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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 ...


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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 is too much confusion here to answer meaningfully. Current always flows through the battery. What goes in one end comes out the other. Some types of batteries can be charged by forcing current thru them backwards. Those are called rechargable batteries. Others don't work that way and can be damaged by reverse current. Those are often called ...


1

Since the circuit is fully symmetrical (left/right symmetry) the potential at C is exactly half the potential between A and B. This means that there is no current flowing across the point (from the "tip of the V" to the middle of the two resistors at point C), and you can break it without changing the underlying equations describing the current flow.


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I assume you mean, how do we know we can transform the circuit this way without changing any of the node voltages or branch currents? Let's use your node "B" as the reference node, and assign it a potential of 0 V. Then we can see that $V_A$ is just the battery voltage. From symmetry, we can tell that the voltage at "C" must be $\frac{V_A}{2}$. ...


5

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 ...


5

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

The electrons are not moving in a curved path. They are moving according to the solutions of the Newton's equation $$ m\textbf{a}=\textbf{F}(\textbf{r},\textbf{r}')=q\,\textbf{E}(\textbf{r},\textbf{r}') $$ As the above being a Cauchy problem, the form of its general solution explicitly depends on the initial conditions for position and velocity and in ...


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The alternative to moving in a curved path is moving in a straight path. Which your electron will only do only if all the forces are parallel to its velocity vector. So an electron starting from rest in a uniform electric field will travel in a straight line - but that is an exception, not the rule. But you could think of that as a "curved path with infinite ...


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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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Voltage is a difference in electrical potential in a circuit. You can compare this to a ball sitting at the top of a hill - as it rolls downwards, it moves through a difference in gravitational potential. Because the ball has lower potential at the bottom of the hill, it will roll there spontaneously. The ball will never roll up the hill by itself, ...


1

Reducing eddy current does not change property of conductor or circuit Eddy currents (also called Foucault currents) are circular electric currents induced within conductors by a changing magnetic field in the conductor, due to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. so ...


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Fringing in cases of electric fields and magnetic fields are mandated by the laws of electromagnetism. For electric current, there is no such law which would be violated in case fringing does not happen. It must be made clear that current in itself is not a 'field'. Therefore, even if fringing does happen, it is not 'similar' to the case of the capacitor or ...


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First a small correction: voltage is a difference in potential between two points, in this case the difference in potential between the to ends of a resistor. I suppose your question comes from the interpretation of $U_R=R.I$ with $U_R$ the voltagedrop over the resistor. Now, the direction of your causal interpretation is wrong. Assuming an Ohmic resistor ...


1

I can verify using Kirchoff's laws that this is solvable. Ultimately, $$R_{equiv} = \frac{I_aa+I_bb}{I_a+I_e}$$ where the subscripts denote which resistor the current flows through. Setting up $I_e = I_d+I_c$, $I_b = I_a+I_c$, $I_ee + I_cc - aI_a = 0$, $I_dd - bI_b = cI_c$, which is 5 unknowns, but only 4 equations. This isn't a problem, as you can soon ...


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Best brightness probably is achieved at 2V - the original setting. The reasons are more complex than may be apparent at 1st glance. Neither 1V or 4V is very good and it depends on the characteristics of your light and your power supply. In most case the higher voltage would be better if the only choices were 1V or 4V. This is because the lamp will "load ...


2

In electric circuits there is concept called duality, which says that for any phenomenom that occurs related to voltage, there is a dual phenomenom related to current. Thus, voltage and current are equally important to understanding electric circuits. As for your light bulb problem, your light bulb requires 16 W of power to operate. Your supply is only ...


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I'd use the Mesh method that is simple to remember and use. on youtube and three simple equations (3 current loops) will solve your problem.


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Current density is defined as electrical charge per unit time for a certain cross-section. Since a cross-section is a two-dimensional entity, it has to be $ A / m^2 $. In some cases it can be simplified to $ A / m $.


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I'm not too sure, what you mean by "Point of Symmetry" method, and the link doesn't make much sense as well. Anyways, I'll list the approach, I usually use, see if it is of any help to you. Suppose the cube is: Kirchhoff's current law, which states that the sum of the currents entering and exiting a node is zero, is essential in the analysis. The ...


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Consider, lets say, a wire of cross sectional area $A$, with charges (each of the same magnitude, $q$) flowing through it. If you consider a section of the wire of some length $x$, the volume of this region would be $Ax$. Because $n$ is the no of charges per unit volume, the charge in this region would simply be $q(nAx)$ ($nAx$ is the number of charged ...


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Current is defined as the rate at which charge flows through a given surface: $$I=\frac{\mathrm{d}Q}{\mathrm{d}t}$$ and in a circuit this surface is any cross-sectional area (perpendicular to the flow). You can often simplify that to charge through a cross-section per unit time and write $I=\frac{Q}{t}$ Answer to the comment: Another but equivalent way ...


0

It's the amount of charge flowing through a surface per unit time.


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Whenever there is a battery connected in a circuit we assume that the positive terminal of the battery is at a higher potential then the negative terminal by convention. Also by convention + charge flows from positive to negative terminal .The driving force here is the potential difference. Correct Electrons flow in the opposite direction. ...



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