# Tag Info

1

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 ...

1

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 ...

1

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 ...

0

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 ...

0

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 ...

1

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 ...

0

Current is not exactly a flow of electrons but rather a flow of charge. Electrons carry charge, so it is related and you might hear it like that here and there. But remember that not only electrons can carry charge, other types of particles can do that to. We still call it current in those cases, because current is just charge per second moving through. ...

1

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 ...

1

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 ...

0

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 ...

0

Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof: Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left ...

0

First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive. Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery: $$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$ So, the ...

0

The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc. If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge ...

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 ...

0

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 ...

1

"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

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.

0

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 ...

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 ...

2

$R(V,I) = \frac{V}{I}$ 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: $r = R_0$.

2

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 ...

0

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 ...

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

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 ...

0

Your question includes both the conversion (since you speak of processing) and light propagation. Conversion involves electronics, as @Nasha mentions, and thus is directly impacted by the slew rate. Light propagation speed is reduced (with respect to that in vacuum) by the refractive index of the material. The physics causing the finite slew rate is also ...

0

Also, when both S1 and S2 are closed - what type of circuit would this be considered to be (as in parallel or series)? Instinctively it seems to be neither. Or, it's both; there are two, series connected, identical pairs of parallel connected bulbs. With S2 closed, the top two bulbs are in parallel (and so have identical voltage across) and the ...

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.

0

It is not that it is just necessary. Electron flow, in the first place, OCCURS due to the potential difference between the 2 terminals - negative and positive terminals. Emf or potential difference is the driving force of the electrons in the electric circuit which causes the electrons to flow from the negative terminal to the positive terminal. If the ...

2

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 ... -5 Electricity actually flows faster than the speed of light. If you think of electrons as each having a little engine, it makes it very easy to understand. I hope this helps. Edit: There seems to be some confusion as to my statement so I'm going to clarify. Okay. Then. Imagine I've got a block. I pull the block. It moves. Now, imagine I've got two blocks, ... 0 Everything you say is correct in the steady state. The problem you run into is that when you remove charge from a charged capacitor to an uncharged capacitor, there is a potential difference. And somehow, you have to remove the energy from the electron that moves from one to the other. It turns out, as you calculated, that you in fact remove half of the ... 1 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 ... 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 ... 0 You are right that the total resistance increases since L3 is removed, therefore the total current also decreases. However, at the point where the current splits for (L1,L2) and L3, the sum of the current between the two legs of the circuit must equal the total current. When L3 is disconnected, (L1,L2) no longer have to share the total ... 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 ... 3 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... ... 0 Emf is the force at the terminals & within the circuit. It is proportional to the energy stored in the cell & it varies throughout a circuit depending on how the energy moves through the circuit components (higher resistance components dissipate more energy within them--which is what causes a proportional voltage drop (higher Vd with higher ... 1 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 ... 0 What you've drawn is an open circuit. If you initially have the switch open and then close it, there may be some slight momentary flow of current due to the fact that there are always very small stray capacitances between various parts of your circuit, including a very small stray capacitance from the negative terminal of your battery to ground which would ... 0 The reason I give is simply that attraction from some opposite charge induces motion in an electron. The resistance to its flow is caused by the medium. If the medium was empty, there would be no resistance, hence empty space should conduct electricity better than copper wires! A metal wire is conductive because metals have lots of energy levels near ... -1 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 ... 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 ... 2 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 ... 0 Based on equation I just found I was able to solve for the phase angle. phi = arctan((XL-XC)/R) Information from http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rlcser.html 2 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 ...

0

The Reactance of a capacitor varies with respect to the frequency of the signal, AC has a finite positive frequency, but DC has a frequency of zero. Higher the frequency lesser the Reactance, Hence A Capacitor allows AC to pass through, but Blocks DC

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