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20

I'll answer the concrete question, because it's one of those fun ones where the units are all wrong and the scales are just absurd. Does this also mean that if I release a million amperes of current into the earth, every living entity walking barefooted should immediately die? It depends on how long you do it and with how much power. And ...


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 ...


4

Use kirchoff's loop and junction law. :)


3

A counterpoint to Schwern's answer (which was instructive, but I believe wrong on some key points - but I will borrow a couple of numbers from it). I think the correct way to pose the question is: If a 300 mA current for 100 ms will kill a human, what should be the rate of change of the electric field around the body to induce that current? Treating ...


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

You should not simply add the errors, you should sum them squared in case of $y=AB/C$. $dy/y=\sqrt{ (dA/A)^2 + (dB/B)^2 + (dC/C)^2 }$ This comes from the partial derivation of the function $y$ by all the components and weighting them by the uncertainty: $dy^2=(dA\frac{\partial y}{ \partial A})^2 + ... $ The square is there because you treat the different ...


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

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

Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...


1

I think the approach given by jaromax is correct (+1, I also get 1.4$\Omega$), whereas the formula quoted in the linked question should not be used if the measurements of $R_1$ and $R_2$ are independent and (slightly) overestimates the total uncertainty. However, I am adding this answer because the approach you adopted based on percentage errors is ...


1

There is a concept of "voltage of a step"* in energy industry - if a high voltage power line is leaking into the ground and isn't shut down, then near that point the ground voltage difference over a single human step (when one feet is closer than the other) can be enough to kill a person; that's why it may be dangerous to approach fallen wires after a storm ...


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

Hard to know why you got a different answer in your calculation, but here are some general considerations. Assume you have a circuit like this: You would think that you can easily compute the time constant $\tau = R_1 C_1 = 10^{-4}s$. However, it is most likely that your real circuit isn't actually like this. First - capacitors are notoriously ...


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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