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But what about the 180° gap in the middle? This implies a negative real part of the impedance, i.e., a negative resistance. Express the impedance $Z$ in polar form $$Z = R + jX = |Z|e^{j\phi}$$ where $$|Z| = \sqrt{R^2 + X^2}$$ and $$\tan \phi = \frac{X}{R}$$ For $R > 0$ $$-90^{\circ} \lt \phi \lt 90^{\circ}$$ but for $R < 0$ $$90^{\circ} ... 0 If the phase angle is greater then 90 degrees the load would effectively become the source and vice-versa. 0 I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its V (notice that V is really just measuring the voltage drop across the ... 1 I think the range of answers you have got from friend, teacher and web reflect that there is not a straightforward response. Without high voltage it would not be possible to drive the dangerous current through the body, but high voltage itself is not lethal - it depends how much current can be delivered at high voltage. Another question is how high does the ... 0 Electrical sources, like current sources are conventional electrical elements that behaves according to a definition. A current source is said to impose a current. For example, when studying the circuit, you do assume that the current in the wire is chosen by the current source only. Thus, when there are two different current sources in series, you have two ... 0 If I understand correctly, the power that power sources output can be constant, but the current and the voltage is purely dependent on what loads you have in the circuit. Say my power source is a coal burner, that burns 5 coals a minute to produce 10 watts of power. Using P=IV, the current and the voltage can both be seemingly a wide variety of variables. ... 3 Actually, I am not sure you should expect a much lower resistance value. For example, with aluminum resistivity of 2.65x10^(-8) Ohm-m, a sample with length of 5 micron and cross-section area of 100 nm x 1 micron has a resistance of about 1.3 Ohm. Of course, you can question the rather arbitrary effective width of the sample - 1 micron, but I guess this ... 7 Another possibility is that you are having to break through the oxide film that all Aluminium has in air. The voltage needed to do that can vary considerably. You might need to use sharp steel electrodes that will punch through the oxide and contact the metal directly. 8 This isn't really an answer, because your question doesn't provide enough information for an answer. However it explains what you need to do. In fact this is exactly what I did (back in 1983!) to measure the resistivity of evaporated silver films. If you have the film formed on some substrate you need to score two lines to leave a long narrow track like ... 3 The "derivation" you describe is valid at a particular moment in time.$$\begin{align}\Delta E &= V \Delta Q\\ \frac{dE}{dt} &= V \frac{dQ}{dt} \\&= VI\\ P(t) &= V(t) I(t)\end{align}$$I added the dependence on time explicitly. To address your comments: W=QV is only true when V is constant; you can't simply take the derivative and ... 3 Power is an instantaneous concept. P=IV gives the instantaneous power at a given instant of time, given V and I at that time. 2 You are asking for a proof of Kirchhoff's voltage law or KVL, that the algebraic sum of voltage differences around any loop in a DC circuit is zero. KVL stems from Faraday's law, which essentially says that (unless there is time varying magnetic field, which, for an ordinary circuit is not true) the E field is curl-less, therefore the integral of E around ... 0 I found the solution to my problem : The equations can be written :$$ I = V \times C_{ij} \times i\omega  Where $I$ is the vector of known currents going threw the electrodes, $V$ the potential of the electrodes and $\omega$ the rotational frequency of the current. In this case the potential of the transmitting electrodes is known and the current ...

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You've begun with this: $P_l = I^2R$ $P_l = IV$ This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop. Your diagram with a single resistance and a power station implies that the current in the line depends on that ...

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Can we say that there has been a voltage drop in the current First, it is true that electrons leaving a resistor have less potential energy than those entering the resistor. Second, assuming an ideal conductor, the electrons that flow along the length of the conductor have the same potential energy. This must be the case since there can be no electric ...

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All batteries have some internal resistance. If you connect the terminals with a wire of zero resistance than there is no voltage drop across the wire and the whole voltage drop occurs within the battery due to its internal resistance. If you go farther and require that you battery has zero internal resistance as well, then when you connect the wire the ...

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The equation for capacitance is Q=CV or V=1CQ. I don't understand what is the physical meaning of this "C": Does the charge in a system changes linearly with voltage under all circumstances? This first part is the statement of the behavior of an "ideal" capacitor. Because it is an idealization, it is easy to characterize as a linear component whose ...

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Worth noticing the difference between capacity and capacitance. You want the latter to have the meaning of the former - when instead it just describes how much the voltage increases when you add charge. It doesn't tell us when the capacitor is "full" - for that you need to know the rated voltage as well as the capacitance. When a capacitor is used in a ...

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The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...

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You're allowed to be confused on this issue - for many students it's a new phenomenon for which no analogy is perfect. Current is very easy to understand. As you mentioned it's basically just electrons flowing past per second. However, as it would be rather time-consuming always saying things like "the current is 1000000000000000000 electrons per second", ...

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First of all I think we are thinking about the free electrons in the metal rather than the ones which are bound to individual atoms. Generally to apply a voltage we attach connections to each end of the metal and apply a potential difference. One way of thinking about what happens when we apply a voltage is that the fermi level gets slightly tipped from one ...

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You should be able to get the same result by using the electrical work formula - but note that you need to integrate since the force changes with position. That's really all the potential is - it is the integral of force for unit charge. That's why force has the $1/r^2$ relationship while potential has $1/r$ (with appropriate signs and constants...). ...

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