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1

Is it possible for electrons to carry more than one charge? If by, one charge, you mean more electric charge than (the negative of) the elementary charge, the answer is no. More specifically, an 'electron' would not be an electron if its charge were not $-e$. However, electric charge is not the only type of charge electrons 'carry'. But this is a ...


2

Each electron has a fixed charge of $-1.602\times10^{-19}\,\mbox{C}$ (where (C stands for coulombs). If you gather $6.24\times10^{18}$ electrons, the total charge will be \begin{align*} \mbox{Total charge} &= \mbox{Charge per electron}\times\mbox{Number of electrons}\\ &= ...


1

There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy Ep as an associated potential field such as $\mathbf{F}=-\frac{\partial Ep}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(Ep)$. That's no more than a definition of the potential energy. ...


-2

electric field strength is $$E=\frac Fq=\frac Vd$$ with $V$=voltage, $d$=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but $Fd$=energy $$\therefore {\rm energy}=qV$$


2

I'll give this a shot though I'm uncertain if this will clarify or cloud the issue. electrons are people, OK current is the number of people, No, that's not the correct analogy. Current is a flow and a people (electron) current is a flow of people (electrons) and the amount of current is the number (Coulombs) of people (electrons) passing a ...


1

Since your question is kind of pictorial, i will try to stick with this and avoid any deeper mathematical / physical explanations. electrons are people ok with that current is the number of people no, that's a mistake. Actually current is is this case the number of people crossing a line on the street per time. Like Steeven already mentioned in ...


1

But what is potential in this example? In your people-in-the-street analogy, the electric potential would be a shop with sale in one end of the street, or maybe an accident or something else interesting. People tend to move towards the interesting end (less potential) and away from the other and less interesting end (higher potential). Just like a stone ...


0

We know that the tension between a perfect current source is 0 That's a common misconception held by those first introduced to current sources. In fact, an ideal current source produces a fixed current regardless of the voltage across. Put another way, the voltage across an ideal current source is determined by the attached circuit which is, in this ...


4

know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality. It's true that we can ask for the average of the magnitude of a sinusoid over time and calculate it. This is a useful quantity if, for example, we want the time average voltage at the output of a full wave rectifier ...


2

Waffle's answer shows you exactly why the RMS isn't the average: Here, the average value of $V(t)$ clearly isn't $2V_p/\pi$ that you've obtained, it's zero. The half-cycle business doesn't make much sense given that if you chose $\pi/2$ to $\pi$, instead of 0 to $\pi/2$, you'd have a negative average, so which would be the true "average" $+2V_p/\pi$ or ...


5

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...


2

The definition of RMS is Root Mean Square and means the square root of the avarage of the square of some quantity, that is $a_{RMS}=\sqrt{\langle a^2 \rangle}$. So for the case of a sinusoidal current, we would get $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$


0

I think that the problem here is that you haven't properly set up the circuit equations. So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit Since the bulb and capacitor are series connected, the current through each is identical. Denote the series current phasor as $I_s$. Assuming the source is an AC voltage ...


0

What is the easiest way to solve this? The easiest way is unlikely to be the most illuminating way. What I propose is that you first work the problem from basic circuit principles and then explore more powerful solution methods such as, e.g., node voltage analysis, Thevenin equivalent circuit etc. Start from the right hand side with Ohm's law: $$I_6 ...


2

This is as much a chemistry question as it is an electricity question, because batteries are chemical devices. A battery is constructed such that there's a redox reaction split into two half-reactions which must move electrons through your circuit in order to complete. The reaction happens at a finite speed, and may also be limited by how fast various ions ...


2

The internal resistance of batteries is caused by a number of different mechanisms. A major contribution comes from the ionic conduction mechanisms in the electrolyte solution. Ions are large and can only move very slowly in electrolytes. Another source for the voltage drop is the concentration and current density dependent polarization of electrodes. To ...


4

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...


3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...


1

If $E$ is the emf (open-circuit voltage) of the battery and $r_s$ is the internal resistance, the equation relating the series current $I_S$ through an external resistance $R_L$ is given by: $$I_S = \frac{E}{r_s + R_L}$$ Now, the voltage across the battery terminals $V_{BAT}$ is given by $$V_{BAT} = E - I_S \cdot r_s = E\frac{R_L}{r_s + R_L}$$ So, even ...


0

When current flows, inside the source there is always some voltage drop, so the voltage on the terminals is lower than in the static situation. We model this behaviour by simple model depicted on your picture; we assume that the real source consists of idealized source -| |- that gives the emf and has not resistance, and of ordinary ohmic resistance (2 ...


1

We always consider than wire has negligible resistance and if the wire is 99% copper wire it usually has resistance too small to cause drop in voltage measurable by common multimeter used in laboratory. Considering that the wire you used was such wire, between a and b, c and d and from a to + terminal and e to - terminal $P.D \approx 0V$ . Though between b ...



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