# Tag Info

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A generalized view about impédance of basic components and combination of components is exposed in the paper "The Phasance Concept" published on Scibd : http://www.scribd.com/JJacquelin/documents

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Wires are in fact resistors, but with VERY VERY tiny amounts of energy being thermally dissipated by the current due to EXTREMELY low drops in voltage over large lenths of the wire. Thus, current DOES flow through the neutral wire, but the drop in potential along a length is literally far too small for your voltmeter to detect. Review Kirchoff's Voltage Law ...

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I think your question answers itself, indeed the area of what? Faraday's law is for a closed loop of wire, thus Faraday's law is inappropriate and we should look for an alternative, as you have done by considering the Lorentz force. If the metal were a closed loop of circumference $l$ then Faraday's law would be valid. The forces on the electrons from the ...

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In the absence of supporting information, it's hard to say where to start. Do you have a circuit diagram, at least? Typically one would identify certain nodes at which the voltage(s) are known, then trace all electrical paths to figure out which capacitors have that node in common (suggesting parallelism), and which capacitors are connected in sequence on ...

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This may not be a direct answer, but it can be shown by an elementary method that the relation between the number of nodes, branches and loops in a well-posed problem corresponds to Euler's polyhedron formula.

2

The analogy is wrong. A voltage source can only shock us if it is able to pass a considerable amount of current through our body ( ~ 250 mA or so, I dont know the exact value but you can Google it ). The circuit that you are trying to discuss, does indeed have 36 Amps of current flowing through it, but once you connect yourself to the circuit, you are in ...

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With respect to power loss concept, when we say that the power is dissipated (or lost as you call it) it means that power was dissipated (or spent) as something else which might be useful (as an example power dissipated in a perfect lamp where all power is converted into radiation) or not useful (example is portion of power lost in heating the motor of fan, ...

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Just to add to the answer a DMM is not considered a good way to measure such a high impedance reading like skin conductivity. A typical Galvanic Response Sensor (GSR) has the ability to adjust filters/rates and is optimized for high impedance measurements to reduce the signal to noise ratio. DMM's will often pick up environmental noise on very high ...

2

It has nothing to do with any capacitance. It's all about skin resistance. The resistivity inside your body is much lower than that of the skin. As a result, your measurement is really showing you the sum of two resistances thru the skin. The main reason skin has higher resistance than the body internally is because the skin is dry. However, the skin ...

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The ground which locates in the non-inverting terminal of an Op-Amp is called virtual ground.

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Actually there is a little voltage on the neutral wire when current is flowing thru it. However, this wire has low resistance specifically so that the voltage accross it remains small. The resistance of the lightbulb is much larger than that of the wire, which is why you see most of the voltage accross the bulb and not the wire. Consider the limiting case ...

2

Voltage is the difference in potential between two wires. You can't say the "hot" wire has any sort of voltage by itself. Just as you can't say the "neutral" wire has any voltage by itself. The voltage is the measure of electrical potential between the two. When you connect the light bulb between the two it provides a path for current to flow from high ...

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The neutral regions have some characteristic finite resistance which decreases if diode is forward biased and vice-versa. in practice we consider a variable resistance in series with depletion region and the metal semiconductor contacts are taken into account. EXPLANATION (Taken from Millman's book) An ideal p-n diode has zero ohmic voltage drop across the ...

3

Calling it a built-in voltage is something of a misnomer. People usually think of "voltage" as "what you measure with a voltmeter". So "voltage" is normally synonymous with "electrochemical potential of electrons" (in stat mech terminology) and with "difference in fermi level" (in semiconductor terminology). Under this definition, the built-in "voltage" is ...

3

Kirchhoff's loop rule is also called Kirchhoff's voltage law (KVL). Which is different from Kirchhoff's current rule which is also called Kirchhoff's current law (KCL). KVL is derived from Maxwell–Faraday equation for static magnetic field (i.e. the derivative of B with respect to time is zero). KCL is derived from charge continuity equation which is ...

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I'd like to build on Lionel Brits's answer and the electronic book reference you cite Sophocles J. Orfanidis, "Electromagnetic Waves and Antennas", Chapter 22. The structure of the relationship is more readily understood if you think of the Laplace or Fourier transforms of all your impedances. A relationship between $I(t)$ and $V(t)$ that is given by a ...

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Your question is difficult to answer exactly because of the assumptions you have to make for the ideal case. In the real world you can not have current flowing without an electrical field due to Ohm's Law where $\mathbf{J} = \sigma \mathbf{E}$. You can try to argue in a perfect conductor $\sigma \rightarrow \infty$ so it doesn't matter what $\mathbf{E}$ ...

2

This question is somewhat nonsensical because it is current, not power, that maintains the magnetic field in a solenoid. Also, you haven't defined P, so we obviously can't say if it changes. What happens in the circuit also depends on how the solenoid is driven, like with constant voltage, constant current, or something else. When you are driving a ...

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Suppose you have a dipole radiator. Then you can look up or calculate it's radiation resistance $R_\mathrm{rad} = \frac{2 \pi}{3} Z_{0} \left( \frac{\ell}{\lambda}\right)^{2}$ , which might be off by a factor of 4. I would say that you can use the superposition principle to just treat the problem in frequency space. There is also reactance to think about.

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An electric field isn't necessarily required to sustain a current. Remember electric charge is accelerated by an electric field. In the case of an ideal conductor, which is assumed to connect the source to the resistor, the current can be any value and the voltage across the conductor is identically zero. This isn't a contradiction. Consider the motion ...

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There's a limit to how an "ideal model" can be used. Actually, that was an unintended joke: sometimes you need to consider how an ideal model behaves in the limit, not at the limit. So, consider a voltage source with $V = \epsilon$ with zero internal resistance and look at the current flow through a resistor as the resistance $R$ approaches zero. The ...

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There is some small potential difference between a and b, because conductors always have nonzero resistance (at room temperature).

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The word "Earth" is sometimes misleading. I think (if I get the sense of your question right), it is more properly called a "Protective Earth". In home electricity supplies, one side of the supply is "tethered" to the same potential as a protective earth circuit. This latter is simply a system of conductors, going through the third "Earth" pin on the socket ...

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Parallel connected circuit elements have identical voltages. Series connected circuit elements have identical currents. Now, fill in the blanks: To measure the voltage across a circuit element, the voltmeter should be placed in __ with the circuit element. To measure the current through a circuit element, the ammeter should be placed in __ with the ...

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P=VI 480A at 120V. I calculated 57.6 J

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Use a Volt and a Ammeter to calculate , P=VI , shouldn't that be simple enough

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This is not correct (dissipated power is depending on operation conditions, it is not invariant). Correct is the value of the current through LAMP2 (it is 1.13A). You need to look up what voltage this corresponds to from the LAMP2 chart. The voltage V1 is then 5V plus this voltage.

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No current flows from C to B or A to F. So if you can find the potentials at C and F, you need only add or subtract the 4V potential. I would start by defining the potential at C, D and H to be zero. 12-8 and 10+5 may help get current for D to H, remember potential from D to H is zero, that's a wire. Looking forward to your final solution.

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This diagram (cribbed from here) shows the voltage current curve for a typical zener diode: Below the breakdown voltage the diode does not behave as a perfect insulator, but has a small leakage current. This means the voltage across the diode is strongly current dependant. The minimum current, $I_{Z(mini)}$, is simply the current at which the breakdown is ...

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$$V = I R \space \space \space \space \space \space \space \space \space \space or \space \space \space \space \space \space \space \space \space I = {V \over R} \space \space \space \space \space \space \space \space \space \space or \space \space \space \space \space \space \space \space \space R={V \over I}$$ The basic equation for voltage ...

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$P= IV$ is fundamental, and applies to any circuit element (they have complex analogues too for time-varying circuits), whereas the equations containing $R$ are specific to resistors. It doesn't matter if the circuit is in series or parallel. You just calculate equivalent resistance in each case.

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They are all equivalent. But if you are given P, V, R and I it doesn't make any sense to use any of the equatios you mentioned. What would you want to calculate?

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