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The best explanation I can think of for you is that on the negative side of the circuit, electrons "pile up", due to the restriction of flow from the resistor. On the positive side of the terminal, there are fewer electrons (the missing electrons were "pumped" to the negative terminal), so it has an effective positive charge. When an electron crosses a ...

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Think of it in terms of conductors and insulators. Metals make good conductors because electrons flow easily through them. On the other hand, electrons flow poorly through insulators like plastic. This is because the crystalline structure of plastics is such that the polymers are bonded together in such a way that the electrons inside the atoms are already ...

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Why it doesn't stay the same since the attraction force from the positive terminal is still there??? I don't understand your reasoning here; an electron moves from the more negative terminal of the resistor to the more positive side and thus, loses potential energy. If the electron were otherwise free, it would have gained kinetic energy equal to the ...

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The electrons accelerate from the source to the resistor, converting their potential energy to kinetic energy in the process. Once at the resistor the kinetic energy of the electron is lost in the form of heat due to the interactions with the resistive material. The total energy of the electrons before and after the resistor is therefore different and is ...

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Your "Rule X" is incorrect. Consider the following network: The rule says: The network is symmetric about the entry point A and exit point B. By symmetry, we mean that if the minimum number of identical resistances along the shortest paths between entry and exit points of the current is the same for two or more paths then those paths are symmetrical ...

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Following on from A Googler's comment to Carl Brannen's answer: But I think $R_1x_1=R_4x_4−R_3$ and $R_5x_5=R_2x_2−R_3.$ What I'm doing wrong? Please explain If you follow this correction through, (ie. swap your subscript 1's and 4's, and 2's and 5's in your opening horizontal consideration - the vertical statements do not need changing), then you ...

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Friction is awfully complicated because you're trying to model a system with infinite microscopic interaction with few macroscopic variables, It ought go haywire theoretically and hence all those "empiric" and "adhoc" stuff you have already found out. It's not possible to find a "unified theory of friction"(Well formally we have one, that is QED). That's why ...

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This problem requires a free-body diagram, where there is a friction force pushing the car forward, and air resistance and rolling resistance resisting the forward motion of the car. If there is a net force on the car (a non-zero force remaining after adding up all of the propulsive and retarding forces) the car will accelerate, either in the forward ...

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Dissolving salt in the water creates sodium and chloride ions which in the presence of the potential of the battery provide a path for current flow, the movement of charge. Thus resistance is decreased and current is increased. While an ideal voltage source would see no decrease in the voltage, a real world battery has its own internal resistance, and so ...

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Yes it does. Classically, the current density in a conductor is given by $\vec j = e \vec v_D \cdot n$, where $n$ is the concentration of charge carriers, $e$ is the charge of the charge carriers and $\vec v_D$ is the drift velocity (this is part of the Drude theory). The drift velocity is the average velocity of the charge carriers, the idea is, that they ...

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I'll give you a hint, since this is a homework question. There are many types of resistors. One is the ohmic resistor (it corresponds to one of your graphs, I won't tell you which one) and it has constant resistance but most resistors have increasing resistance as temperature increases. That means the current decreases as temperature increases. (the ...

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"Current takes the path of least resistance" is just a phrase people say but it's not totally accurate. When one path through the circuit has 0 resistance (a short), it is true that current follows that path only. It isn't true when you have multiple paths, with nonzero resistance, though. A better way of saying it would be "current flows through all paths ...

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$V = iR$ but $P = iR^2$, so if the current, $i$, stays constant but if the resistance, $R$, increases, then the power, $P$, increases too.

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If you are comparing two voltages with identical currents, you cannot be talking about the same bulb in both cases. This means that you are comparing two different bulbs, and there is no way to tell which will be brighter, since different bulbs can be designed for different luminous efficacy, which is light per unit power. For instance, a bulb can be ...

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The circuit CANNOT be solved as a series and parallel circuit as resistor 2 cannot be covered in the way suggested by Ulin Lathrop. Sorry but the answer was not up to the mark.

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This circuit can be redrawn into a simpler version. The circuit so formed will exactly resemble as the "Wheatstone's bridge" with a resistor in place of a galvanometer(which is commonly used for checking slightest of the current passing through it). First mark the points at the junctions as "1","2","3","4" respectively from left to right. Then you can ...

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A resistor is defined as the circuit element for which the voltage across is proportional to the current through and the constant of proportionality is the resistance $R$: $$V_R = R\cdot I_R$$ Clearly, for this linear relationship, it is also true that $$\frac{dV_R}{dI_R} = R$$ However, for general circuit elements, the derivative of $V(I)$ is not a ...

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$R(V,I) = \frac{V}{I}$ by definition, it is not a gradient. $r = \frac{dV}{dI}$ is called the fractional, differential, dynamical or small-signal resistance. It just happens that for resistors $R(V,I) = R_0$ is a constant, thus the two quantities are the same: $r = R_0$.

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My argument was that because the resistance is higher, there must be less voltage going through at that point. This is probably the cause of the confusion. In spite of the usual formulation $V=IR$, in an electrical circuit Voltage and Resistance are the "inputs" to the equation and Current is the result or output. As an analogy, think of Newton's 2nd ...

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The problem is really with the parallel diagram on the far left. It shows a 6V drop across the parallel combination of resisters. That is determined backwards from how we read left-to-right. In order to solve for the voltage drop you must: (1) Solve for the equivalent resister to the pair, the 20 Ohm resister shown in the middle diagram. The equivalent ...

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Use a table of the resistivity of wolfram, ie: http://hypertextbook.com/facts/2004/DeannaStewart.shtml If you MUST use the quadratic form, then calculate A and B from the values in the table.

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resistance will be R=rho(l+x)/a. it wont increase with a factor of x as in you are not multiplying the additional turns. though the resistance increase but the magnetic field strength increases more.

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Why and how does a resistor limit the current flowing through the entire circuit? doesn't it limit only the current that is flowing past and after the resistor? First, this is a DC circuit (ignoring the switch) which is to say that the circuit voltages and currents are constant with time. Since that is the case, by conservation of electric charge, ...

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Two questions: How can the ammeter tell how much current is flowing the resistor? since it's "behind" the resistor? There at least several means that current can be measured using different technologies. The early ammeters used galvanometric technology where a coil in the galvanometer becomes part of the current path. The coil generates a magnetic ...

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We know that electricity cannot pass through glass and wood There is no such thing as a perfect insulator. There is always some minute current flowing through any insulator, including glass and wood. There are ways in which the resistance of an insulator can be reduced. One way is the one you mentioned: a very wide block of it will conduct more than a ...

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Lets do the arithmetic, as suggested by Energizer777 $$R= \frac{\rho L}{A}$$ $$\rho_{copper} = 10^{-8} \Omega m$$ $$\rho_{glass} = 10^{11} \Omega m$$ How wide a piece of glass would I need to have resistance (per meter length) equal to a very fine copper wire with a radius of 0.1 mm? The area of my copper wire is $\pi r^2 = 3.14 \times 10^{-8} m^2$ The ...

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This is actually much simpler than you think - Kirchoff not needed. If you have a known voltage on the terminals of a resistor, you can compute the current directly from Ohm's law. This is the case for $R_A$ where you have a voltage of (12-5)V. You need to know the nature of the COM terminal to calculate the other two. If COM == ground, then the voltage ...

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Perhaps this is what you are looking for: Screen capture: http://www.falstad.com/circuit/ The default circuit, as shown, is an LRC circuit. On the Schematic: Gray is 0V Green is Positive Voltage Red is Negative Voltage The yellow dots are a visualization of current: positive holes. The graphs along the bottom, from left to right, are for the ...

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Because conventionally we assume constant temperature, and length and density are also assumed to be constants for a given resistor. Of course, this is not true. In some circuit designs I have to pay very careful attention to resistance changes with temperature, and indeed this is sometimes used to provide temperature measurements in the form of RTD ...

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Electrons (charge carriers in a wire) move from high electric potential (high voltage) to low electric potential(low voltage). While electrons are travelling, it is the resistors which pick the amount of electrical energy they want (per their electrical capacity) and it is not the electrons that determine how much they should drop off at each of the ...

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The question is ill-posed; the electrons "know" nothing, and voltage is not a property of the electron (other than e.g. charge, which is a property). In fact, voltage is a pretty abstract concept; it is energy divided by charge. And that means explaining an abstract term by another abstract term. Let's be more fundamental: nature shows that charges exert ...

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BEWARE THE SANDWICHES!!! :) In the spirit of math-avoidance sandwich-juggling, here's a better analogy, a visible one. The movable charges within conductive circuits are like silver bead-chains, like those little chains which attach the pens to desks in old-school banks. (Growing up I always played with these when mom was in the teller line. Do those ...

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The easiest method to determine series vs parallel connectivity is to do the following: 1. If one end of R1 is connected to one end of R2 and the other end of R1 is connected to the other end of R2, then the resistors are in parallel. R1 & R2 are in parallel. 2. If only 1 end of R2 is connected to 1 end of R3 and the other end of each resistor is not ...

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The trick is to look at the nodes in the circuit. A node is a junction in the circuit. Two resistor are in parallel if the nodes at both ends of the resistors are the same. If only one node is the same, they are in series. So, R1 and R2 are in parallel and R3 is in series with R1||R2.

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Basically, you simplify, simplify, simplify. For instance, your $R_1$ and $R_2$ are simply in parallel, so you can replace them with a single resistor. Then, depending on which terminals you're measuring from, the merged resistor and $R_3$ will be in parallel or series.

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