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Repeating what @Dale said in a slightly different way, if you want relativistic laws, it is better to work with relativisitic quantities.
The response of matter to applied electromagnetic field is generally expressed through charge density ($\rho$) and current density ($\mathbf{J}$). The 4D equivalent is the four-current:
$J^\mu=\left(c\rho, \mathbf{J}\...
12
So basically I have three variants, but professor that I need to find one result. Any ideas?
Circuit theory is inherently nonrelativistic and so there is no relativistic circuit theory. This should actually not be surprising since circuit theory does not use the concept of space at all and relativity is a theory of spacetime.
The issue is in the ...
4
The Ohm law is approximately(*) valid for water in the local form $J=\sigma \cdot E$ for the free liquid, but not in non local form $I=R \cdot U$, if electrodes and DC current are applied.
The latter is due involved electrochemistry, mainly but not limited to factors as:
Equilibrium potential of electrodes
Potential difference for electrolysis
Kinetics of ...
4
As modeled, it's totally undetermined. $I$ could be any finite value, positive or negative, with no limits.
In the real world, it would be determined by the non-ideality of the two voltage sources. They won't both be exactly 10.000000 V, and they will both have non-zero equivalent internal resistance. Exactly how much would depend on how these sources are ...
3
To reduce the current through the device to be protected, the voltage across the device must be reduced. Since an ideal DC voltage source has a fixed voltage across independent of the current through, the voltage across the device can only be reduced by adding an additional device in series forming a voltage divider.
It's true that adding a parallel ...
2
Since the resistor is kept at $T=300K$(which is wierd since its heating up), the entropy change of the resistor is
$$\Delta S_{resistor}=\frac{-\Delta Q}{T}$$
(heat flow out of the system taken -ve)
Since the surroundings stay at $T$(since no other temp is given, assuming that),
$$\Delta S_{surroundings}=\frac{\Delta Q}{T}$$
therefore
$$\Delta S_{universe}=\...
2
As suggested by Alfred Centauri:
I believe what you need is the concept of electrical impedance. The complex impedance of a capacitor is
$$ Z_C= \frac{1}{j \omega C}, $$
where $j=\sqrt{-1}$, and $\omega$ is the angular frequency at which the circuit is being driven. The derivation for the $\Delta$-$Y$ formula will be identical but with $R$ replaced by $Z_C$...
1
$\def\Ip{I_{\rm pr}} \def\Is{I_{\rm sec}}$
You're doing a very common mistake. It's totally false that to
increase the voltage supplied to a building from a constant power
source substantially by putting a transformer[...] would reduce the
current substantially
The effect would be to burn every electrical equipment connected to
that building's mains. ...
1
I'm assuming you've been asked to find the the voltage across the capacitor. Is that the question?
If so, re-arrange the circuit so that R and 4R are actually a voltage divider. Same for 2R and 3R. Now you've got a capacitor across the mid-points of two voltage dividers. You can then calculate the relative voltage at each side of the cap and so get the $V_{...
1
It is not necessarily symmetric. The conductivity tensor $\boldsymbol \sigma$ is given by:
$$\mathbf J = \boldsymbol \sigma \mathbf E$$
And its inverse $\boldsymbol \sigma^{-1}=\boldsymbol \rho$ is the resistivity tensor. If you use matrix notation you have:
$$\begin{pmatrix} J_1\\ J_2\\J_3\end{pmatrix}=\begin{pmatrix} \sigma_{11} & \sigma_{12} & \...
1
I think that the point that you are missing is that the ideal circuit diagram that you have drawn is an approximation of a real life situation and so you should be wary of going outside of the limits of the approximations which you have (unknowingly?) made.
Consider a very simple circuit of a cell of emf $\mathcal E$ connected in series to an open switch ...
1
Does the current instantly have a value of $\frac{V}{R+r}$ where V is the emf of the cell and R and r are the external and internal resistance respectively? Or does the current initially have the value $\frac{V}{R}$ and then it decreases (at a rate too fast for humans} and reaches the value $\frac{V}{R+r}$ at a steady state?
If there is a substantial ...
1
So the resultant voltage across the 30 ohms resistance just at the
instance of closing the switch S2 is 6V so the current just at the
moment of closing the switch is $\frac{6}{30}=0.2$.
This isn't a valid solution (in ideal circuit theory).
First, it is typical to consider the solution just before and just after the switch is closed rather than at the ...
1
This is more of a mathematical question than a physics one.
In principle you approach these kind of problems by writing down an equation that models the system in a recursive way. And then solve for the unknown(s).
You start by an appropriate definition. This is often the key. Lets say
$S_n = \text{<resistance of the network when the "first" ...
1
The cell and wires it connects to produce electric field, not only inside the cell, but also outside. The terminals of the cell are like two points of an electric dipole and their field outside the cell has lines of force that are similar to the wires.
The charges on the wires tune the electric field so that inside the wire, the field is teh same all over ...
1
Ohm's law indeed isn't valid at the conductor surface. This is because electric field has non-zero normal component on the surface, but electric current density's normal component there is zero, as charges won't jump out of the conductor into vacuum (unless the electric field is extreme). Ohm's law is usually valid inside the metallic conductor, where the ...
1
There are several "impedances". In general, we can say that it relates an input with an output for a particular system of interest. Some of them are "intrinsic" and some are "extrinsic".
Extrinsic
Acoustic impedance, that relates pressure with volume flow rate.
Electrical impedance: relates voltage with current.
Mechanical impedance, that relates force to ...
1
I agree that e-e scattering does not contribute to resistivity. When the medium is conducting a current the electrons system has a net momentum in the direction of the current. Resistivity occurs when this net momentum is transferred to the lattice. e-e scattering conserves the momentum of the electron system. Only a coupling between the electrons and the ...
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