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A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have Q(t) = CV ...

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Since lurscher thinks he is so smart but obviously doesn't care to help you with the physics of this problem, I will. Get a square tub and fill in a couple of inches of saltwater. This will simulate a dense conducting mesh. Now make yourself two flat electrodes from copper or steel that are approx. 5-10% of the width/length of the tub. Connect them to an ...

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If your power supply is sourcing a positive current toward the ground, that corresponds to a flow of positive charge from the supply to ground. This is equivalent to a flow of negative charge from the ground to the power supply. In a real wire, only negative charges can flow, so the second thing will happen: electrons (which have a negative charge) will ...

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I don't know if this answers your question, but whenever I talk about in-series and in-parallel resistors I like the water analogy. In this analogy the battery is a pump that lifts the water from low (potential) to high (potential). The electric current is the water current, and the resistor is a wheel or a constriction in the pipe. So I have In-Series ...

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First of all, if you are dealing with a network of finite number of resistors, try redrawing it in some form in which you'll be able to recognize the parallel or series connections. Secondly, take a look at Delta-Y Transform which might be really helpful in some cases. If these fail, turn to Kirchoff's laws i.e. put a test generator between the points ...

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In the following situation you have the voltage source ensuring a potential difference V = 60 Volts between its terminals. The source's upper terminal is connected to the switch's upper terminal, so they have the same electric potential. The switch's lower terminal is connected to the resistor's upper terminal, so they also have the same electrical ...

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The current flows through both resistors. What works for me is to use the analogy of fluid through pipes under pressure. Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The ...

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Theoretically, these requirements arise from the way you connect the measurement devices to the rest of the circuit. A voltmeter is connected in parallel, as you said. Say that you are trying to measure the voltage drop across a resistor $R$ through which passes a current $i$. If the internal resistance of the voltmeter is comparable to $R$, then the ...

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No, it will rotate at a speed determined by the load. Witness that the current in, and thus the magnetic field produced by the stator coils is either in-phase with or anti-phase with the rotor current, with the $\pi$-phase change triggered by the split ring commutator. So the torque in each half of the rotor's rotation will throb at twice the AC line ...

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In the body of the question, you mentioned the limit of R approaching 0. Let's begin from there. In this case we have what is called an L-C Oscillating Circuit. For convenience I will assume that the left plate of the capacitor has charge $q(t)$ such that $q(0) = Q$. Similar to the simple discharge of a capacitor, the upper plate of the capacitor begins ...

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Let's say we hook up a capacitor to a battery (and maybe toss a resistor in there). The battery will pump a charge difference between the plates, which creates a potential difference between the plates. When the potential difference reaches the potential difference of the battery, current that takes and adds charge to the plates stops flowing, as current ...

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Both holds true. If you use Ohm's law, you can easily see that $$i_1 R_1 = i_2 R_2$$. So, $$10 \times 1 = i_2 \times 0.2$$ gives $$i_2 = 50\,\mathrm{ampere}$$ Again by power conservation, $$V_{left} i_{left} = V_{right} i_{right}$$ And current in left loop will increase to be $i_{left} = 500\,\mathrm{ampere}$. As you can see, here both Ohm's Law as ...

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An ideal capacitor never "dissipates" energy, it merely stores it. The amount of energy stored in a capacitor is given by the formula you mentioned: $U = \frac{1}{2}CV^2$. In the case of the LC circuit, the energy stored in the capacitor moves into the inductor in form of magnetic field energy and then goes back and forth from them. In the case of an ideal ...

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But how can there be current without electron potential (voltage)? In the case where there is no resistance, current (once flowing) does not require any voltage to continue flowing. If you start a current flowing in a superconductor, then even with no applied voltage, it continues to flow. It doesn't take any force to keep a ball rolling if there is ...

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When thinking about inductors on a conceptual level, the thing to remember is that they oppose change in current. In other words, if the current, $i$, is dropping, they provide voltage in the direction of that current; if $i$ is increasing, they provide voltage in the other direction (this can be very loosely thought of as resembling "inertia" in the ...

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In class we learned about point charges, electrostatic force, voltage, current etc. and discussed circuits along the way. And then you hopefully learned that voltage isn't a general concept, and that the scalar field is an entirely gauge dependent concept in electrodynamics. I first thought about an electrical circuit as a 1-dimensional "restricted ...

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