# Tag Info

1

So what exactly happens to the potential inside the resistor ? Unlike the ideal conductors, for which an electric field cannot exist inside, there is an electric field through the resistor body when there is a current through. And, as you may know, the rate of change in electric potential is related to the value of the electric field. Thus, the ...

3

Let me first take a little detour away from this circuit to particle accelerators. If you have some electrons in vacuum and a potential set up between two points (exactly the same as saying you have an electric field set up) you can accelerate your electrons. If you move a single electron through $1V$ of potential the electron gains $1eV$ of energy where ...

0

I think this is really about which way you count current and voltage to be positive. For every element in a network you can define a a current and a voltage. If voltage and and current point the same direction, it's called "receptor". If they are in opposite direction its' called "generator". The most common convention is to use "receptor" for resistors, ...

1

The bleeder resistor is across the capacitor so they have identical voltage across.

2

I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...

2

This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is ...

0

As DanielSank said, I cannot say anything quantitative about the question without a diagram. That being said, your formula, in general, is correct if the topology of the circuit is symmetric. Without any details about how mechanical energy is coupled to the circuit, I can only guess the general reason for the apparent non-conservation of energy. This ...

0

Are you sure you mean the internal resistance "r"? The internal resistance typically can not be adjusted, often this question is phrased in terms of the load resistance "R". The power dissipated in the external load is: $$P=\frac{V^2 R}{(r+R)^2}\;,$$ which you need to maximize. If you maximize with respect to R, you find R=r... If you maximize with ...

0

Edit: I had previously misread which node was a and which was b in the question. V_ab = 108.75 - 15(4.25) = +45V If a is on the right, and E = -108.75, then you should have $$V_{ab} = -108.75 + (15\Omega)(4.25\mathrm{V}) = -45\mathrm{V}$$

0

Technically the resistance of a wire is never 0. Typical wires of copper have electrical conductivity of $10^{7} S/m$ at room temperature. So there is indeed a very slight potential drop but highly negligible compared to the one that would be experience at a proper resistor. As for the problem of electrostatic, indeed from a single charged particle the ...

3

At equilibrium, the field inside an ideal conductor is zero. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c2 A charge moving through such a conductor neither gains nor loses energy. We can't attach an ideal conductor to an ideal voltage source. Something has to give. There will be a voltage drop along a real wire due to non-zero ...

2

I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that ...

1

Conductance is the extrinsic property while conductivity is the intrinsic property. This means that conductance is the property of an object dependent of its amount/mass or physical shape and size, while conductivity is the inherent property of the material that makes up the object. No matter how the object changes in terms of shape/size/mass, as long as it ...

-1

The fact that the conductivity $\sigma=\frac{1}{\rho}$ of a metal scales like $\propto\frac{1}{T}$ is due to elastic electron-phonon scattering, i.e. the coulombian interaction between the charge density fluctuation induce by a phonon and an electron. In an incoherent transport theory of electron in solids (i.e. avoiding corrections due to interferences ...

-1

In my point of view, it is independent thing. Dependence of resistance on temperature is determined by Nernst-Einstein equation. $$R=\frac{l k_B T}{S D Z^2 e^2 C}$$ where $T$ -is a temperature of resistance, $k_B$ is a Boltzmann constant, l- length, S - cross sectional area, $D$ - diffusion coefficient, $C$ is a charge carrier density, $Z$ is a amount of ...

0

The law of currents says that the sum of currents in any node/junction is zero. The law of voltages says that the sum of emf=sum(i*R) around any closed loop and equals the sum of voltage sources; sum(e) in the loop. According to this your equations on the right of the figure are correct. The rest is simple mathematical manipulations. For complex circuits all ...

0

In mathematics we have boundary value problems or initial value problems or a mix. For the first you need steady state and specification of the values on All the boundary points. Your case belong to this I guess. So the answer is no, you have to know all the boundary points. For initial values problems and this you need if you have storage elements like ...

0

The question was not very clear. If $V_2$ was not there, you simply combine $R_2$ and $R_3$ to $R_x$ as $1/R_x=1/R_2+1/R_3$ and sum $R_x$ with $R_1$ and you get simply $I_1$. Inserting $V_2$ creates another current competing with $I_1$. Concerning the boxed statement, I believe that it reflects the sums of voltages, not really it combines the resistors, ...

0

They aren't. Two resistors are in parallel if they have the same voltage drop across them; two resistors are in series if the same current flows through them. In your problem, because of the presence of $v_2$, it's entirely possible to have different currents in and voltage drops across all of $R_1,R_2,R_3$. I believe your boxed statement, \begin{align} ...

-1

Nichrome is used to connect the two terminals of the battery or any source that is connected to the circuit. It is not used for making filaments. Therefore, it strong text offers low resistance to the flow of the electric current.

0

The $\frac{1}{I}$ term is the analogue of $y$ in the equation for a straight line, $y=mx+b$. I'll guess that $R$ is a variable resistance you control and $r$ is some internal or fixed, possibly unknown, resistance in a circuit. When you plot the controlled resistance $R$ on the horizontal axis, that is the analogue of $x$ in the linear equation pattern. ...

2

For your circuit, $V = I\cdot R$. You are plotting (unusually) R along the X axis and $\frac{1}{I}$ along the Y axis, so the slope is $\frac{1}{V}$. Now the fact that this slope is a straight line tells you that the voltage is constant. This means that (over the range of your experiment) your voltage source has a low internal resistance. Imagine for a ...

1

If it's a simple circuit where Ohm's law applies, then we should get $$V=IR$$ so we see that $$V/I = R$$ $$1/I=R/V$$ $$1/I = (1/V) \times R$$ The gradient should then be $1/V$. Seems like a slightly bizarre plot but if you got a straight line then that makes the maths simple at least!

2

A good reference was given in an answer to a related question: Cserti 2000 (arXiv preprint, whose numbers I'll be referring to) solved a number of generalizations of the 2D lattice problem. For a $d$-dimensional lattice, the resistance between the origin and the point $(l_1, \ldots, l_d)$ is given by eq. 18 in that paper:  R(l_1, \ldots, l_d) = R_0 ...

0

Each branch containing resistors acts as a potential divider, dividing the voltage across the whole branch in the ratios 40:50 and 50:40 (top and bottom respectively), ie the ratios 4:5 and 5:4. This means that each resistor will have either 4/(4+5)=4/9 or 5/(4+5) = 5/9 of the total voltage (18V), which as you have correctly calculated is 8V for both the ...

2

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...

0

There may be a confusion between enery and power. While the current, which is the number of charges per second, and the energy of the ,electron do'nt change at the output of the resistor, the power does. Power is the amount of energy per unit time, and that does not affect the current which, again, is the amount of charge per unit time.

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