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0

In this case, the magnitude is telling you how to scale your input signal, and the argument is telling you how to phase shift it. Complex numbers usually represent 'amplification' and 'twist'. So, say, 1 means 'leave it the same', 2 means 'double it', 0.5 means 'halve it', i means 'one quarter turn', -1 means 'one half turn', -3i means 'triple it and give ...


1

Imaginary components in physics often mean phase shifts. In this case the impedance is sort of like a resistance, but it kicks in when there's a changing current by messing with its phase.


13

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


5

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = ...


0

To be clear on the setup, we have an ideal battery (DC voltage source) and an ideal wire (zero resistance. In ideal circuit theory, asking what happens when one connects an ideal wire across an ideal voltage source is essentially asking what happens when $1 = 0$. However, if we allow that the wire has non-zero radius and non-zero length, then, when the ...


0

It depends on inductivity: I suspect the circuit would explode, with or without capacitor. With just the battery - which also has zero inner resistance - the current would get infinite during the first 0 seconds. Except - while no energy is lost in heating wires - that would make the whole capacity of the battery.going into creating a magnetic field ...


0

A few pointers (since this is homework): A capacitor of capacitance $C$ charged to a voltage $V$ has charge $Q$ such that $$Q=CV$$ The positive side of capacitor 1 has charge $+Q_1$, the negative side has $-Q_1$ etc. Then the total charge (on the two capacitors) after connecting the capacitors depends on the polarity of the connection you made. If you ...


1

When a wire with no resistance is connected to the terminals of an ideal battery, will a current exist in the circuit? yes, it would be infinite If a capacitor is added to the circuit, will it be charged by the battery or will it remain uncharged? yes, the capacitor would be charged.


0

The battery would be shorted in first case. In the second case: Although, resistance is never zero. You see that while charge on a charging capacitor varies with time as $$q=CV(1-e^{-\frac{t}{RC}})$$ When you set $R=0 \Omega $, the result is undefined. Although you can talk of $R\mapsto 0^{+}\Omega$ In that case, the capacitor will be charged instantly. ...


0

I will give you hints how to do it. If you do it yourself, you will gain confidence. Charge will be conserved separately in the two plates you connect The charges will flow till potential difference across both capacitors is same. Note that equation $1$ will matter on which $2$ plates are connected. When you connect a resistor, the steady state will ...


2

If you want to include "all real world effects" in your analysis, you need to make sure you include all effects. At the very least, include parasitics. And include the fact that your "real world voltage source" has finite impedance, output capacitance, inductance in the leads, ... So when you state Say it starts of at a voltage V when you connect it to ...


0

A partial answer to where the simple linear no-Doppler model has a gap: in the coupling of the speaker cone to the air. There is a boundary condition imposed on the acoustic wave equation at the location of the speaker cone. However, the speaker cone itself is moving around, meaning that this boundary condition is imposed at different locations at ...


2

The Doppler shift for small speeds is $\Delta f/f = \Delta v/c$, where $\Delta v$ is the (signed) speed of the source relative to the detector, and I'm using $c$ as the speed of sound. So let's plug in some numbers. I'm going to use numbers that will produce a large effect to see how larger an effect is plausible. Let's take a woofer operating at $f = 200 ...


1

Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals. If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then ...


3

Think of it in terms of current, V, W, and Z are in series, so each is equally bright. X and Y are in parallel, so each gets half the current of the others. If you assume each bulb is a constant resistance R (not true for incandescent bulbs, by the way), then V,W and Z will each dissipate $i^2R$. For X and Y, since each has a current i/2, the power will be ...


1

The conceptual problem here is that of EMF, $\mathcal{E}$ vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ...


0

If the three resistors are connected in series, the total effective resistance is $3\times 2\Omega=6\Omega$ and the current is $I=U/R_{\rm tot}=12/6{\rm A}=2{\rm A}$, and the potential across each resistor is $IR=2\times 2{\rm V}=4{\rm V}$. However, an easy way to do it is, since we know in this case the current through each resistor is the same, so the ...


0

Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator: $$ i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t $$ $v_L$ would be $$ v_L(t) = ...


0

There are two kinds of "earth" being talked about here. There is: The kind that aims to use the ground itself as a "return path"; and A protective "earth", which is actually a separate conducting line laid throughout a building. For the "return path" earth of 1. there are several ways wherein this will work: There is actual conduction (i.e. drift of ...


0

Actually according to my opinion it give you a shock but it is negligible. If you give a high energy and try to make contact with earth it will neutralize by giving a little to you. The reason is, if there is a good conductor like body it goes through the body other than only earth. Just take a lightning application and think of that.


3

Earthing something means dumping the electron flow into the earth. Since the earth is so big, it can absorbe/give a practically infinite amount of charge without changing potential, this means that you can treat earth as a reservoir of ready to use electrons. If you plug the phase of your home power line into the ground (without safety devices in the ...


-1

Your problem is clearly and comprehensively treated by Hans De Vries in: http://chip-architect.com/physics/Magnetism_from_ElectroStatics_and_SR.pdf The quintessence is that a current carrying wire appears electrostatically charged to an observer in relative motion to that wire, even when the same current carrying wire appears uncharged to an observer at ...


1

The voltage across $R_1$ and $R_2$ will be the same, $V$. You are right that the net resistance will decrease to $(1/R_1+1/R_2)^{-1}$ and this change will be compensated by an increase in the current $I$ (Ohm's Law). $V$ will stay constant. Why? Because each component in a parallel circuit has two common nodes with each of the other components in the ...


0

$Electromotive$ $force$ abbreviated as E.M.F and denoted by $\varepsilon$ is not a force. It is defined as the energy utilized in assembling a charge on the electrode of a battery.Simply, it is the work done per unit charge which is the potential difference between the electrodes of the battery measured in volts. Mathematically, $\textbf{V} = ...


0

it doesn't seem to have anything to do with phasors But it does since this is the way one adds phasors. Assume you have two series connected circuit elements with phasor voltages that differ in phase by the angle $\phi$. $$\vec V_1 = V_1$$ $$\vec V_2 = V_2e^{j\phi}$$ Since the voltage across the series combination is the sum of the individual phasor ...


1

$V_L$ and $V_R$ are phasors which are 90 degree out of phase. Like two vectors, phasors follow the same rule for addition that's why they have used pythagoras theorem.


0

try taking a look at this page it goes in to good detail about inductors in a dc circuit. http://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_15.html


4

Thare is only one node connecting elements B, D, E, and G. The dots on the diagram are just to indicate that the lines do in fact connect, so that all of the wires are part of the same node. In this kind of schematic diagram wires are considered as ideal, and all points on the wire are considered to be at an equal potential.


1

See here. There's instructions on how to calcuate $V_{th}$ and $R_{th}$. Let $I_3$ be the current through $R_3$, $I_1$ through $R_1$, etc. With $V_{AB}$ open, by KVL, we have $V_1 - R_1I_1 - R_3 I_3 = 0$ and $V_1 - R_1 I_1 - R_2 I_2 -V_{AB} = 0$, but when $V_{AB}$ is open we have $I_2 = 0$ and so $I_1 = I_3$, so $V_1 = (R_1 + R_3) ...


1

Both examples are using the same methods. In the first example, the break in the circuit removes the 6 ohm and 5 ohm resistors altogether. All of the current then flows through the 2 ohm resistor, causing the Thevenin resistance to be 2 ohms. In the second example, the current first passes through $R_2$, then is split between $R_1$ and $R_3$ in parallel. ...


0

Suppose first that $C_{5}$ is absent, that the voltage at the connection point of $C_{2}$ and $C_{3}$ is zero and that of the connection point of $% C_{1}$ and $C_{4}$ is $V(\omega )$. The impedance of a capacitor $C$ at the angular frequency $\omega $ is $$Z(C)=\frac{1}{i\omega C}.$$ Then \begin{eqnarray*} V(A) ...


0

See at bottom. A formula for unbalanced Wheatstone bridge. (Lazy as shikamaru to write the stuff in latex). sorry for quality The book moved while scanning.


0

No, potential difference across each resistor need not be different always. In series, you should remember that potential difference across any resistor is proportional to its resistance. In other words, potential difference can be same across each resistor if each resistors value is same. If resistance values are different, say $R_1$ and $R_2$, then by ...


0

Your reasoning is correct. If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law: $$I = \frac{E}{r + R_L}$$ or $$E = (r + R_L)\cdot I $$ The output voltage $V$ of the solar cell is the voltage across the load ...


0

You need to assume how the current gets divided across the whole circuit. For example, take the current coming out of the main battery as $I$, then when it reaches the loop that you have selected, it breaks into $I_1$ and $I-I_1$. Do the same for all the loops, then apply Kirchoff's Rule. For capacitors, apply it exactly the same way you do it for batteries. ...


2

There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements. You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves. The magic is dealt with here http://oeis.org/A000669 . It talks ...



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