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11

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power ...


5

When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ...


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 ...


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 ...


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 ...


3

Let us crudely imagine the voltage source as a pump pumping water up to the top of a water slide, and the resistor as the slide itself with water flowing down through it. The height difference between the top of the slide and the bottom of the slide is the same as the height difference between the top of the pump and the bottom of the pump. The voltage ...


3

This is solved using start-delta transformation and Delta-Star transformation like shown in the below picture when you rewrite the circuit diagram it looks like this and you can notice delta network and star network in it. Applying proper transformation reduces the complex network to simple network, then you can solve simple series/parallel combination ...


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) ...


2

The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say, $$ v_{drift}=\eta E $$ where $\eta$ ...


2

They are absolutely connected. In both cases, you have the following components: an "inertial term": this can be mass, or inductance. Something that resists change (in velocity, in current) a "linear force term": the spring ($F=-kx$) or the capacitor ($V=Q/C$). a "displacement term": this is $x$ for mechanical, and $Q$ for electrical. Their derivatives $v$ ...


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 $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$


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 ...


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 ...


2

The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ...


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

A motor-generator can be used to convert electric power from one voltage x current combination to another voltage x current combination. Such systems have been used, and are sometimes still used, for exactly this purpose. One advantage is that a large flywheel can be added to the shaft, effectively low pass filtering the average power. This can be useful ...


2

A couple of suggestions: (1) the EE stackexchange site a better home for this question (2) simply solve for the voltage across the capacitor and the current through the inductor. Once you have those, the energies stored, as a function of time are just $$W_L(t) = \frac{L}{2}i^2_L$$ and $$W_C(t) = \frac{C}{2}v^2_C$$ Since this is evidently a DC circuit ...


2

How do I identify which ones are parallel or series? If all of the current leaving one resistor enters another resistor, the two resistors are in series. The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g., $$R_{eq} = R_1 + R_2 $$ If all of the voltage across one resistor is ...


1

Isn't it just a convention you have to choose once and for all. It's all about how to pass from Maxwell to lumped element circuits. Especially, I can choose the two different conventions (all quantities are vectorial in the following) $$E=\pm\nabla V$$ since I didn't choose my field-to-potential rules yet. Usually one chooses to conform to the classical ...


1

The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the ...


1

DC bias is $A e^{i\theta} e^{i\omega t} = A e^{i\cdot 0} e^{i\cdot 0 t} = A$, however, quoting this, Phasors and Complex impedances are only relevant to sinusoidal sources.


1

Your question seems to contain two parts. First, you're asking how to set up the equations of motion for this coupled system. Second, you are asking how to use symmetry considerations to find the normal modes and frequencies. Let's first answer the bit about symmetry first Normal modes - symmetry Your observation about the reflection symmetry is spot on. ...


1

Low voltage sources don't have enough potential to conduct through your skin or body so touching either the positive or negative doesn't make a difference. For you to feel it or get tissue damage, current must flow through you. This won't happen with very low voltage sources unless you're covered with something more conductive like wet salt water. But 12V ...


1

Your method for evaluating the electric field assumes it is appropriate to model it as spatially constant within the wire since you're basically taking a spatial average. You'll have to decide wether or not this is accurate.


1

It comes from the fact that the voltage in any loop is required to sum to zero. Here is a description of the rule (Kirchhoff's Voltage Law). Your instructor may have told you to follow an electron for the current law, but it is not ideal if the current flows in the opposite direction on one leg of the loop. Determining the voltages on each leg is needed, so ...


1

Here the capacitors look like in series but they are not actually. Capacitors can be said to be in series only if they carry same amount of charge which is not the case here.If you calculate charges will come out to be different.Look at the time constants(R*C) for the first branch and second branch, they are 1ms and 10sec respectively. As switch is on ...


1

After the switch is opened, we have a series circuit - the current through each circuit element is identical and equal to $i(t)$. Choosing a clockwise reference direction for the series current, KVL clockwise yields $$\frac{1}{1\mu F}\int_{-\infty}^t i(\tau) d\tau + i(t) \cdot 1k\Omega + i(t)\cdot10k\Omega + \frac{1}{1000\mu F}\int_{-\infty}^t i(\tau) ...


1

Your question has two parts. First addressing why electron needs a closed loop to flow- Consider a wire which is not closed for e.g in the diagram. Here Due to the electric field(marked in green) the electrons move to the right.But after some time the electrons accumulate in the right forms their own electric field with the positive charge(marked in ...


1

Why does current need a closed loop to flow? Not in every case you need a closed circuit. A good example that current in an open "circuit" can flow is the antenna rod. A generator generates an alternating current and the electrons in the rod moves and there is an electric flow.


1

What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of ...



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