# Tag Info

1

You are right in principle, but by considering the field of a battery; you are considering something, which is considered negligible by your book. In real world problems, one always try to ignore effects which are negligible, to solve problems to a reasonable level of accuracy and highlight the principles. If your book starts to talk of all these effects, ...

2

The capacitor will indeed be charged a little -- but the charge will be so low that we may as well call it uncharged. Here is why: the open switch is another capacitor (two conducting terminals, although not quite in plate form, separated by a dielectric). Its capacitance is extremely low, though: the terminals' cross section will be on the order of a ...

0

Your argument is: If the circuit is open, the current must be zero. Consequently the field must be zero. For the field to be zero, the capacitor's field must cancel out the battery's field. Therefore the capacitor must be charged. Replace the capacitor with a resistance. Following your argument in the same way, there must be a voltage drop at the resistor ...

0

A capacitor in an open circuit $may$ be charged: It could be totally discharged, or it could be that the switch was opened while the capacitor was fully charged. It is also known that capacitors leak. That is, they lose their charge over time. This is a bit more of an argument about the semantics of your book, but I mostly agree with you. In your ...

2

In Nilsson and Riedels textbook: Electric circuits, it is actually stated on page 28 that " when represented in a circuit diagram, copper or aluminum wiring isn't usually modeled as a resistor; the resistance of the wire is so small compared to the resistance of the other elements in the circuit that we can neglect the wiring resistance to simplify the ...

1

What might help (or not) here is to distinguish between reference polarity and absolute polarity. When you label one terminal of the inductor with a plus sign, you're simply choosing a reference polarity. You are, in essence, choosing which end to place the "red lead" of your voltmeter. If you calculate that the voltage across the inductor is negative ...

0

You don't need to worry about the convention you choose because the solution of the corresponding differential equation will automatically adjust the polarity. For the sake of simplicity, I usually use the passive convention as follows.

3

You are absolutely correct, the electric field does fall off with distance from the battery. However, this is only true during the transient state (the state of the field when the battery is first connected). In fact not only are the magnitudes inconsistent, but so is the direction of the field. The field doesn't always point in the direction of the wire. ...

-1

The electric field of a collection of static or stationary electric charge(s) would indeed diminish as the inverse square of the distance from the charge(s) outside of the conductor, but not inside a conductor in which an electric current is flowing. Examples of stationary electric charges would be a metal sphere atop a Van de Graff generator, or charges ...

1

Well, if you look at a practical circuit which produces that graph, there must be a sudden drastic change in the circuit at that instant of time to cause the capacitor to abruptly shift mode from charging to discharging - quite possibly a switch/switches were turned on/off effectively putting the capacitor in a different circuit. If you are asking the ...

2

In figure (a), the voltage is continuous but the time derivative is not; the capacitor current would discontinuously change sign from positive to negative. In figure (b) however, the voltage is discontinuous. It is typically said that the voltage across an ideal capacitor is continuous since, for the current to exist, the time derivative of the voltage ...

6

If you have just given the voltage signal with $$\def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V$$ then the current at $t=2\rm s$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that $$i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ... 0 There never is any confusion or contradiction because, unlike charge, the current does not have to be continuous, that is i(t-0) = Cv'(t-0) and i(t+0)=Cv'(t+0). Since time's flow is unidirectional, from  -  to  + , the time derivative of the voltage is well defined to mean f'(t-0) and f'(t+0). 0 A much simpler answer to this question is that the current induced in the transformer secondary windings (assumed not an 'autotransformer') derive from an application of Lenz's law, which is to say, when the magnetic field in the common iron core of the transformer is forced to collapse by current being reversed and transitioning through zero in the primary ... -1 bcause the emf induced in the secondary is known to be back emf...which opposes in such a way to show it lik a spring when gets compressed..it bounces with opposite phase.yoga.... 0 To set the record straigth, conductivity cannot be infinite, an ultimate limit will be given by the velocity of light which can never be over shot. It can be very large as in superconductors but even there, it is limited. You are talking of conductors at electrostatic equilibrium., static electricity. For your vector conclusions to be true there should be ... 0 The charging current warms the battery as it charges, and because of mass-energy equivalence, an unmeasurably small amount of mass is indeed added to the phone in the process. However, this is not what is measured by your electric utility in order to bill you for energy. The insignificant mass added to your phone when charging is lost partially when the ... 0 A single resistor wired in parallel with voltage source like a battery may give confusing reasons for a novice, so a few other things need to be taken into account. Any battery also has an internal resistance (it is like a resistor in series with the voltage source), and this resistance also varies with the state of charge of the battery, which is to say, ... 12 \def\vE{{\vec{E}}} \def\vD{{\vec{D}}} \def\vB{{\vec{B}}} \def\vJ{{\vec{J}}} \def\vr{{\vec{r}}} \def\vA{{\vec{A}}} \def\vH{{\vec{H}}} \def\ddt{\frac{d}{dt}} \def\rot{\operatorname{rot}} \def\div{\operatorname{div}} \def\grad{\operatorname{grad}} \def\rmC{{\mathrm{C}}} \def\rmM{{\mathrm{M}}} \def\ph{{\varphi}} ... 0 Use delta star transformation to G , R1 and R2 That makes circuit simple and solve it 0 It may be pointed out that the word electromotive force is a misnomer. It does not represent force on the carriers of electricity. Instead, it represents the potential difference between the two poles in an open circuit (when no current is drawn from the cell). 0 There is no creation of anything, but it can be assumed that a circuit creates a voltage when the power, combination of that voltage and any current which would flow from it, has been gained from "outside the circuit" - e.g. through chemical processes (batteries), or electromagnetic processes (dynamo that converts mechanical power to electrical). This very ... 0 I dislike the term EMF (Electromotive force) as it is very confusing. Electromotive force, also called emf (denoted \mathcal{E} and measured in volts), is the voltage developed by any source of electrical energy such as a battery or dynamo. Which means that all EMF are voltages but not all voltages are EMF. A voltage is only an EMF if it is a ... 0 1) With a constant and DC power source eventually the solenoid will become fully 'charged'. At that point its 'resistance' term vanishes because it no longer produces an emf against the battery. At this point, the \frac{di}{dt} term will be zero, because the current isn't changing. 2) When you cut power, the magnetic flux is no longer maintained by the ... 1 I think you have some misconceptions about voltage. You mention "net voltage" but voltage is always a difference in electric potential. In a circuit, that means you never talk about a "net voltage" or a voltage at a certain point. Voltage is always meant to be read as "the voltage between two points" or "the voltage at A with respect to B." It is never ... 0 The net voltage around the entire circuit is zero. However, the voltage across the resistor is equal in magnitude and opposite in sign to the voltage inside the battery. You could view this conceptually as electrons flowing from the minus to the plus terminal (through the resistor) while the internal mechanism of the battery produces more electrons at one ... 0 It is doubtful if Thevenin's eqv also supports transients/dynamics. Suppose the ïnternals of a battery include a resistance as is usual and also an inductance in series with it. Concept of ïmpedance does not hold as supply is DC, and the same cannot also be converted to an eqv Norton's. If the supply is AC and the "source" internals include a series ... 0 It is showing the direction of current flow reversed through the meter with the reversal of the leads. 0 A Short answer is here, Firstly the capacitor gets charged.Though it takes infinite time to reach the battery potential the current is reduced to considerably low values. So obviously bulb won't glow. Note:As capacitor gains more and more charge its potential increases finally reaching the value same as the battery.So potential across bulb tends to zero. ... 1 You can't talk about "correct" without a definition of what you are trying to achieve. However, you can rule out some obviously bad arrangements if the placement of the meters keeps the circuit from working, even though we have to guess at what the intended "working" is. This is a poorly specified problem, or there is context surrounding it that you ... 0 The ammeter actually creates a short across the battery in the second figure (so that one is certainly wrong). Very likely, this would cause damage to the ammeter, or at least result in a blown input protection fuse. I think the other diagrams are OK, depending on what it is you are trying to measure (which is open to interpretation). This answer assumes ... 0 It's quite opposite. All are correct except the second one. Ampere-meter should be positioned in series configuration and voltmeter should be positioned in parallel configuration to the measured element. Notice that voltmeters should have very high resistance, so that most of the current will flow through the measured element. The bigger resistance of the ... 0 Actually is the other way round, the wrong one is the second. Ampere-meter has to be connected in series and Voltmeter in parallel. Think that Ampere-meter measures current so it has to be "a part of the circuit" as opposed to the voltmeter which measures differences of potential, so it has to be connected to two point of the circuit. Ideally Ampere-meters ... 2 A general strategy for these questions is to start at the battery and trace the current through the circuit. So, starting from the batter, we can see that the entire current passes through R_1. After that, the current hits a split (at the top of the circuit in your drawing), where part of it goes to the left through R_2 and the other part of it goes to ... 2 The parallel of R2 and R3 is in series with R1. The current that flows through R1 gets divided between R2 and R3. 1 Yes, until energy is conserved. Kirchhoff's law is amongst great basic laws which help in constructing electronic physics. Kirchhoff's law is based on the Law of Conservation and we also know that energy is always conserved according to our latest experiments and analysis. If in future it gets false then we must once have to think about the applicability ... 0 Use Thévenin's theorem and find the Rth (thevenin equivalent resistance) and Vth (Thevenin voltage) across the two terminals of Galvanometer. Just apply ohms law to get your answer. 0 Many times this kind of problems becomes very simple just by redrawing the circuit in a more standard way. Have a look at this: Now I think you will have no problem to solve it! 0 The net resistance depends on the point where you have applied the potential difference. Indicating the direction of current is very useful. You can Use [ ( R_3 series R_4 ) parallel R_2 ] to be R_7 Now Redraw the circuit and by naming the point with same potential as one point as I have done in my diagram. Helps a lot: Seems like I have ... 0 The back emf voltage actually gives the motor its mechanical power and it allows to control the motor speed. Let us take a look at the formulas. The terminal voltage of the motor is$$ v = Ri + k_V \omega $$with the electrical resistance R of the winding, the voltage constant k_V of the motor and the shaft speed \omega. The torque delivered by the ... 0 You yourself have answered your question. ...into an ammeter by connecting a low resistance (called shunt resistance) in ... The galvanometer has it's own resistance R (of course it will unless it is ideal). If entire current flows through it, it will have a lot of contribution to potential difference (i*R) across it and affect circuit working. ... 0 First things first, the resistance is connected in the circuit parallel to the galvanometer to minimise the total resistance of the ammeter. Whenever 2 resistances are connected in parallel, the equivalent resistance is always less than the lowest resistance. You know that: \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$\implies R_{eq} = ...

1

Would there be any current flowing though an ideal voltage source? Yes. An ideal voltage source will produce whatever current is needed to maintain its voltage. This current will flow in a complete circuit through the source and through whatever network is connected to the source. I am asking this more specifically when you are analyzing a circuit ...

2

The internal resistance is 0. An ideal voltage source is there in order to supply current at a constant voltage. The amount of current flowing through it is entirely determined by the outside circuit. Imagine a short circuit with an ideal voltage source. The current would skyrocket to infinity (Obviously any real voltage source would soon run out of ...

0

The effect of CRs on electronics will increase with time. As the electronics shrink, we will be fitting more bits per unit area. As such, there will be less charge necessary to define a bit (how many electrons or holes signal that a bit is 1 or 0). Since one CR event will deposit a given amount of charge, the probability that a bit will flip (or the number ...

2

If you just plug in your suggested solution, you get $$\frac d{dt} A\cos(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)\\ -A\omega \sin(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)$$ Now you should be able to use the function sum formulas to solve for $\phi$ and $\frac A{V_{in}} ... 1 Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field ... 1$20V$is not normally dangerous. You may not even feel it. It is not the voltage that causes danger, but rather the current it generates through your body. Anything over about$10mA$will be unpleasant, above$50mA$it gets dangerous. The current is determined by both the voltage and the resistance of the body. That resistance depends on how you touch the ... 1 I don't understand the question and the diagram associated with it, partly because it contradicts with something I learned. I was told that if I have a DC current in a straight current, by Ohm's law$\mathbf{J} = \sigma \mathbf{E}$, the electric field lines must be all directed along the current. So I don't see a way by which surface charge can build up on ... 0 The PF of a series RL circuit is found from$\tan(PF) = R/\omega L$while the PF of a parallel RL circuit is found from$\tan (PF) = \omega L/R\$. Thus the tangents of the power factors of the 2 circuits are reciprocals of each other. For example, if the series RL circuit has a power factor of 0.4 (or 21.8 degrees), the parallel RL circuit has a a power ...

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