# Tag Info

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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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Two identical hoses with the same pressure difference between their ends will carry twice as much water as each does individually. So you get twice as much flow (current) for the same pressure difference (voltage), which is another intuitive way of thinking about Dave's Answer that it is more enlightening in this case to think in terms of conductances rather ...

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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 ...

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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.

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This is niether series nor parallel circuit, because of resistor 4. We have to use Kirchoff's laws. We will assume that both ends are connected to a battery to simplify our analysis. First, reformulating the loop rule, we get that since the potential difference between the top and bottom end is the same, all paths from the top to bottom end must end up with ...

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You are correct that the resistance in some materials will increase as the temperature increases, and that the temperature can increase as a result of current flow. It still obeys Ohm's Law, however, as the relationship that Ohm's Law describes still holds. In other words, as the resistance increases, the current will decrease, just as Ohm's Law says it ...

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The reason for restricting temperature change is that some materials exhibit a change in resistity when the temperature changes. If the resistivity is constant versus temperature the resistance won't change. In that case, there is no need to restrict the temperature. A resistor is ohmic if it exhibits a constant slope V vs I curve. That resistor obeys ...

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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$E=\frac{V^2}{r}t$ 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 ...

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 ...

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In general the effective resistance of resistors in parallel can be calculated with, $$R_{eff} = \frac{1}{\sum{\frac{1}{R_i}}}.$$ When the ratio between any of the resistors, $R_i$, is an irrational number it will not be possible to use the method you mentioned. For methods, which can be used to calculate something, the most robust method is usually ...

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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 ...

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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.

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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. ...

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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 ...

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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 ...

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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 ...

1

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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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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... 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 ...

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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 ...

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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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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 $Z_{nm} ... 2 Vacuum tubes can conduct hundreds of amps of electricity quite readily. The effect depends on heating the negative terminal so that electrons can leave the metal surface which otherwise keeps them in the surface owing to a phenomenon called work function. 1 Microscopically, in a resistor electrons are scattered by lattice impurities and defects (e.g. surfaces), other electrons and phonons. An approximate description of the process: In the classical approximation of the Drude model the scattering processes are reduced to a scattering time$\tau$, in between the scattering processes the electrons move ... 1 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 ... 0 Model the current through the resistor as $$i(t) = v(t)/R,$$ the current through the capacitor as you have above, that is $$i(t) = C \dfrac{dv}{dt},$$ and the turning on/off of the switch can be modelled using the Heaviside function which would pre-multiply the above expressions. The current has to be continuous throughout so we can equate the two ... -1 You can substitute the switch with a resistor$R_2$, that has a value of either$0$or$\infty$. 0 If the switch is open, then no current can pass through the circuit. Your ODE should then take the form $$\frac{di}{dt}=0\tag{1}$$ When the switch closes, then current can pass through, leading to $$R\frac{di}{dt}+\frac{i}{C}=0\tag{2}$$ When the switch opens again, we revert the ODE back to Equation (1). Thus, the time range$0\leq t\leq2$gives you ... 0 The maximum of $$\left|\cfrac{V_2}{V_1}\right| = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}$$ is$1$at$\omega=\pm \infty$, and you find the half power frequency by solving: $$\frac{1+(\omega RC)^2}{4+(\omega RC)^2}=\frac{1}{2}$$ which gives$\omega=\pm \sqrt{2}/RC$0 My answer to the question will be "No, it doesn't have to be constant to obey Ohms Law." This is because the resistance itself has a temperature dependence as,$R(T) = R_0 [1+\alpha (T-T_0)]$. Where$R_0$and$T_0$may be taken to be the resistance and temperature at room temperature, respectively. And,$\alpha$is the temperature coefficient of the ... 4 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: $$I = ... 1 Consider the left branch of the bridge. The total resistance is P + Q, so the current is:$$ 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 ...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 ... 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 ...

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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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@Jerry Schirmer is right, the best way to think of a superconductor is like a skier being able to ski at a constant speed without a slope. The power output is zero since no power is dissipated through any heating affects of the conductor so no energy is transferred except when resistance is met at the other end (like a bulb).

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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 ...

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