# Tag Info

## Hot answers tagged resistance

17

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 | ...

11

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? ...

10

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

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. ...

8

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

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 ...

6

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 ...

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

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 = ...

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

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

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 ...

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

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 ...

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. ...

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: $\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, ... 4 The answer is "yes", if you take for granted that$R$is defined by the relation$\Delta V=IR$. In fact it is derived from (the real) Ohm's Law. Ohm's law states that, for some materials (the so-called "Ohmic" materials) the current density vector$\vec{J}$(current per unit area) is parallel to the electric field$\vec{E}$, i.e., ... 4 If you are wondering about causality, then I think that voltage difference$\Delta V$is fundamental as it is the cause, and the current$I\$ is the consequence. If you want to have current, you need movement of the charges. The most obvious way to move charges is to act upon them with electric field, and each electric field is accopmained with voltage ...

4

Why does this overall resistance decrease? A more elegant, sophisticated way to see why is through the notion of duality. In electric circuit theory, conductance (the reciprocal of resistance) is dual to resistance. Other dual pairs are: voltage - current series - parallel inductance - capacitance Thevenin - Norton and so on ... For example, ...

4

Typically, yes: current will flow as long as it has a path with finite resistance (even zero), a voltage difference, and a supply of charge carriers (e.g., electrons). If there really were no resistance in the circuit, the electrons would go around the circuit, and arrive back at the beginning of the circuit with as much energy as the potential difference ...

4

Basically, a "12 V" battery is not a perfect voltage source and the starter is a large load. Due to the very large current (100 A or more) a starter motor can draw, it causes the battery voltage to sag a bit. Dropping 2 V or so during the brief period the starter motor is on would not be out of line. You can think of the battery as being a perfect voltage ...

4

Yes there is. Basically, you want to look at all of the available paths from A to B. If, while taking a path, you must cross two resistors no matter what, then they are in series. If there are different paths so that you can cross one resistor, get to B, and then another path where you can cross the other resistor and get to B, then they are in parallel. ...

4

All the way it means that charge transferred through the resistor 1(Q1) is less than charge transferred through the resistor 2(Q2) over the same time T. I don't understand how this follows from the preceding two statements. If R1>R2,resistance offered by resistor 1 is greater than resistor 2,so the amount of charge transferred through the ...

4

Is this allowed? It is often the case that resistors in the milli-Ohm and micro-Ohm range are required especially for current sensing applications. One need look no further than Digi-Key for examples.

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