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It is perfectly reasonable that on increasing the voltage across one arm of the transformer, the current decreases in that arm. A transformers has the two arms linked by a flux linkage, which in our ideal case, we assume to be 100% (means we transfer the entire magnetic flux in one arm to the other). That means the power input is the same as power output (Or ...


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The photocurrent is the steady state current. Let's take your example and suppose your light source is emitting 10 photoelectrons per second. You can use two frequencies - at the lower frequency electrons travel at 1km/s and at the higher frequency the electrons travel at 2km/s. When you first turn on the light there is a delay before the first ...


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Well, it depends. The photoelectron can have a kinetic energy between zero to Max. Kinetic energy = hv - (the work function of metal) It depends on the electrons. You never know how many collisions did it underwent before leaving the metal. So if you strictly say that all the electrons will have maximum kinetic energy then theoretically the photocurrent ...


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In some cases, it DOES increase: about one ion per 36 eV, is the rule of thumb for X-rays generating free charges, as for instance in an ion chamber detector. That only happens when the amount of energy in the photon is so high that it generates a cascade of ions when absorbed, though, and the simple photoelectric effect one-photon-produces-one-electron ...


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First off, what's dangerous about electricity isn't current — it's energy. The power $P$ flowing through a circuit element with current $I$ undergoing voltage drop $V$ is given by $P=IV$. For your test circuit that is $\rm 0.2\,A \cdot 2\,V = 0.4\,W$, which is a quite modest power. However, that's the power flowing through the resistor in the circuit ...


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Exactly you have a higher resistance than the other components thus as for same voltage current is inversely proportional to resistance, no current flows if resistance is very high and takes the path with low reistance and therefore even a voltmeter is made on the same principle.


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The flux is defined as integal {if you don't know calculus, take this as a sum} of $\int_\Gamma \vec B \dot \, \mathrm d \vec A$ where dA points out the surface $\Gamma$. if you had a circular shape of coil, the flux would not change, and hence, no Electric field or EMF would be induced, but here, as the field of solenoid may not be perfectly symmetrical ...


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There is an alternative version of Ohm's law $\vec J = \sigma \vec E$ where $\vec J$ is the current density (current per unit area), $\sigma$ is the conductivity of the material and $\vec E$ the electric field strength. If $\vec E$ did become zero then no current would flow. If for a fixed $\vec E$ the conductivity increases (resistivity/resistance ...


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It all depends by what you mean by the word "flow". Let the charged body which is assumed to be a conductor produce an E-field. In a conductor which has mobile charge carriers then the charges can be made to flow within a conducting body which has no net charge. If you subject an uncharged conducting body to an external E-field then the mobile charge ...


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It sounds like you're asking "If two conductive materials are brought in contact, and one of them is electrically neutral, and one is positively charged, which direction will charge flow?" If that is indeed your question, then the answer is that negative charges (electrons) will flow from the neutral object to the positive object until they are at the same ...


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I think that your reasoning is correct: A standard circuit breaker should not trip if the resistance of a circuit is increased. However, in recent years there is a new kind of house circuit breaker called an AFCI circuit breaker which may trip if there is a loose or poor connection in a circuit. That may be the type of circuit breaker that they were talking ...


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The net effect of the charging process is the movement of electron from one plate which then has a net positive charge to the other plate which then has a net negative charge. The battery facilitates this by creating an electric field in the wires and it is this electric field which applies forces on the electrons which makes them move. The movement of ...


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If the circuit used alternate current, it might happen that a wrong terminal would introduce additional series capacitance or inductance; these could cause the breaker trip if the original load was mostly inductive or capacitive respectively. But it is not probable that the effect of a termination would be sufficient for this. Instead I conjecture that ...


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You set up a circuit which contains the galvanometer and adjust the circuit so that the deflection on the galvanometer is a maximum (for greatest accuracy). A calibrated resistance box is connected in parallel and adjusted until the deflection on the galvanometer is half of the deflection with no resistance box. In this condition the resistance of the ...


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The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance ...


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The choice of a non-metallic sample is to get the resistance high enough to measure. The resistance of the rheostat, the voltage supply, and the hookup wires can make the measurement inaccurate if you do not account for them. If they are much smaller than the resistance of the sample they will not matter much. You also will not try to draw too much current ...


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You are correct, electric current consists of electrons travelling from one place to another. Some materials conduct electricity better than others. Copper is one of the best and that's why our conductors are usually made of copper. Aluminium is also very good (so is silver) and high-voltage cables are usually made of aluminium. However, everything conducts ...


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Ground is at 0 potential,so,it accepts electron from negative terminal. And at very far place any positively charged electrode accepts electron from ground and current flow.


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Your confusion is between two related concepts. Power dissipated in total = internal power + external power. If that is the power you are talking about, then an external short circuit will maximize the current and therefore maximize total power, $V\cdot I$. Power delivered to the load. That is the thing addressed by the maximum power transfer theorem, and ...


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Electron density is a property of the material - same as the mass density. If you have a wire made out of copper, there will be more copper atoms in a thick wire than in a thin one - but the same number of electrons per atom. So unless you change the composition of your wire, the electron density (number of electrons per unit volume) stays the same.


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Yes you are right. We have made an assumption that (charge flown in) = (charge flown out). There will surely be an accumulation of charges at least in some part of elements.In the case of accumulation of charges we can find the other parameters like Voltage drop etc., using Maxwell's equations. But by making this small assumption we have greatly reduced all ...


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It is possible for a capacitor to have a net charge. However, most components have a small capacitance with respect to infinity. Unless you're dealing with a huge metal ball, it will take a large voltage in order to get any appreciable charge there.


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1.) When you first close a circuit, there is a very brief period of time in which the electrons push each other forward "one by one", so to speak. This very brief period of time is probably on the order of nanoseconds, so we usually ignore it. After that, a steady state condition is established, and the electrons move all at once ... more or less. Don't ...


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You've asked some really good questions here. Before starting, I want to first mention that the traditional picture of particles moving through a wire in electostatics is missing some physics; for instance, it ignores the quantum mechanical nature of electrons. The reason we still teach this model is because it captures the main effects (the phenomenon of ...


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Sign conventions aren't "wrong", but they can be misleading. For example, we could re-define work done in a gravitational field so that escaping earth's gravity well would require negative work. That's an equally valid convention, but we associate positive work with effort, so reversing the convention would hinder our physical intuition for no discernible ...


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Current is the motion of charged particles, not necessarily electrons and not necessarily in a conductor. For instance the motion of ions in a solution creates current, or holes in a solid. Currents and charges are the sources of Maxwell's equations. These sources create the electric and magnetic fields. When the sources behave in certain ways (like ...


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Current is defined by the movement of charges from one point to another, and the concept is independent of a medium, or the charges being electrons. In this context the current is constituted by the uniformly moving charge. And that is why an uniformly moving charge doesn't create any radiation, as what happens to be a current in one frame is stationary ...


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Is this part of a homework problem? Low resistance - think about how you would connect the resistors to the bridge. High resistance - think about how you note when the bridge is balanced. The answer to your question is in all three Physics textbooks that I have opened as well as a couple of webpages.


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To be clear, you can use Ampere's Law: $$\oint \vec B\cdot \mathrm{d}\vec\ell=\mu_0I_\text{enc}.$$ Specifically, it is the form without the displacement current, and it works because you are in magnetostatics. And Rob Jeffries' answer is totally satisfactory. But to specifically address your concern with the charge build-up lets look at an example of a ...


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This example can be done using the simple form of Ampere's law. Rather than get an accumulation of charge at the end extend the conductor out at right angles and off to infinity. You then have an infinite L-shaped conductor.


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in the problem that you describe the electric field is indeed conservative as you suggest. We have here a steady-state current, which means that $\text{div} \;\mathbf J = 0$ and therefore, according to the continuity equation $\frac{\partial \rho}{\partial t}+ \mathbf {\nabla} \cdot \mathbf{j}= \sigma$, the charge density is constant in time; this is an ...



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