Hot answers tagged resistors
14
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
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 $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.
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
$$Z = ...
6
For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of ...
5
Why is current not 0 in a regular resistor - battery circuit immediately after you closed a circuit?
In real life, the current can't jump instantaneously because there is always some finite inductance in a circuit. However, this is just a typical idealized textbook problem where the inductance is assumed identically zero, so the current can jump instantaneously according to the assumptions of the problem. Note the current also jumps in their solution for ...
4
Potential for 2D problem
Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk. I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$.
Starting with basic electrodynamics:
$\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, ...
4
While mmc's answer is correct, the result of applying Kirchhoff's laws is that you get the Graph Laplacian problem.
Given a graph G with each edge E_i given a resistance R_i, the weighted graph Laplacian is given by considering the operation on functions which takes $\phi(V)$ to
$$ \nabla^2 \phi = \sum_{<W,V>} {1\over R_i} (\phi(W)-\phi(V)) $$
Where ...
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
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 ...
4
Regarding what you consider to be a contradiction:
How can the total current be the same if a resistor is reducing current at some point in the circuit?
The current at any point in the circuit is the same because the current distribution in the circuit has reached a steady state (i.e., charge buildup is forbidden). Your intuition is telling you that ...
3
Under assumption that three bulbs are connected to constant voltage, brightness actually changes. Brightness is very loosely proportional to power $P = U I = R I^2 = \frac{U^2}{R}$, so it is necessary to calculate the change of current/voltage through the remaining two bulbs, after the first breaks.
Considering your very case, if all three bulbs are the ...
3
Here is how I would do it, following the method outlined by kleingordon in a comment. This method is less cool but more general than Carl Brannen's answer, because it will work even in the case where there are crossing wires and you can't rearrange it into a single sheet of resistive material.
Let the electric potential at $A$ be $V_A$ and that at $B$ be ...
3
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]
| ...
3
Hint: Use Ohm's law $I=\frac{V}{R_p}$ and the formula $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R_0}$ for parallel resistors to derive a straight line in a $I$-$\frac{1}{R}$ diagram
$$ I~=~V \left(\frac{1}{R}+\frac{1}{R_0}\right). $$
Here $R$ and $R_0$ are the variable and the fixed resistor, respectively. To answer OP's two questions:
The slope is $V=1{\rm ...
3
You can connect a unit current source between these two vertices. Then you can solve the circuit using nodal analysis, in essence:
Setting the potential of one arbitrary vertex to zero.
Applying Kirchhoff's Current Law at each vertex to get a relationship between edge currents.
Applying Ohm's Law to express each edge current in terms of the potential of ...
3
Resistors are generally used to dimension electrical devices to the ranges in voltage, current, time constants, what have you, that are needed. In this specific example the resistor is used to dimension the voltage drop in case one of the inputs has low voltage (lower than $V$), so that a current flows from $V$ to the input (it can only flow in this ...
3
I'm not sure why the resistor to ground from B is there, but you are incorrect at point D, the capacitor doesn't pass the DC level as you've indicated. It's a high-pass filter with C and R, so basically you need to move the DC-level on the Vd plot to ground - but keep the two transients like you've plotted them. That is, the curve should start at ground and ...
3
No, they do not all have the same voltage drop. If they were in series, however, they would.
By Ohm's Law, the voltage drop is proportional to the current flowing through a resistor. (So in several series resistors with the same resistance, the drop across each one is the same, since the current across each one is the same). However, because B and C are in ...
3
I mean, can I replace this configuration by one capacitor with one
resistor in series such that this resistor is equivalent to the other
two?
The answer is actually no.
For a single resistor and capacitor in series, the real part of the impedance is independent of frequency, i.e., the real part acts like a resistor.
$Z_s = R_s + \frac{1}{j \omega ...
2
A Wheatstone bridge is only a very sensitive way of finding a match between two resistances. You have got to have a nice, well calibrated variable resistance on the opposite side to the unknown resistance. The accuracy with which you can judge the unknown is limited by the calibration accuracy of the variable resistance.
Using Ohm's law to find an unknown ...
2
The reasons can be found here:
Even in the most simplistic case when all Rs (they are reference resistors with low tolerance error) are equal the sensitivity to any deviation is very large (last eq in the page). See the structure of the eq put some numbers to see that the numerator goes near zero. It is temperature balanced because the effects in the ...
2
As far as I know, the first solution to the general problem is given by Cserti,
József Cserti. Application of the lattice Green’s function for calculating the resistance of an infinite network of resistors. Am. J. Phys. 68 no. 10, pp. 896 (2000). doi:10.1119/1.1285881, arXiv:cond-mat/9909120 [cond-mat.mes-hall])
using lattice Green's functions (and ...
2
Do not follow these suggestions!
A typical car battery has a voltage around 12V and a very small (less than 0.1 Ohm) internal resistance. If you would connect the ammeter in parallel to the battery it will hopefully trip the internal fuse or just blow up. There are very few specialized ammeters than can measure currents above 100A. So you have to approach ...
2
Your equation/setup is essentially correct as long as we are talking about idealised resistors, ie they have no complex impedance.
To deal with time varying currents and/or voltages we generalise the concept of resistance to impedance. Impedance takes into account any delay/lag between resistive components and the voltage across them. In an idealised ...
2
One fundamental
Well, if you consider $V=IR$ as fundamental (and not $\bf J=\sigma \bf E$--IMO this is the actual Ohm's law), then you can derive it by what I call "discrete calculus"
Taking the definition of resistivity as "resistance per unit length and unit area":
Take a cuboid of dimensions $L,W,H$ along coordinate axes $x,y,z$. Current flows along ...
2
Yes, if you write $V \propto I$ it means:
$$V = kI$$
for some constant $k$. If you rearrange this equation to:
$$I = \frac{1}{k}V$$
this is the same as:
$$I = k^'V$$
for a new constant, $k^'$, and therefore $I \propto V$.
Response to comment: in some situations it makes sense to think of the current dependent on the voltage, but in others you ...
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