# Tag Info

1

Since the circuit is fully symmetrical (left/right symmetry) the potential at C is exactly half the potential between A and B. This means that there is no current flowing across the point (from the "tip of the V" to the middle of the two resistors at point C), and you can break it without changing the underlying equations describing the current flow.

0

I assume you mean, how do we know we can transform the circuit this way without changing any of the node voltages or branch currents? Let's use your node "B" as the reference node, and assign it a potential of 0 V. Then we can see that $V_A$ is just the battery voltage. From symmetry, we can tell that the voltage at "C" must be $\frac{V_A}{2}$. ...

0

An intuitive answer could run along the following lines. When any two dissimilar electrical conductors (say, A and B) are brought into contact, the distribution of the charge carriers in A and B at the junction gets altered so as to assume a new equilibrium distribution. This new distribution of charge carriers changes the potential drops from A to air ...

0

If resistor is an ohmic one, on applying a higher volt battery, power will increase as power for a resistor is $V^2/r$. If it is non ohmic one, the answer may depend upon other factors as well.

0

Although there are several sources of contact resistance, the main source of contact resistance is the oxidation of the contact surfaces. For the electrical case, the oxides of the materials have a much lower electrical conduction (higher resistance) than the materials, therefore the contact area (that is not cleaned and protected) will have a higher ...

0

First a small correction: voltage is a difference in potential between two points, in this case the difference in potential between the to ends of a resistor. I suppose your question comes from the interpretation of $U_R=R.I$ with $U_R$ the voltagedrop over the resistor. Now, the direction of your causal interpretation is wrong. Assuming an Ohmic resistor ...

1

I can verify using Kirchoff's laws that this is solvable. Ultimately, $$R_{equiv} = \frac{I_aa+I_bb}{I_a+I_e}$$ where the subscripts denote which resistor the current flows through. Setting up $I_e = I_d+I_c$, $I_b = I_a+I_c$, $I_ee + I_cc - aI_a = 0$, $I_dd - bI_b = cI_c$, which is 5 unknowns, but only 4 equations. This isn't a problem, as you can soon ...

-1

The proof is based on considering a 2-loop resistive circuit with a resistance of R1 in series with R2 and R3 in parallel from the left and R3 in series with R1 and R2 in parallel from the right.let us assume the circuit is driven by a voltage V1 in series with R1. There will be a voltage across R3.If V1 is replaced by a virtual V2 in series with R3 but ...

0

There is not a truly simple answer that can be posted on a web site in a few minutes to the last question that was asked by the OP: "How is contact resistance explained?" This is easily shown by reading, for example: Heinz K. Henisch. Semiconductor Contacts: An Approach to Ideas and Models. Oxford Science Publications, 1984.

5

This is a very interesting question, especially considering the very recent history of scholarship on electrical contact resistance (a term first coined in 1964 by William Shockley, one of the inventors of the transistor), as well as thermal contact resistance. For the following explanation, I will use this research paper on electrical contact resistance ...

13

Another term is thermal resistance, This is incorrect. Thermal resistance is something that prevents heat flow. It is an entirely separate concept from electrical resistance. How is contact resistance explained? To obtain very low resistance in a material like most metals, the electrons must be delocalized from the individual atoms, and free flow ...

1

I suppose that in this type of circuit the current and the voltage are in phase, This question is asking for a transient solution, not an ac steady-state solution. So we can't really talk about the phase of the voltage or the current. meaning that immediately after the switch is shut the current should be I=0? This conclusion is incorrect. ...

2

Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current. If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the ...

0

For a given constant resistivity of the material $\rho$, its resistance is $$R=\rho \int \frac{1}{A} dl$$, where $l$ is the differential length element along the current flow at an area $A$.

0

Not that I am aware of. The best I could do is to make a numerical simulation. Make a mesh of points that fill the object, compute the resistance between each neighbor pair, and do a numerical relaxation to determine the potential throughout the object.

1

OK. Let us start with the inital equation $$Z=Z_1\left[1\pm\sqrt{1+2Z_2/Z_1}\right] ;\ (1)$$ and consider lossless circuit. It is a good idea to determine the value of $Z_1$ as $Z_1=ix_1$ and similarly, $Z_2=ix_2$ where $x$ denotes the reactance. For the the capacitive reactance $x<0$ and for the inductive reactance $x>0$. We have: ...

1

Always remember that you are free to deform and stretch a circuit without changing its topological properties, i.e. no deleting or adding nodes, or popping them over circuit elements. The easiest way to see that R1, R2, and R3 are all in parallel is to pull R1 to the left along its wire until it is vertical, and similarly pull R2 to the right along its wire ...

3

Assume you have four waterfalls, and they connect one river to another. The waterfalls drop by five feet. If you put the waterfalls end to end, then the second river must be 20 feet below the first river. If, however, the waterfalls get put side by side, and still connect the two rivers, the second river is just five feet below the first river. Now, this ...

0

since voltage increases with resistance if we connect resistors in series the resistance too will increase with increase in voltage which is usually undesired so the votmeter is arranged in such a way that it gives least resistance and maximum current and that's why voltage gets divided In series. For parallel resistors since change of voltage of one ...

0

I'd use the Mesh method that is simple to remember and use. on youtube and three simple equations (3 current loops) will solve your problem.

1

The standard trick is to split off the circuit after the first link, and treat the 'tail' as another copy of the circuit itself. This means that the impedance $Z$ of the whole circuit must satisfy $$Z=2Z_1+\frac{1}{\frac{1}{Z_2}+\frac{1}{Z}}.$$ This gives a quadratic equation in $Z$ which is easy enough to solve.

1

In simple circuit is current in wire different from current in resistor as wire also opposes the flow of electrons so wire should also be added in series connection of resistor IF you follow this answer step by step the situation should be MUCH clearer to you. IF you just skim this answer it will completely confuse you. Short summary: Read my ...

0

After this comment "See if we think of a series circuit cant we say wire as also a resisitor as wires also oppose the flow of electrons" I think your question is this: If I say that the system has a resistor of $5\Omega$ but there is also the resistance of wire so the total resistance is greater than $5\Omega$? Then the question is of course yes. The ...

1

The current will be the same in the wire and the resistor if they are in series. This is ensured by conservation of charge - you cannot have more electrons leaving a component than entering it. (Kirchoff's current law)

0

I'm not too sure, what you mean by "Point of Symmetry" method, and the link doesn't make much sense as well. Anyways, I'll list the approach, I usually use, see if it is of any help to you. Suppose the cube is: Kirchhoff's current law, which states that the sum of the currents entering and exiting a node is zero, is essential in the analysis. The ...

1

I can't see your graph I supose it has been taken down. I have never seen the exponential fit to the experimental data of resistivity. Could you show me the reference? As I'm searching for experimental data on resistivity of ultrathin metallic layers I can show you few theoretical results. I don't know how thick is your film but I can give you ...

2

Half of the energy is lost to the battery's internal resistance (or other resistances in the circuit).if you try to consider an ideal battery with 0 internal resistance, the notion of charging the capacitor breaks down.since the capacitor and the battery are connected by a (0 resistance) wire, their voltages are the same the instant they are connected, no ...

0

At the moment the circuit is completed, the capacitor has zero voltage, while the supply has $V$. This voltage difference creates an electric field that accelerates charges. This acceleration sets up a current. As the current flows, the capacitor charges until the voltage reaches $V$ as well. At this point there is no voltage difference. But the ...

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