# Tag Info

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Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric ...

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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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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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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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$R(V,I) = \frac{V}{I}$ 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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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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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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"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 ...

1

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 ...

1

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 ...

1

It sounds as though you're on the right track: if I understand correctly, you're saying that once you have found the voltage across the pair $R_1 and R_4$ (equal to the voltage across $R_2$ and $R_3$), presumably by lumping $R_1,\,R_2\,R_3\,R_4$ together through parallel addition of $R_1 +R_4$ and $R_2+R_3$, then you simply work out $V_o$ thinking of ...

1

As you may know, it takes infinite time to charge a capacitor. So, the time when the capacitor is 100% charged never comes. Thus, we require a Time Constant to help us understand the time when the capacitor has got a decent amount of charge and after which the rate of charging becomes really slow and thus charging further is not of much use. You may also ...

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Just adding $0.02 for clarity: The formulas$I^2R$and$V^2/R$describe the power dissipated in a resistor. If you're considering the power lost in power transmission lines, the "resistor" is the power line, not the appliance in the home where the power is wanted. The$V$in the$V^2/R$formula is the voltage beetween the two ends of the resistor so, in ... 1 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 ... 1 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 ... 1 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, ... 1 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 ... 1 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 ... 1 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 ... 1 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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