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0

But isn't $I^2R$ equal to $V^2/R$, therefore if $R$ is constant doesn't the power depend on the square of the voltage so surely it doesn't matter whether it is high voltage or high current. Consider the wires connecting the power plant to the appliance; let the effective amplitude of oscillating current flowing through all wires be $I$ and let $V$ be ...


1

for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


0

Without going too deep pick for example the first circuit (from left to right) and call the first resistor $R_1$, the other $R_2$, same with the capacitors. Remember the definition of capacitance: $ C = Q / V $, and how its charge is related to the current through it: $ i = dQ / dt $. Then label the currents, for example call the input current $i$, then by ...


1

Like all steady state circuit analysis things in parallel share the same voltage, and things in series share the same current. The question then really boils down to what the current vs. voltage curve looks like for a solar panel. A quick search turned up this: from https://www.folsomlabs.com/modeling/module/module_model So in this case it looks like you ...


0

What does this mean? It means that your model isn't valid. Only in the context of ideal circuit theory does a voltage source produce a voltage across independent of the current through. In this context, connecting two (ideal) voltage sources in parallel leads to a contraction, e.g., $$1 = 2$$ But physical voltage sources cannot produce unlimited ...


0

Easyest anser is a battery with more volt than the other will drain until they are equle voltage unless you put a diode from low battery + to full battery +. Sorry for spelling errors


2

How quickly discharge will occur in the situation you sketch depends entirely on the surface properties of the negative electrode. For current to flow, electrons need to be released from the negative electrode; once they are free, they will accelerate unimpeded to the positive electrode. They will arrive there with 1.5 eV of energy, causing a small amount of ...


1

Electrons (and other charge carriers, e.g., ions) in vacuum travel without resistance. However, as pointed out, correctly, in the other answers, there are no charge carriers in vacuum. Nevertheless, electrons can escape from the terminals if they have a kinetic energy which is bigger than the potential barrier of the terminal surface, i.e., the work ...


1

Lines of electrostatic force exist between the positive and negative poles of the battery, even though they're separated by a vacuum. Vacuum permittivity is ε0 = 8.854 * 10^-12 farads per meter. By convention, this is called the dielectric constant of 1, a baseline against which the dielectric permittivities of other materials are compared. ...


1

In vacuum there are no charge carriers like ions or electrons. With nothing to carry charge, i.e. current, such a battery would discharge much, much slower than when the battery poles are connected by something that can carry charge like a conductor or an imperfect insulator.


1

NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.


1

No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.


0

We only consider the current from $0$ to $\pi$ since the current from $\pi$ to $2\pi$ is effectively 0. Consider a sinusoidal A.C. current $I = I_0\sin t$ (assuming the period is $2\pi$ as in your question), when there is a half wave rectifier, current can only flow in one direction. Thus any part of $I$ that is negative becomes zero. The current in the ...


0

Without more detailed data, further experimentation, I cannot provide a definite answer. But with regards to your supply transformer that feeds the primary - could it possibly be a GFI protected transformer? Many of the later date neon light transformers (Franceformer for example) built GFI protectors into the transformers for safety. Using these ...


2

You are taking a shortcut when you say, "The voltage is zero." Voltage is always measured between two points. In electrical engineering, when we say the voltage at point X is V, we actually are measuring the voltage between point X and an implicit other point called "ground". In the electric power grid, "neutral" is ground, by definition. So the voltage ...


0

The key thing to remember here is that we are talking about an ideal wire. This means that there is (effectively) no resistance in the wire, and therefore no voltage drop along the wire. Now if there is a voltage difference between the two ends of the wire, that induces a constant electric field along the wire, causing current to flow.


1

Well, yes you can, but it is usually very hard. Here are the steps: Solve the Laplace equation: $$ \nabla^2V = 0 \, .$$ In your case, find the general solution in spherical coordinates. Try to use every simplification you can. You might wonder why you don't solve Poisson's equation: $$ \epsilon\nabla^2V = \rho \, .$$ That's because a conductor is an equal ...


0

The electrical resistance of a sphere is calculated in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance


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As pointed out in some comment, electrons are being accelerated in the process of charge. This generates electromagnetic radiation. Try doing the calculation using Poynting vector, the way Maxwell defined electromagnetic energy.


1

My brother-in-law faced this EXACT problem, as he worked on high tension lines. There is a corona discharge from these lines due to the very high voltages involved. From experience, the linemen learned that this corona discharge is injurious to internal organs. To prevent injury, the linemen wear the Faraday cage suit, because such a suit keeps the ...


1

You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$ Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$ \frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...


1

Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ...


0

This is how ideal and impossible circuits elements behave, but it's a starting point for a simple analysis: At discontinuous changes in circuits, 1) inductors have the same current immediately before and after the discontinuity, but can have discontinuous voltage changes. The current will then change exponentially/sinusoidally/both. 2) capacitors have the ...


-2

if we consider a current of 6A flowing through circuit we will then get a equation E=6I*R 6IR = 5Ir = 5/6 ohm total current is 10/(5/6)= 12A


0

It is not affected at all. There is no net potential difference across the sandwich whether it is part of a circuit or not.


0

Connect the battery to this circuit. At this point, the charge doesn't know that there is a hole in the circuit. Negative charge therefor flows away from the negative battery pole, since it is repelled by this same charge, as far away as it can along the attached wire - this means that the charge will pass through the light bulb, and the light bulb will ...


1

The voltage of a cell is caused by a chemical process, and the details of this process determine the behavior of the cell when a current flows. Simplistically, that behavior can be described as the cell having an internal (series) resistance - although the real behavior is more complex (an internal resistance would not explain that a battery goes flat after ...


1

I think this picture answers the question There battery's terminals are not charged. It's just a chemical reaction that starts the charge. There's a Zn and a CuSO4. When they meet each other via a cable, the Zn gives 2 electrons to ChSO4. Cu gets rid of SO4, leaves 2 extra electrons to SO4 and takes the 2 electrons it got from Zn. Then we have SO4 ...


0

This means that you are creating a short circuit. Suppose you connect the + of cell 1 to the - of cell 2. So far everything is OK. When you close the loop by connecting the + of cell 2 to the - of cell 1, you will be applying a potential difference $V_2-V_1$ to the wire connecting both. This wire has a very low resistance, so the current will be very high. ...



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