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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

Of course If we have more than two states we would be able to hold more data per bit, ... But only because you are redefining the meaning of "bit". According to Information Theory you are not changing the information content but merely the unit you use to measure it. In fact, you change it from Bit to Ban, or about 3.32 bits :)


0

ENIAC was base 10, so not only is it possible to use bases other than 2, base 10 was used first. According to "50 Years of Army Computing: From ENIAC to MSRC", EDVAC was base 8 and ORDVAC I and II were base 16.


1

Actually, there are many computers that use many states per bit since the turn of the century. They are called analog computers. Actually, the slide rule is considered an analog computer and it's been around for centuries. Just search the internet for information.


0

A brief but helpful description of whether it is possible to implement ternary, decimal, or other base-$n$ computing schemes can be found halfway through this article by Mark Chu-Carroll. More importantly, he explains why there is virtually no advantage conferred by using larger $n$ in base-$n$ representations. The reasoning that using base-$n$ for large ...


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 ...


0

Why would it be any more dangerous than, say, wearing a metal wristwatch, a ring, or metal body piercing? The answer is not at all, in fact less, since most devices with a wireless device in them are surrounded by plastic, which is an insulator.


1

The two currents $i$ and $i_1$ could have been chosen to have whichever direction the author wanted. The choice is arbitrary. The equation $i+ i_1=0$ is just what KCL tells us: the net current leaving any node in the circuit is 0. Since $i$ and $i_1$ are both defined to be leaving the node that connects them, their sum must be zero. This means that one of ...


0

but it isn't the case with resistors in parallels (1) parallel connected resistors have identical voltage across $$V_{R1}=V_{R2}$$ (2) the voltage across a resistor is given by Ohm's law $V_R = I_R\cdot R$. Applying Ohm's law to parallel connected resistors yields $$I_{R1}\cdot R_1 = I_{R2}\cdot R_2$$ Thus, conclude that the parallel connected ...


0

Most basic concepts that must be understood here are "Voltage" and "Current" . Voltage: Voltage is difference in electric potential between two points/nodes. Current: Current is simply flow of electric charge and is defined by Ohm's Law as: $$I =\frac {V}{R}$$ Now, in a series circuit, all resistors are placed between two nodes. Voltage among these two ...


0

In series circuits, resistances add can be simply added and treated as one big effective resistance: $R_{\text{eff}}=R_1+R_2+R_3$.... So the current remains same. In parallel circuits the current splits up so each branch has a different effective resistance (in each of the separate branches one can use the series rule again). Due to this, the current isn't ...


1

Perhaps you meant intuitively? Well, if you imagine the roads are tracks and the flow of cars the current, and if you had the choice between a very bumpy narrow path on your left and a very nice motorway on your right, which would you choose? Nearly everyone would go right (there are some dirt enthusiastic or people that don't think), and overall the current ...


0

Kirchhoff's first law says: "At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node". When there is only one current flowing into the node it is simply equal to the sum of currents flowing out. The node or junction is the place where conductor splits to form couple ...


0

By Ohm's law, knowing the current running through a resistor is the same as knowing the potential difference across the across the resistor $V_1=IR_1$. Connecting two components in parallel cause the two components to be in the same potential difference $V_1=V_2$ Thus, $$I_2 = \frac{R_1}{R_2} I_1 =0.3A$$


2

Since the resistors are in parallel, we have $$I_1 R_1 = I_2 R_2$$ thus $$I_2 = I_1 \frac{R_1}{R_2}$$


1

Calculate the voltage on $R_1$, voltage on $R_2$ should be the same, since $R_1$ and $R_2$ are parallel. Then You can calculate current through $R_2$.


0

The magnetic field from current carrying wires is given by Ampere's Law, which itself can be derived by considering the magnetic field from isolated moving charges. In the limit of having many charge carriers, you can express $q\vec{v} \rightarrow I\vec{l}$, a current along a wire segment of length l. You should keep in mind that currents are the more ...


0

In most cases of magnetic driver design, the construction volume $V$ for the winding (including the fill factor) is restricted. At the first shot one does not care about the number of windings. One must get rid of the heat. Therefore, the power $P$ is limited that can be dissipated as heat at the maximal admissible temperature. From this constraint you get ...


0

Let's model the real wire as two infinitely thin wires carrying 1 mA of current 1 mm apart. $$ B = \frac{\mu I}{2 \pi r} $$ $$ F = q v B $$ Plugging this in, and considering the speed of the current being one third of the speed of light (close to the actual value, depending on the wire) we get that the acceleration exerted on each electron is of the order ...



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