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18

Nerd Sniping! The answer is $\frac{4}{\pi} - \frac{1}{2}$. Simple explanation: http://www.mbeckler.org/resistor_grid/ Mathematical derivation: http://www.mathpages.com/home/kmath668/kmath668.htm


13

It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery: ------------------------------------------ | | | Battery Bulb 1 Bulb 2 | ...


13

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


12

Yes, it is possible. For example Kevin Brown did here and here including this table. so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.


12

There is actually a student-friendly microscopic model how to derive the real Ohm's law $$\vec{j} = \sigma \vec{E}.$$ After its derivation you can transform it into the more common form using the answer by Nesp. The idea goes as following: We must start with the definition of current: $$I = \frac{\Delta Q}{\Delta t}.$$ So where does current come from? ...


11

An ideal resistor is defined as the two-terminal circuit element where the voltage across is proportional to the current through: $V_R = R \cdot I_R$ and the constant of proportionality, $R$, is, well, constant. A physical resistor has at least series inductance and parallel capacitance and can be modelled with ideal circuit elements as follows (for ...


10

instead of thinking your body is empty and that a charged wire has to push electrons one by one through you and into the ground (blood is actually full of charge carriers), a better analogy would be a very long queue of pushy people. if the entrance to the apple store doesn't open, it doesn't matter how hard the guy at the back pushes--nothing moves. ...


10

Hmm your experiment sounds like a good idea but I think it'll be much harder than you're imagining. The resistance of wires is very low. After all, they are designed to conduct! Check out this table. 30 gage wire has a resistance of $0.1\: \Omega/\mathrm{ft}$ which is well below what a typical multimeter can read. Also, because the resistance is so low, ...


9

Perhaps I can clarify what I'm trying to get at with the famous waterwheel analogy 99 years ago, Nehemiah Hawkins published what I think is a marginally better analogy: Fig. 38. — Hydrostatic analogy of fall of potential in an electrical circuit. Explanation of above diagram In this diagram, a pump at bottom centre is pumping water from right to ...


8

At sufficiently high voltages almost everything conducts due in part to quantum tunneling of electrons. An insulator has a breakdown voltage which is the field strength required before it will start conducting. Related to the breakdown voltage is the dielectric strength which is the minimum voltage over distance ($\mathrm{V}/\mathrm{m}$) before a material ...


8

I find this sort of thing becomes much more intuitive if you can think of an analogy in terms of water. In this case, we can think of it like this: Here we have water flowing through a hole in a bath tub, into another tub underneath. The stick figure has been given the task of keeping the water level constant, by lifting water back up into the top tub ...


7

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3 $$Z = ...


7

I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ...


7

Alfred got in before me, but I have a diagram! I've marked all continuous bits of wire in the same colour, and marked the corresponding colours on the ends of the resistors. A quick redraw later and I get: which is a lot simpler!


7

Gregsan's and Kieran's answers are insightful analogies and the pushy electrons are certainly part of the answer. There is another aspect to the "decision" process and that is the propagation of electromagnetic waves. There is a chapter in the second volume of the Feynman Lectures on Physics - I don't have it with me but the relevant section will be just ...


7

As others have noted, the big issue is getting the resistance of the wire into the range where your multimeter can measure it accurately. For that, the simple approach is to make the wires as long and thin as you can in order to maximize resistance. That said, another thing you can do is improve the precision of your measurements, e.g. by using ...


6

For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of ...


6

I will do the case where the material is homogeneous and isotropic, $\rho = \sigma^{-1}$ is a constant proportional to the identity matrix. We are interested in the steady state, where none of our variables depend on time. We have $\nabla \times E = 0$ from Faradays law and, $\nabla\cdot J = 0$ from the equation of continuity, where $J$ is the current ...


6

This is your circuit: The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the ...


6

The voltage across either horizontal resistor is zero so they can be removed from the circuit without changing the solution. This is most easily seen by simply removing the two horizontal resistors and then it's clear that the nodes the horizontal resistors connect to each have the same voltage. Thus, by Ohm's law, there is no current through either ...


6

Wire typically has very low resistance, so what you'll most likely end up measuring is the contact resistance.I.e., the resistance between your multimeter probes and the wire itself might have more resistance than your wire and overshadow it. My suggestion for the temperature dependent part would be to measure the resistance of the wire with it in boiling ...


5

In real life, the current can't jump instantaneously because there is always some finite inductance in a circuit. However, this is just a typical idealized textbook problem where the inductance is assumed identically zero, so the current can jump instantaneously according to the assumptions of the problem. Note the current also jumps in their solution for ...


5

We can't remove the resistor between the two points we've chosen because they're not at the same voltage. OK, let's unpack that a little. Imagine that you actually have a resistor network (any resistor network) built and want to measure its resistance with an ohmmeter. To do that, you need to choose two of the points in the network and connect the leads ...


5

Your problem is assuming that the charge transferred through the resistors is different. I don't know where you got that from, so I don't really know how to refute it other than by saying that since the currents must be the same, so must be the charge transfer in a given time. Edit in response to your comment: What you said is plainly not true. ...


5

I did a similar experiment this year for my final science assignment. We used a transformer which had two long coils of copper wire inside, of different lengths (10m and 750m) and thicknesses. Pop open the transformer and take out the coils. Set your multi meter to ohms and touch the contacts to either end of each of the coils (We got 2.2 ohms and 1500 phms ...


5

I think the short answer is, you don't. The reason we call the unit of force a Newton and not a kg m/s$^2$ is because it is convenient and it expresses the relation you want to convey when used elsewhere (e.g., $F=-kx$ for a spring). Similarly, it is convenient to "hide" the MKS base units into a single term, the potential $V$ in this case, so that the ...


5

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = ...


4

Potential for 2D problem Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk. I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$. Starting with basic electrodynamics: $\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, ...


4

As suggested by Manishearth, one can perform a $Y$-$\Delta$ transform from $Y$-resistances $R_1$, $R_2$ and $R_3$, to $\Delta$-conductances $G_1$, $G_2$ and $G_3$ (using a $123$ symmetric labeling convention), cf. Fig.1 below. A x----x------x-----[3]-----x------x----x B | | | | [4] [2] [1] [5] | ...



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