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Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...

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Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...

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Any wire circuit will have inductance and capacitance between the "outbound" and "return" wires - this immediately follows from very basic laws of physics, and in fact is intimately related to the finite propagation velocity of the electrical signal. The expression $$u=\frac{1}{\sqrt{LC}}$$ would give an infinite velocity if either $L$ or $C$ was zero... ...

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Notice We know that the reactance of a capacitor is given as $$\color{blue}{R_c=\frac{1}{2\pi fC}}$$ Where, $C$ is electric capacitance & $f$ is the frequency of source For an A.C. source, frequency, $\color{red}{f>0}\implies \color{blue}{R_c=\frac{1}{2\pi fC}>0}$ which means that a capacitor offers a constant resistance in A.C. circuit i.e. it ...

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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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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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Let's think about the circuit you drew. It contains a battery, switch, bulb, and a very large inductor. In fact, the inductance of a wire that goes 10 times around the Earth can be calculated (I am going to assume an air core - in fact there is a piece of iron in the middle of the Earth which makes the resulting inductance greater). L\approx N^2 R \mu_0 ... 2 Using the "drop-in-a-bucket" trick, we find that an LC laddar has impedance \begin{align} Z &= Z_L + Z_C||Z \\ &= Z_L + \frac{Z_C Z}{Z + Z_C} \\ Z^2 - Z Z_L - Z_L Z_C &= 0 \\ Z &=\frac{1}{2} \left( Z_L \pm \sqrt{Z_L^2 + 4 Z_L Z_C} \right) \, . \end{align} The impedance of an inductance L is Z_L = i \omega L and the impedance of a ... 2 The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the ... 1 Steady state means, in this context, ignoring transients due to initial conditions. For a capacitor, the steady state current due to a DC voltage source is zero. The steady state current due to an AC source is simply the (constant) amplitude of the sinusoidal current. To be clear, the current is changing in time and so is not constant. But, the amplitude ... 1 The symmetry is that the lamps in the top half are the same as the lamps in the bottom half, albeit with a left/right switch. When S2 is closed, it is clear that the left/right placing makes no difference, since a and b are at the same voltage. 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 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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"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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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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As long as the DC component does not saturate the core of the transformer, the (lower frequency) components of the waveform should be induced in the secondary. Consider, for example, the output transformer of a single ended class A triode audio amplifier Image credit In this case, the primary current is 'pulsating' DC, i.e., the primary current varies ...

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Yes surely, The pulsating DC is impure dc. Each pulse will be creating a change in magnetic flux in the transformer core. If you see the normal ac diagram the wave from 0 to T, It is similar to your pulsating DC diagram, there is a change in flux in transformer in this case. But it interesting to note that the transformer will give the increased or ...

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Does the the EM wave follow the same path of that of the drift velocity? I'll assume you're asking about a case where the wire loop is large enough to radiate effectively. Meanining you're asking about a resonant loop antenna, with a circumference approximately equal to the wavelength of the signal being applied. In this case, no, the EM wave (or at ...

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Current in the wire is because at each point there is a net Electric field along the wire. This E is because of the variation of surface charge density on the surface of wire as we move from Anode to Cathode. This E is continues and of same strength at a given instant of time at any point inside the wire under consideration. The variation of E(t) depends ...

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The short answer to your question is that EM waves travel in the same direction as the wire and current, guided by two opposite conductors, and flow into any device that consumes power (has a voltage drop across it and current flow through it). So for your light bulb circuit, the wave flows from the battery to the lightbulb between the wires. Here's an image ...

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Not completely clear from your description, but do you mean that you tuned your frequency to achieve a current resonance and then inserted the iron core? Did you not consider that the resonant frequency is completely changed by the increased inductance, so you would then be far from resonance at the same frequency? Depending on the resistance in 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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Capacitance is about stored charge - more electrons flowing into something than flow out. This can happen in a piece of wire, although it can take a large amount of applied voltage to accumulate a small amount of excess electrons. In other words, a simple piece of wire has very low capacitance. Even a straight piece of wire will have inductance because any ...

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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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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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First - simplify the circuit. You have $L_4$ and $L_5$ in series (on opposite sides of the battery but that doesn't matter for calculating the current through them) and $L_1$ and $L_2$ in series. The simplified circuit looks like this: Now we consider what happens to the current in $R_1$ if we remove $R_2$: clearly, the total current through the circuit ...

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