point of view. $I = V/R$. Since resistance is constant, this means that voltage must have decreased in the observers point of view.


1

For a classical point charge, the field is divergent at $r=0$, and if you were to take the potential to be zero there, it would be infinite everywhere else. Meanwhile, you can approximate $r=\infty$ as the region with no interaction, so it's reasonably naturally to treat it in the way you would treat ground.


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You evidently understand that any constant can be added to a potential without affecting the physics -- or equivalently, any place can be taken to have zero potential. You also suggest, rightly, that there are really only two "natural" places to define the zero of the potential: either $r=\infty$ or $r=0$. For example, there's no particular reason to ...


1

This is accurate, and it ultimately comes down to the fact that we can get arbitrarily close to an electric point charge in classical E&M. That means that the field right up next to the point charge could be arbitrarily large. So you get these huge, singular potentials close to point charges, which is really more-or-less fine. For instance, that huge ...


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Lets get something out of the way first: The threshold, or turn-on voltage, is not really an intrinsic device property per se. It originates more from a desire by circuit designers to have a rule of thumb about how much a diode has to be forward biased to get it into conduction mode. As such, one takes the inherently non-linear current vs voltage response ...


0

But how can the circuits be of absolutely any shape? Circuits can be any shape because you can bend wires. But don't worry I'll explain how the charged react to the weird shape so as to have a nice flow of current. If I have an extremely long, convoluted wire, how can we analyze voltage as simply as we can? A lumped matter model means you treat the ...


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The drop in potential energy is dissipated as heat in the resistor, mechanical energy of the charges is not conserved. In the water analogy, the current is equivalent to Moved volume per unit time, not velocity of the molecules.


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It must be mentioned that there are a lot of variables at play here. The amount of voltage generated by the static shock varies on the method used to obtain the static shock and the materials used. This post is an interesting look at the situation. The resistance of a human is very hard to quantify and varies depending on conditions such as moisture, ...


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The energy in a typical static charge from walking across a carpet is too low to kill a human. It may be a few milliJoules. The energy available is approximately Voltage x Charge. The current that passes through your body depends on that voltage divided by the effective resistance of your body (including any clothing, shoes and other material in the ...



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