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Let's assume that the power company is supplying a neighborhood with 1000 A of current at 120 V. Since P = IV, the neighborhood is receiving 120 kW of power, which is the "load" seen by the power company. To maximize efficiency, the power company wants to minimize the losses involved with transmitting power to the neighborhood, which occur due to ...


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Yes the diagram in the book is not to scale. It should look like this: Each transition increases by 1V. The outer edge would have a potential of 1V, the transition between green and yellow would have a potential of 2V. The final transition has a potential of 6V.


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Should it? Lets say the 6V mark is at distance at 1, then 4V is at 1.5 and 2V is at 3 meaning that 4V should be in in the middle between the origin and the 2V mark, which could work as far as i can tell on my screen


2

The short answer This is not 100% true since it assumes DC transmission, but it gives the simplest form of the idea: even if the transmission lines are themselves at high voltages, that doesn't directly mean anything, since voltages are not defined relative to anything special (they're defined relative to some other line which is in parallel with your ...


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There are two different $V$'s here. Suppose the power station outputs at 10,000 V. By the time the wire makes it to your house, this may have dropped to, say, 9,000 V. The $V$ in the first equation refers to the voltage difference you can use, which is 9,000 V (between the wire you receive and ground). The $V$ in the second equation refers to how much ...


0

Voltage is a measure of the electric potential difference across two points in a circuit. It may be considered the work done to transport an electric charge. Power lines are made of thick easily conductive material in order to minimize resistance and power loss to heat. But resistance within power lines is fixed, and power is delivered through the line ...


2

In my opinion, the mathematical equation we call Ohm's Law is best taken not as a “law”, a fact about the universe, but as the definition of resistance. $$R \overset{\mathrm{def}}{=} \frac{V}{I}$$ Given this definition of the quantity $R$, we can then make (as other answers have mentioned) the empirical observation that many materials have approximately ...


1

Ohm's Law actually follows the definition of power, current and voltage. Let's begin by defining power $P$, current $I$ and voltage $U$ as $P = \displaystyle \lim_{\Delta t \to 0} \frac{E}{\Delta t}$, $I = \displaystyle \lim_{\Delta t \to 0} \frac{Q}{\Delta t}$ and $U = \frac{E}{Q}$. We then find for a constant current $I$ with a constant voltage $U$ that ...


1

You are partially correct. You are correct about the part that the total voltage in the conductor is the addition of the voltage (S) and induced electromotive force, assuming that: 1. The applied magnetic field doesn't affect the flowing current due to the voltage (S). 2. The induced current due to the time varying magnetic field doesn't affect the magnetic ...


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I'm assuming the bottom of the conductor is grounded as you've indicated a flowing current? EMF occurs in a conductor when submitted to a time varying magnetic field as governed by Faraday's law, is the field in your problem static or time-varying, or is your conductor in motion? In this case the solid conductor would experience eddy currents, circles of ...


-1

Ohm's law isn't fundamental and holds true only under certain conditions, like constant temperature for example. However, there is a simple way to think about it. Imagine the flow of massive objects through a wide water pipe. This is like a current. The water pressure causes the objects to flow quickly, that's your voltage. If the pipe is narrow then the ...


23

You could start from Drude in zero magnetic field, that equates the derivative of the momentum $\vec p$ by the electrostatic force $\vec F_{el} = q \vec E$ as a product of charge $q$ and electric field $\vec E$ minus a scattering term (with time constant $\tau$; compared to Newtons second law that does not feature the latter, crystal term): $~~~~~~\dot ...


33

Ohm's Law is not a construct which can be derived. It is essentially a generalized observation. It is only useful for a few materials (conductors and medium resistivity), and even then virtually all of those materials show deviations from the ideal, such as temperature coefficients and breakdown voltage limits. Rather, Ohm's Law is an idealization of the ...


0

Once you add a changing magnetic field, the electric field no longer is conservative -- i.e. there is no longer a consistent definition of voltage! A voltmeter measures $\int_a^b \mathbf{E} \cdot d\mathbf{s}$ along a path from $a$ to $b$. Normally this value doesn't depend on the path, so you can speak of "the" voltage drop between $a$ and $b$. In this ...


0

The amount of work done by unit charge between any two nodes of current carrying circuit is called the potential difference between those nodes. The amount of work done against the electric field by displacing (without acceleration) a unit test charge from one terminal to other terminal in an open circuit is called the electromotive force. Obviously when ...


0

First of all power will alway be same i.e VI (primary side of amplifier) = VI (secondary side of amplifier). However V and I individually can change. So in your case: Case 1: Voltage= 20 current will become I = 0.5 amp so net power = 10 watt Case 2: If current I = 8 amps V will be = 10/8 Volts so net power = 10 watts


2

The voltage becomes the same as Earth, but this doesn't mean that the charge goes to "zero". By "zero" here, I mean that the positive charges (nuclei) are perfectly balanced by the negative charges (electrons). You can call the voltage of Earth 0 Volts, but this is a relative measure. Charge, in the usage here, is not a relative measure because it is a ...


0

Power(P) = Voltage(V) * Current(I) That law describes the relationship between power, voltage, and current in a conductor. It means that, if you measure the current flowing in the conductor, and you measure the voltage difference from one end of the conductor to the other at the same instant, then the product of voltage and current will be the rate at ...


1

After twisting my own brain for a LONG time about this, I have come to this conclusion: Resistance is a measure of how FAST a load is able to absorb ALL the potential/kinetic energy of a given number of electrons passing through it. (I say all, because with 1 load the V drop will be equal to the V of the source. Which tells me that all energy will be gone ...


2

Why is resistance NOT calculated simply by looking at how much voltage drop is created by a certain amount of charge passing through, as in R=V/Q. Your statement in bold is one way to define resistance. But your words do not match the expression $V/Q$. We're interested in the voltage drop per charge passing through, right? Well, how do you measure how ...


0

Resistance is defined as $$ R(i) = \frac{dV}{di} $$ and, as all definitions, is a matter of terminology. Also, as you stated, since it has to measure the drop against the flow of electrons it makes sense to take the derivative with respect to the current, which is exactly what the flow of electrons is. Since the difference of potential must be calculated ...


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There are really two questions here (I think): why is the voltage drop so different for the cold LED (notice it ranges from 3.5 to 4.5 V at LN temperature, but from 2.0 to 3.2 at room temperature) why does the LN curve exhibit the strange curvature? CuriousOne already hinted at the answer - this has to do with the temperature of the LED. In particular, ...


1

Be careful because that formula is only valid for a very limited set of field geometries. It is always better to derive EMF from the change of magnetic flux. To answer your question, the induced voltage at zero current does not depend on the resistance of the conductor. As soon as a load is connected to it, the effective voltage measured on the conductor ...


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The temperature of the LED increases. Do the same experiment with short pulses of current (1ms repeated once a second) and you will see that it's a temperature effect.


1

Point particles as the electrons (which are the charge carriers) move according to Newton's law $\textbf{F}=q\textbf{E}=m\textbf{a}$. Whenever an electric field is present it generates a difference of potential between two points $A$ and $B$ given by its differential form calculated between the two points $$ V_A - V_B = \int_A^B \textbf{E}\cdot d\textbf{s}. ...



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