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17

In a superconductor, the current can keep flowing "forever" since there is no resistance. But since conductors have inductance (in fact, superconductors are used most often to create magnets like for an MRI scanner), applying a voltage would not (immediately) cause an infinite current to flow. It is instructive to see how an MRI magnet is "ramped" (turned ...

6

If they have 0 resistance then I (V/R) should be infinite? According to Ohm's law, the voltage and current associated with a conductor are proportional: $$V = R \cdot I$$ where the resistance $R$ is the constant of proportionality. This equation holds for an (ideal) ohmic material. We can rearrange this equation to be $$I = \frac{V}{R}$$ except ...

5

The voltage is zero. That's the point. The main way current gets started, like in an NMR magnet, is by inductive coupling.

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The equation: $$\frac{dV}{dI} = R$$ is a definition of $R$. Ohm's law is the statement that $R$ is constant over all voltages and currents (with $I = 0$ when $V = 0$), thereby giving: $$V = IR$$ With this definition, it is all but impossible to say $V = IR$ for any electrical component other than Ohmic resistors. Consider the Shockley equation: $$I = ... 3 Heat losses (aka I^2R losses) occur because charge has a hard time getting through the conducting element. In a light bulb the filament is purposely made with a higher resistance, R, compared to the resistance of the metallic leads on which it is welded to. Heat is energy and power is the rate at which energy is transfered$$P=I^2R$$and so with a ... 3 ... but what happens if you connect it in series? Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-\Omega and a 200-\Omega resistor (series resistors). The voltage across the 100-\Omega resistor will be 1 V, and across the 200-\Omega, 2 V. Next, connect a 1 M\Omega resistor (a voltmeter) in parallel with the ... 3 First of all, notice that (for the initial 4 terminal resistor), the port potential V_i is equal to the v_i in you first image, but the port current I_j is equal to -i_j or i_{(j+1) mod 4} and what you call resistance matrix R is generally said impedence matrix Z (because impedance parameters are calculated under open circuit conditions Z_{nm} ... 3 The resistance of a lamp filament is not constant. As the current increases and the filament heats up the resistance increases. That's why the statement you quote is phrased that way. It means that the resistance is 4 Ohms when the current is low enough that the heating and resistance change are negligable. More verbosely it could be phrased: The limit ... 3 Resistance is Voltage per Current. R = V/I or V = RI if you like. So if you put two resistors in series, the voltage is that due to the first, plus that due to the second. You understand that perfectly. Now, what is the inverse of Resistance, Current per Voltage? Give it a name, call it Conductance, perhaps. C = 1/R = I/V Then I = CV if you like. ... 3 It perhaps is not as rigorous as you want, but it is simple and intuitive:$$ R = \frac{\rho L}{A} $$Where \rho is the resistivity, L is the length, A is the cross-section area. When you plug resistances in series, you "are" increasing L, and thus R increases. If you put in parallel, you "are" increasing A, and thus R decreases. 3 Use kirchhoff's first law, so for two resistors in parallel:$$ I_\text{total}=I_1+I_2 $$Then just use I=V/R$$ \\\frac{V}{R_\text{total}}=\frac{V}{R_1}+\frac{V}{R_2} $$The voltage across any component, whether it be across resistor 1, 2, or the whole parallel portion of the circuit, is the same. It just cancels out so you can divide both sides by V. 2 This is more intuitive if you think in terms of conductance, which is the inverse of resistance (1/R). When you put two equal resistors in parallel, you double the overall conductance. Why? You are adding a second path for current to flow, so you double that flow. Unfortunately, we tend to speak mostly in terms of resistance, which makes the math a bit ... 2 The power rating given on lightbulbs always refers to the power at a specified operational voltage (which is always given together with the power or implied by the type of socket). The power at different voltages is not easily predictable as the resistance of the filament will vary strongly in dependence of temperature (which depends on the dissipated ... 2 Since I understand that power = V x Current, the power for a bulb can not be a constant if its resistant is assumed a constant. A normal mathematical thinking can confirm that. If the voltage and current don't change, then the power is constant. The electricity supplied from the wall is at 115V (more or less). If the resistance of the bulb is 1322 ohms, ... 2 Actually, we don't know that "filament bulb has straight Volatage vs Current graph": "The actual resistance of the filament is temperature dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold ... 2 Ohm's law assumes the temperature remains constant. An Ohmic conductor is one in which the current flowing through it is proportional to the voltage applied across it. A non-ohmic conductor is one in which the voltage and current are not linear. A) The resistance of most conductors increases as the temperature increases, however being ohmic and not ohmic ... 2 Your analysis doesn't apply to what you are applying it to. P=V^2/r is the power dissipated in the internal resistance of the battery. It is the power dissipated by the internal resistance if the battery is shorted. What you want is "the power supplied by the battery", which is the power dissipated in the external resistor: P = V^2/R, where care ... 2 Vacuum tubes can conduct hundreds of amps of electricity quite readily. The effect depends on heating the negative terminal so that electrons can leave the metal surface which otherwise keeps them in the surface owing to a phenomenon called work function. 1 Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire. R = pl/A, where R is resistance, p is the material's resistance in ohms, l is the length, and A is the cross sectional area in m^2. As a wire gets longer its resistance increases, and as it gets thinner its resistance also ... 1 Microscopically, in a resistor electrons are scattered by lattice impurities and defects (e.g. surfaces), other electrons and phonons. An approximate description of the process: In the classical approximation of the Drude model the scattering processes are reduced to a scattering time \tau, in between the scattering processes the electrons move ... 1 Heaters are one of the very few devices that are 100% efficient. All of the energy we put into them ends up as heat (though not all that heat may go where you want it to). So to a first approximation the energy used by your shower is determined by how hot you run the water and for how long, and it doesn't matter what heat setting you use. I say to a first ... 1 ...and V is zero so there is no energy ? The other answers all touch on this part of your question, but none of them explicitly says, that there is no energy dissipated in the superconductor itself. Ohm's Law applies to a conductor. The I in Ohm's Law refers to the current flowing in the conductor, the V refers to the voltage difference between ... 1 Consider the left branch of the bridge. The total resistance is P + Q, so the current is:$$ I_{left} = \frac{V}{P+Q} $$The voltage drop across Q is just V = IR, so the voltage at the PQ midpoint is:$$ V_{PQ} = V_{in}\frac{Q}{P+Q} $$We argue in the same way for the right hand branch to get the corresponding equation:$$ V_{RS} = ...

1

Most resistive materials have a nonzero temperature coefficient: the resistance changes with the temperature of the device. An incandescent lamp filament, which glows because it is hot, might have an temperature of several thousand kelvin. On the other hand, if you probe the lamp with an ohmmeter while it is off, you'll typically use a current of a few ...

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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An ohmic conductor is a conductor that obeys Ohm's law. That's all. Ohm's law are expressed as following: $$V = RI$$ That's simple like that.

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What's the effect of a resistor? It's a component that dissipates energy end thus lowers the voltage. So what is voltage? It's the strength of the field that moves the electrons, while current represents the number of electrons flowing through the wire. Free electrons can be stopped all together or slowed all together, but it's not possible to select only ...

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then shouldn't the variable resistor be connected to the negative terminal of the battery? It makes no difference at all where the variable resistor is connected. The current will be given by Ohms law. viz i = V/R. Regardless of where the resistor is located in the circuit shown the electrons pass through it in the same direction, and the same voltage ...

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$E=\frac{V^2}{r}t$ only holds if $I=\frac{V}{r}$, which only holds if the internal resistance $r$ is the only resistance in the circuit. Taking into account the load resistance $R$ we would get $I=\frac{V}{r+R}$, and hence $E=\frac{V^2}{r+R}t$ (assuming the battery's voltage is constant; otherwise it would become an integral). If you want to take the ...

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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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