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To be clear on the setup, we have an ideal battery (DC voltage source) and an ideal wire (zero resistance. In ideal circuit theory, asking what happens when one connects an ideal wire across an ideal voltage source is essentially asking what happens when $1 = 0$. However, if we allow that the wire has non-zero radius and non-zero length, then, when the ...

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It depends on inductivity: I suspect the circuit would explode, with or without capacitor. With just the battery - which also has zero inner resistance - the current would get infinite during the first 0 seconds. Except - while no energy is lost in heating wires - that would make the whole capacity of the battery.going into creating a magnetic field ...

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When a wire with no resistance is connected to the terminals of an ideal battery, will a current exist in the circuit? yes, it would be infinite If a capacitor is added to the circuit, will it be charged by the battery or will it remain uncharged? yes, the capacitor would be charged.

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The battery would be shorted in first case. In the second case: Although, resistance is never zero. You see that while charge on a charging capacitor varies with time as $$q=CV(1-e^{-\frac{t}{RC}})$$ When you set $R=0 \Omega$, the result is undefined. Although you can talk of $R\mapsto 0^{+}\Omega$ In that case, the capacitor will be charged instantly. ...

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What you wrote is a lagrangian of a free theory. All free theories have infinite-dimensional symmetry, called higher-spin symmetry. The current you gave is a standard Noether current corresponding to one of the generators of the higher-spin symmetry. Usual space-time symmetry is a subalgebra of the higher-spin algebra. As Maldacena a Zhiboedov showed in 3d, ...

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The Lagrangian density reads $$\tag{1} {\cal L}~=~\frac{1}{2}\sum_{i=1}^2\left( \partial_{\mu}\phi^i \partial^{\mu}\phi^i -m^2\phi^i\phi^i\right).$$ Consider an infinitesimal transformation $$\tag{2} \delta \phi^i~=~\epsilon Y^i_{\nu_2\ldots \nu_n},$$ with generators $$\tag{3} Y^1_{\nu_2\ldots \nu_n}~:=~ ... 3 Think of it in terms of current, V, W, and Z are in series, so each is equally bright. X and Y are in parallel, so each gets half the current of the others. If you assume each bulb is a constant resistance R (not true for incandescent bulbs, by the way), then V,W and Z will each dissipate i^2R. For X and Y, since each has a current i/2, the power will be ... 1 The answer is yes; this can be shown by evaluating the variation of the action. The action consists of three terms, we will consider them separately:$$\delta_\epsilon K(\Phi,\bar{\Phi})=\frac{\partial K(\Phi,\bar{\Phi})}{\partial\Phi^i}\delta_\epsilon \Phi^i+\frac{\partial K(\Phi,\bar{\Phi})}{\partial\bar{\Phi}_i}\delta_\epsilon\bar{\Phi}_i=i\epsilon K_i ...

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Current is produced in a metal when the free electrons in the metals acquire a drift velocity due to an electric field. But when these free electrons travel through the metal, their path is hindered by other atoms and particles and their electomagnetic pull. More the resistance, higher is this hindrance and lesser is the drift velocity. Hence, the current ...

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The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed. In a basic Drude model for electrical transport both, $n$, the charge carrier density and $\tau$, the time between collisions determine the resistance: $$\mathbf{J} = \left( \frac{n q^2 \tau}{m} ... 1 Your understanding is correct. From V=IR if voltage stays the same while resistance is increased, the current should be decreased. But if you have heard of another equation I = \frac{Q}{t} If current(I) is increased and the charge Q is fixed (Charge is fixed if the power supply is from power cells like battery). Time will be decreased. Which means ... 0 According to Ampere's Circuital Law, ∫B.dl=μi. [i.e. circular integral of product of magnetic field along imaginary circulating path and path length is equal to permeability of medium multiplied with current flowing through the imaginary cross section made by circulating path] So, ∑i must be zero, hence net current comes zero, i.e.Jb+Kb=0. This implies that ... 3 Lightning gives off high current when it hits a car/vehicle, and since the electrical components of the car is grounded on the chassis, are there possibilities that these high currents can destroy or damage the electrical components,battery, antenna etc through the 'grounding'? Lightning also has a massive electromagnetic field that is created ... 1 The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ... 0 Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator:$$ i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t $$v_L would be$$ v_L(t) = ...

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The switch really has 2 positions: on and off. However, when you move the switch very slowly, it may leave the closed position slowly. When the switch is just barely open, the field may cause the air to break down and start conducting, to form a spark (as @anna v explained). To rephrase, the reason why sparks happen is because the switch may only be open a ...

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Air is a bad conductor up to a certain value of the field generated by charges and the distance between them. After that air breaks down and a discharge happens, i.e. sparks. So below this level charges can accumulate by rubbing for example , positive ions left on one surface and negative on the other. When brought close a spark occurs. Why does holding ...

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In lamination, sheets of metal are separated by insulators. Charge can't flow between plates (they're insulated from each other), so there can't be large eddy currents perpendicular to the plate. Obviously large eddy currents can still flow on the plane of the metal sheets. If you properly align the source of the emf and the plate, you can rid of most of ...

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Modern processors are built from CMOS technology. These digital circuits consume relatively little current when sitting in one state or the other. However, there is some inevitable capacitance on every node. When the output of a digital gate changes state, that capacitance is charged or discharged, which means current has to flow. The total average ...

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Yes, it is possible. The simplest qualitative answer to this is that, at the microscopic level, the electrons in a conductor are dictated by quantum mechanics, which is inherently probabilistic. Velocities and positions are rarely ever totally excluded from a given value; it's just insanely unlikely for a single electron to attain that given value. ...

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Your reasoning is correct. If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law: $$I = \frac{E}{r + R_L}$$ or $$E = (r + R_L)\cdot I$$ The output voltage $V$ of the solar cell is the voltage across the load ...

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Electrical current can be carried by conduction electrons, or by 'holes'. For ordinary matter, there is roughly one electron per two daltons of matter, which is to say, $\frac 12 N_Ae$. This roughly works out at $480*10^6$ coulombs per kilogram. Many of these electrons are bound in the inner orbitals, but there are still plenty of conduction electrons ...

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I will be very difficult to write it down here. And I don't know LaTEX Please check out the link. http://www.youtube.com/watch?v=tvfu9S_GQY8

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A reference current is just a stable source of current that does not fluctuate over temperature, supply voltages or loads. This "reference" is known to be at a specific value and very stable. This way it can be used to compare other currents to (as a reference). For example if you want to drive your electric motor at 100A and you know your reference ...

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but it isn't the case with resistors in parallels (1) parallel connected resistors have identical voltage across $$V_{R1}=V_{R2}$$ (2) the voltage across a resistor is given by Ohm's law $V_R = I_R\cdot R$. Applying Ohm's law to parallel connected resistors yields $$I_{R1}\cdot R_1 = I_{R2}\cdot R_2$$ Thus, conclude that the parallel connected ...

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Most basic concepts that must be understood here are "Voltage" and "Current" . Voltage: Voltage is difference in electric potential between two points/nodes. Current: Current is simply flow of electric charge and is defined by Ohm's Law as: $$I =\frac {V}{R}$$ Now, in a series circuit, all resistors are placed between two nodes. Voltage among these two ...

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In series circuits, resistances add can be simply added and treated as one big effective resistance: $R_{\text{eff}}=R_1+R_2+R_3$.... So the current remains same. In parallel circuits the current splits up so each branch has a different effective resistance (in each of the separate branches one can use the series rule again). Due to this, the current isn't ...

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Perhaps you meant intuitively? Well, if you imagine the roads are tracks and the flow of cars the current, and if you had the choice between a very bumpy narrow path on your left and a very nice motorway on your right, which would you choose? Nearly everyone would go right (there are some dirt enthusiastic or people that don't think), and overall the current ...

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Kirchhoff's first law says: "At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node". When there is only one current flowing into the node it is simply equal to the sum of currents flowing out. The node or junction is the place where conductor splits to form couple ...

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