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0

Let the node between R2 and R4 be a, and node between R3 and R1 be b. Then By voltage Divider Rule: $ Va = (R2)/(R2+R4) * E $ $ Vb = (R3)/(R3+R4) * E $ And then subtract them. You will get your answer.


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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.


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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

In electronics , a voltage divider (also known as a potential divider) is a passive linear circuit that produces an output voltage ( V_out ) that is a fraction of its input voltage ( V_in). Voltage division is the result of distributing the input voltage among the components of the divider.Consider a point A on a rheostat,or simply take a potentiometer.Now ...


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 ...


1

Point particles as the electrons (which are the charge carriers) move according to Newton's law $\textbf{F}=q\textbf{E}=m\textbf{a}$. Whenever an electric field is present it generates a difference of potential between two points $A$ and $B$ given by its differential form calculated between the two points $$ V_A - V_B = \int_A^B \textbf{E}\cdot d\textbf{s}. ...


0

Voltage is a difference in electrical potential in a circuit. You can compare this to a ball sitting at the top of a hill - as it rolls downwards, it moves through a difference in gravitational potential. Because the ball has lower potential at the bottom of the hill, it will roll there spontaneously. The ball will never roll up the hill by itself, ...


3

As you slide the switch open, it doesn't instantly transition from zero resistance to infinite resistance. As the contact area decreases the resistance rises, which acts upon the current to produce a voltage operating against the current. Even when the contact resistance becomes zero, there is capacitance across the gap, which produces a rapidly rising ...


0

The electric field around high voltage transmission lines (or "high tension" lines) is extremely high, and can be close to the breakdown threshold of air. That's why the highest voltage lines use multiple (often three) parallel conductors, to increase the effective conductor radius and reduce the peak electric field. Now, introduce a human into that field, ...


1

When a motor moves it also acts as a generator and the current trough the windings is given by the difference of the external voltage and the induced voltage. When the motor stands still, though, the generated voltage is zero and the windings will draw the max. current they can based on their DC resistance. In other words, the faster the motor runs, the ...


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.


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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 ...


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As stated in the comments, any loss in the system will contribute to the Johnson noise, so you are right about the skin effect and the Eddy current. I want to add that, interestingly enough, this apply not only to electric circuits, but to other linear dissipative systems. A very interesting paper from 1951, Irreversibility and Generalized Noise, proves it ...


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I don't think there is a "preferred" type of LC circuit. How to wire your circuit depends on how you want your circuit to work. Notably the impedance is different in the two cases, while the resonant frequency is the same according to Wikipedia. If you have a look at the equations they present you can quickly see that in terms of what the two kinds circuits ...


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 ...


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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 ...


2

When one says that bulb is 100-W, does that mean it is 100-W at 120V, which would tell me the resistance of the bulb? Somehow I have to find the resistance of the bulb which is not given. Recall that, for a resistor, the AC power dissipated is $$P_R = \frac{v_{rms}^2}{R}$$ Assuming the voltage across the bulb is not significantly reduced by the ...


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You didn't add the resistors for the light bulb and cord in series. $$I_{system} = \frac{120V}{R_{cord}+R_{bulb}}$$ Edit: Your math seems to be correct on the resistance of the cord though. I get 0.167 Ohm Now, multiplying out by the current to get power, since the energy drop across the bulb is 100W: $$I_{system}^2(0.167 Ohm) + 100W = ...


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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.


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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

Why does it maintain the status quo? There is energy stored in the magnetic field, and the magnetic field is proportional to the current. In order for the current to change, the magnetic field must change. By conservation of energy, that means the magnetic field energy must be transformed to some other kind of energy. Until you provide a mechanism ...


0

The information about inductors you were given is not quite correct. Inductors resist a change in current flow, just like capacitors resist a change in voltage. When an inductor is switched into the circuit, the current starts to increase quickly, but the increasing magnetic field impedes the current. As the current increases, the magnetic field gets ...


0

Start with thinking of the case that you have lossless inductive loop in which the stationary (dc) current flows indefinitely, so then no loss and no current decay. Now assume that the sinusoid current is so large that the dissipation per cycle is insignificant when compared to the magnetic energy ("high Q") case, so that you have a constant amplitude ...


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. ...


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Another way to look at this is to realize that whenever you move charge, you produce a magnetic field. Even a straight wire will have a certain amount of self-inductance. Initially, this will provide a limit as to how fast the current can increase. As the voltages equalize, the interaction with the magnetic field will actually cause the current to continue ...


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

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 ...


2

This is more of a math question, but it's just a trigonometric identity. $$\cos\Big(\frac{\pi}{2}-x\Big)=\sin(x)\\$$ Or $$\cos\Big(x+\frac{\pi}{2}\Big)=-\sin(x)$$


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

According to an Intel study, soft-error failure rate at 16 nm is expected to be more than 100 times that at 180 nm, because with scaling of operating voltage, the critical charge required to flip a stored value has been decreasing. Also, in atmospheric radiation, particles of lower energy occur far more frequently than those of higher energy and hence, with ...


2

The temperature of the circuit can probably be crudely modeled by assuming that you have two channels: one for heat absorption, and one for heat loss. Heat loss Suppose you have something like a light bulb. The hot filament loses heat via radiation and conduction. Let's focus on radiation. The rate of heat radiated by an object of temperature $T$ depends ...


0

Resistive materials in your capacitor dielectric will totally distort the field pattern and essentially "guide" the e-field so it extends outside the plates. Replace your wire with a large number of hundredth-ohm resistors in series. Mark the voltages at each node, then sketch in the field lines. You'll find that the chain of resistors is acting like a ...


1

First, do not put them "in series"! Each is powered from the mains ie they are in parallel. Second, earthing is typically applied to any external metalwork that anyone might come into contact where there is any danger of the live coming loose and touching that metal. Think worst case fault conditions. Finally, the fact you have to ask these question means ...


1

The main requirements are that Current flow into or out of "static sensitive" components be limited to below a level where damage is caused by I^2 x R heating. Devices which are damaged by voltage without current flow (such as Field Effect Transistors gate to drain voltages) do not have excessive differential voltage applied. If all "bodies" concerned ...


0

The charge on the two plates will get distributed in such a manner: (+1C)|||(+3C) (-3C)|||(+1C). There is a particular behaviour that I noticed(while doing many questions of the same type) in which the charges on the outer surface of the plates is equal to the sum of all the charges present on the combination of plates... This is what I have used ...


0

It's not entirely clear from your question, but it appears you have 6 coulombs charge on a capacitor C, which is then distributed to be on a capacitor 2C. The rest should be obvious application of the voltage on a capacitor equation.


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You're exactly in trouble! Wh is a symbol of ENERGY, not the POWER!


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You should look at an electric circuit a like a river or a water slide attraction in an amusement park (see my little artist impression below). The resistors are the steep parts: that's where the potential energy is lost. The wires are the horizontal parts, so there no potential energy is lost. But as the water is already moving, it doesn't stop moving in ...


0

Technically you are correct and there is a voltage drop with distance but since the voltage drop is current * resistance and your resistance on a PCB track is usually measured in the milliohms it's ignorable unless your doing something really hairy with LOTS of current or ultra sensative.


1

The unit Wh, shown on the battery, is Watt hours, which is the total energy (Joules), not the power.


0

You are right for an ideal voltage source and an ideal wire Ohm's law doesn't work anymore. You can easily see that mathematically for fixed $V$ and $R=0$ the equation $V=IR$ has no solution for $I$. So Ohm's law doesn't work here, but since there is no ideal voltage source in the real world why bother looking for some generalized law that includes this ...


1

A magnetic field is determined by the current and a changing electric field. And it has energy just for existing. It takes energy to make the magnetic field, for instance to increase the current, and you get energy back when magnetic fields decrease in strength. For a common inductor the magnetic field and associated stored energy are due solely to the ...


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

With a perfect conductor, there is no resistance in the wire. However, that does not mean there is no resistance in the circuit, as the battery itself has some internal resistance. The current through the wire will then be limited by that resistance, i.e. $$I=E/R_i$$ The internal resistance is quite low. It has to be as otherwise its voltage would drop too ...



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