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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 ...


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 ...


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 ...


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 ...


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. ...


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 ...


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

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 ...


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)


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 ...


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

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

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.


1

Why is resistance NOT calculated simply by looking at how much voltage drop is created by a certain amount of charge passing through, as in R=V/Q. Your statement in bold is one way to define resistance. But your words do not match the expression $V/Q$. We're interesting in the voltage drop per charge passing through, right? Well, how do you measure how ...



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