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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 ...


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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


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


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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 ...


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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 ...


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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 ...


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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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You are mistaken about your fundamentals. Your initial assumption was that the potential difference across the resistors was nE, not the whole circuit. You implicitly assumed the difference across the whole circuit was zero when you connected the components in a closed circuit. One of Kirchhoff's laws says the sum of the voltages in a loop in zero.


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I do some work with magnetic fields in tissue as well for wireless power applications, though we don't typically deal with fields that strong hopefully I can help. First of all human tissue is largely magnetically transparent at low frequencies. While modeling the electromagnetic properties of tissue is very difficult problem (this is why you largely see ...


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Bulb H is connected to second battery directly. As long as this bulb is not short circuited, there will be some applied potential difference across its terminals and hence it will be ON


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There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...


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I'm assuming $V_R$ is taken to be the voltage across the resistor in a series RC circuit. The transfer function comes directly for the voltage division rule: $$\frac{V_R}{V_\text{in}} = \frac{Z_{R}}{Z_\text{series}} = \frac{R}{R+\frac{1}{i\omega C}} \, .$$ In this equation $V_R$ and $V_\text{in}$ are phasors, meaning that the actual time dependent ...


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C2 and C6 are AC coupling capacitors. An AC signal on their input appears on the output. If you just connected these two capacitors together, they would act as a voltage divider - the voltage at the connecting node is the mean of the voltages at the other side of the capacitors. The "weak coupling" comes about from the fact that since the point connecting ...


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For slowly moving charges, magnetism is just a relativistic correction, so the relative size of its effect is $O(v^2/c^2)$. Since $v$ is very small for charges in a wire (less than 1 cm/s), the effect will be insignificant. Since parallel currents attract, the current will be attracted to the center of the wire a tiny, tiny bit.


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Does it take the same amount of time to charge a capacitor as it does to discharge to get to the stable state? Yes, in an RC circuit both of these are $\infty$. will it take 5Tau to discharge a charged capacitor to its stable state like it would to charge a capacitor when the capacitor is completely uncharged? Well, that's not "completely ...


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First of all, it will never get fully (dis)charged, i.e., its voltage drop will never match the supply. The time constant is always the same, because it does not depend on the boundary conditions of your equations. To answer your question, all time scales remain the same. That means if you define the system to be stable after $5\tau$ (which is reasonable), ...


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The battery does create an electric field near itself that forces electrons away from it. As they travel they begin to build up on the bends of the wire. This is known as feedback. The next electrons coming by are repelled away from this build up allowing them to travel around instead of just directly away from the battery. This is how electrons can travel ...


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The ammeter measure the current flowing through itself. If you want to measure the current flowing through another component, then you must make the current through the ammeter equal to the current through the component. If you wire it in series, that's true. If you wired it in parallel, the current would be unevenly divided between the component and the ...


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You seem to be asking two questions: why is the current on either side the same, and what happens to the energy. To clarify where the energy goes, the kinetic energy is transferred entirely into electrical potential energy of the capacitor. This is the energy stored as a result of all the similar charges being close together on either capacitor plate. There ...


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The amount of work done by unit charge between any two nodes of current carrying circuit is called the potential difference between those nodes. The amount of work done against the electric field by displacing (without acceleration) a unit test charge from one terminal to other terminal in an open circuit is called the electromotive force. Obviously when ...


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There are broadly three classes of materials: conductor, semiconductor, insulator. The conductor contains a LOT of electrons per unit volume. If you were to charge it, you would add a few more electrons. How many? Let's take copper. It has roughly $8.5\cdot 10^{28}$ electrons per $m^3$. If you have a wire of radius $r$ the number of electrons scales with ...


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To do this experiment, you will need: 1> a grounded, sheilded cable, (to prevent magnetic interference) 2> an ability to apply a voltage from the cable to ground, (say a battery, or better a variable DC source, with the postive terminal grounded and the negative one attached to the cable) and 3> a sensitive ohm-meter. Then compare the ohms through the ...


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The energy is stored in the magnetic field. I usually think of it as "magnetic field lines repel" but that is not very precise (useful for intuition though). But along the same lines as your capacitor example (moving the plates to infinity takes work), if you look at a simple current loop there is a force on the wires from the magnetic field generated. This ...


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Your answer would be correct if there was no voltage drop across the resistor on the right; but since a voltage will be developed there because of the return currents, the voltage across the resistors is less than the 30 V, 20 V of the respective batteries. It might be easier to see this is so if you drew the circuit in a slightly different form:


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Your awnser is not correct because in the equation V=RI , V is the potential difference of the resistor but in the question The potential difference of the resistors are not 30 and 20 volts. 30 and 20 volts are the potential difference of the batteries.



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