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16

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 ...


10

Short answer - yes, everything in the circuit can contribute. But usually, an ohmmeter is zeroed with the probes in place - in other words, whatever resistance the probes represent is taken out by the meter. There are two other factors that play a role, especially when you try to measure small resistance. The first of these is contact resistance: it is ...


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.


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 = ...


4

Use kirchoff's loop and junction law. :)


3

should the voltage be the input voltage or the voltage across that specific resistor The latter. generally when you want to know something about a particular component, you work with the conditions applying at the boundary of that specific component. I have a desk lamp with a LED. The other end of the lamp plugs into a 240 V AC power outlet. If I want ...


2

The magnetic force term in Ohm's law is often ignored where it would complicate description and bring nothing essential - often magnetic force is much smaller than the electric force. In cases magnetic force is essential, like the Hall effect, moving conductor in static magnetic field, current - carrying plasma (pinch effect), it is common to take the ...


2

Ok so I first have taken the diagram from the wikipedia page for reference and put it here. Now if you are happy with the idea of how the potential divider works... .... then I hope that you can see that $R_1$ and $R_2$ in the Wheatstone Bridge diagram form a potential divider and there is another potential divider with $R_3$ and $R_x$ - and the points ...


2

By the very functioning of an open switch, there can be no current through the switch, and there is no constraint on the voltage across the terminals of the switch. This of course means that there cannot be a current through the entire branch containing R4 (and therefore the resistance R4 plays no role in this case). The current then flows through the rest ...


1

Ohmic conductors: that follow v=ri, Non Ohmic: that donot hold for this equation


1

In response to your comment, there is a way to find the voltage without finding equivalent resistance. You can write the potentials at each junction point. Taking the points on the lower line to be zero potential, the potential gain on going up the last branch is $1$V. Then going left, add another $1$V. Then, current through the last branch is 1A and the ...


1

The total resistance seen by the voltage source is: $$ R_{tot}=[(6//12)+(12//6)]\Omega = 8\Omega $$ then: $$ i_1=\frac{20V}{8\Omega}=2.5A $$ Now let's calculate the voltage of the intermediate node, i.e. the node where you measure $i_2$: $$ V_{mid}=20V\frac{12//6}{(12//6)+(12//6)}=20V\times\frac12=10V $$ The current $i_2$ can be computed as the current in ...


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} = ...



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