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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} = ...


0

Rate absorbing of heat by ice is equal to the rate of supply of heat by the wire if no heat is lost to the surrounding. You can calculate rate of supply of heat by, $P = \frac{V^2}{R}$ Where $P$ is power (rate of supply of heat) V is the voltage across the conductor R is the resistance Therefore we have $P = \frac{210^2}{20}$ $P = 2205W$ So ice ...


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 ...


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 ...


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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.


0

@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).


0

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 ...


0

As with all such problems, you will have to apply Kirchhoff's voltage and current laws at each loop and node respectively, then combine the equations and solve them. In this case, the time dependence comes from the inductor, as the voltage across it is given by $V_L = L \frac{dI}{dt}$, where $I$ is the (time-dependent) current flowing through it. Thus, you ...


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 ...


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try re-drawing it like this: then work out the total resistance in the circuit and use V-IR to calculate i1. then find the percentage of current goung through each branch of the wire.


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Use kirchoff's loop and junction law. :)


1

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


0

When the switch is open, there is no current through the middle branch consisting of the switch and R4. The 5V above R4 doesn't have any effect on the circuit at this configuration. Just forget it when the switch is open.The 5V above R4 will contribute to the voltage in the circuit when there is closed path for the ions of this 5V battery to circulate and ...


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 ...


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in the absence of the voltage V2, R2 and R3 will be parallel so R(equivalent)=(R3*R2)/(R3+R2) ,and R(equivalent) which im gonna call R' is in serie with R1 so you circuit will contain just the voltage V1 and a resistor R''=R'+R1. in the presence of the voltage V2, R1,R2 and R3 are in wye "Y" ,get a look to this ...


0

As noted by Floris, the best way to measure small-value resistance is to use a four-point Kelvin connection; unless the current drawn by the voltage meter is significant (in which case there are other problems) then provided that current-source probes are either the inner two or the outer two (as opposed to being interleaved with the voltage-reading ones) ...


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 ...


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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 ...


-1

Stainless steel is a relatively good conductor of electricity, as are all metals. The body is also a good conductor, due to water. The skin is a relatively poor conductor of electricity because of dead skin cells. Any break in the skin, i.e. cut, greatly reduces its resistance.


0

In this case,if they are connected one at a time to the battery then, your reasoning is correct. But if both are connected at same time in series, then current will be same and electric field will also be same.


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 ...


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 ...


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 ...


1

Presumably it is the linear temperature coefficient of resistivity (or of resistance as they come down to the same thing).



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