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A power station or generator is not the thing you are asking about here - Your main question is why 25kV is the output. When designing a power station an engineer tries to get the most power out of the system as possible for a given input of energy, or peak efficiency. After the generator distribution and use becomes the main concern. It tends to be the ...

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The generators in power stations consist of coils rotation in a magnetic field produced by an electromagnet. Some of these electromagnets are energised by the generator itself but the larger generator have a smaller generator associated with them to provide the current which passes through the electromagnet. The induced voltage depends on the magnetic ...

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The definition of resistance of a component is $\text{resistance of component (R)} = \dfrac{\text{potential difference across component (V)}}{\text{current passing through component (I)}}$ This is not Ohm's law it is the definition of resistance. It so happens that for some components it is found that $V \propto I$ which is called Ohm's law. This is an ...

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I'm not sure exactly what you're asking for. But let's say there's an electric current flowing through a straight wire segment of length $l$, then the change in $\Delta\phi$, or $V$, would be defined by $$\Delta\phi = \int \mathbf{E}\cdot d\mathbf{l}$$ Because it is a straight wire, $$\Delta\phi = E\cdot l$$ But we have a definition of current $$I = ... 2 Georg Ohm's original experiments, 1825, established that for a set temperature, the current through a specific length of a conductor was proportional to the potential difference applied. Ohm's law is empirical; it cannot be derived directly from Maxwell's equations as it depends upon material properties. It is violated by many materials, and even then ... 0 The first thing to note is that you must never use the word voltage unless you are sure that you and your audience are aware of the fact that is shorthand for “a difference in voltage” or as you have used “voltage drop”. I would prefer to use the term “potential difference”. All this has to do with energy and a type of energy that a system has which makes ... 0 You are correct about inserting lightbulbs at X and Y. The same current flows through each branch, so the lightbulbs will be equally bright. However, the question is not whether there is a difference in the currents flowing through X and Y, but whether, if you connect a wire between them, any current will flow through it. Your objection, of course, will ... 0 What is the difference between the potential difference and potential energy of an electron? If I understand your question right, these terms are describing the same thing - one is just in a "per charge" version. Electric potential energy U_e is the potential energy associated with one spot in the circuit. Electric potential or just potential V is ... 0 Batteries are made up of one or more electrochemical cells, arranged in series. In these cells an electrochemical Redox reaction takes place: R+ze^{-} \to R^{z-} O-ze^{-} \to O^{z+} Where R and O resp. are a reducing agent and oxidising agent. This transfer of electrons provides the EMF of the cell. As more current is drawn from the cell, the ... 1 Let X be a sphere of radius R with charge +Q on it. The potential of sphere X is +\frac{kQ}{R}. Let Y be another sphere of radius 6R with charge +5Q on it. The potential of sphere Y is +\frac{k5Q}{6R}. The potential of sphere X is larger than that of sphere Y so if the spheres are connected with a wire electrons will flow from ... 2 Maybe it is worth bringing a comment into an answer: Batteries have protective circuits. The most basic safety device in a battery is a fuse that opens on high current. Some fuses open permanently and render the battery useless; others are more forgiving and reset. The positive thermal coefficient (PTC) is such a re-settable device that creates high ... 0 The credit to this answer goes to @knzhou, who provided it to me in the chat. What earthing does is simply equalizes the potential of the Earth (as in the planet, or the ideal earth, whichever is relevant). Now, we see that to calculate the potential difference between the plate (or whatever it is to which the earth is connected) and any point on or in the ... 2 The current doesnt pass through the voltmeter so the the resistance of load R is not affected. 1 As Giorgio says, you are overcounting the terms when multiplying by 4. I would like to elaborate a little bit, just to show how this is more than just a math error. It is also a conceptual error: You mention in your reasoning "The total potential energy = the sum of the PE of each of the 4 charges." But this is false. Not only that, but there is no such ... 1 This time you're wrong. You are overcounting the terms when multiplying by 4. Potential energy is given for two particles at a time, so you should be multiplying by 2, which is the same as multiplying by 4 and then dividing by 2. -1 You can think of the arrangement as two parallel capacitors with a common connection B which has a charge +Q on it for all time. The capacitance of BC is twice that of AB. Equal numbers of positive and negative charges are induced on plates A and C. When both A and C are connected to earth they are then at the same potential and so the ... 0 What actually happens on earthing something in electrostatics? Grounding a charged rod means neutralizing that rod. If the rod contains excess positive charge, once grounded the electrons from the ground neutralize the positive charge on the rod. If the rod is having an excess of negative charge, the excess charge flows to the ground. So the ground behaves ... 1 It may well be that the cross-membrane voltage difference is largely homogeneous, but not because intracellular protons are very mobile. To the contrary, the cytosol mobility (rate of diffusion) of protons, and ions in general, is very different from the high mobility measured in water or dilute solutions of small molecular species. According to this ... 1 The battery is an energy source that supplies the electrical energy to the electrons in the conductors. There is no actual flow of electrons. It's the energy that is transferred. A conductor contains large no. of atoms tightly packed with plenty of availability of valence electrons that are ready to move out from the atom if you supply a little bit of ... 0 A filament of a light bulb can be thought of as being composed of a lattice of positive metal ions which are vibrating about fixed positions and a sea of mobile electrons which are responsible for the metal being an electrical conductor. With no external circuit present a chemical process within a battery moves mobile electrons within the battery to produce ... 0 I try to think of everything in terms of chemical potential. Many batteries utilize lithium ions to create a chemical gradient. This creates a driving force across the circuit, called voltage. Keep in mind that the electrons are not moving that quickly - it acts more as a wave travelling through. If you are interested, you can learn more about the actual ... 0 Okay, it's partially wrong. A battery is a device that maintains a constant potential difference between its terminals (generally through chemical stuff). Circuits are conductors. Conductors have electrons that can easily flow across their atoms. It's like conductors have a "sea" of electrons. The potential difference causes the circuit's electrons to move ... 0 The electrical power delivered by a battery is \mathcal{E} I where \mathcal{E} is the emf of the cell and I is the current passing through the cell. This "input" power originates from a chemical reaction within the battery. The electrical power dissipated in an external circuit connected to the terminals of the battery is VI where V is the ... 0 Suppose that initially the capacitor was uncharged. A current would flow in the circuit which would charge the capacitor. The maximum voltage that the capacitor or the resistor can have across them is the voltage of the battery which is 40 V. The sum of the voltages across the capacitor and the resistor must equal the voltage of the battery. No matter how ... 0 Let us write down the equation of motion for the given RC circuit. Using Kirchhoff's voltage law we have U=iR+\frac{q}{C}, where i is the charging current in the circuit at time t, q represents the charge stored in the capacitor at time t. Initially(at t=0), q=0 and i is a maximum.The charging current decreases with time because of a build-up ... 0 A capacitor is meant to store charges. The ability to store how much charges is measured as the device's capacitance. It is related to the voltage across the capacitor plates and the charge on the plates: C=Q/V Consider a parallel plate capacitor as shown: Suppose initially the charge on the plate is zero. Now you connect it to an emf. The charges ... -1 Regarding your first question, I suggest you look up "parallel plate capacitor" and see how the capacitance C relates to the inter-plate distance d. To say more would risk violating the homework policy. I cannot be sure what scenario your teacher was describing in which the voltage decreased. In order to determine exactly what occurs when the plates of a ... 0 Potential difference is a more general term because the interaction is not specfified. It could be electromagnetic, gravitational... . Voltage is however specified as the potential difference of the electromagnetic interaction. Therefore voltage and potential difference are equivalent if we deal with the electromagnetic interaction only. You also have to be ... 0 Electric field lines are always at right angles to equipotential lines or surfaces. The electric field is minus the potential gradient. So in the diagram showing a uniform electric field a positive charge would experience a downward force in the direction of decreasing electric potential. In this case the magnitude of the electric field is \frac ... 0 The electrical field is related to a force concept: it describes the force per unit charge. The potential is related to a potential energy concept: it's the added electrical potential energy per unit charge. So, just as the force is the negative gradient (or in 1-dimension, the negative slope) of the potential energy function, the electric field is the ... 0 In electrical circuits impedance Z = \dfrac {V_{\text{peak}}}{I_{\text{peak}}} or \dfrac {V_{\text{rms}}}{I_{\text{rms}}} If I = I_{\text{peak}} \sin \omega t and V_L = L \dfrac {dI}{dt} then V_L = \omega L I_{\text{peak}}\cos \omega t = V_{\text{peak}} \cos \omega t This gives Z = \omega L However look at the ratio of the instantaneous ... 2 The difficulty is that there are three voltages involved. The voltage at the power station end V_S, the total voltage drop across the cables V_L and the voltage at the consumer end V_C. The voltages are related as follows. V_S=V_L+V_C So you have power supplied by power station is equal to the power lost in the transmission cables plus the power ... 1 It is indeed the ratio of V_{max} to I_{max} but only when you are talking about a sinusoidal voltage source and a "linear" component. For example, consider the charging curve of a capacitor: It would be incorrect to assume the impedance of this circuit is just V_{max}\over I_{max} as this circuit behaves in a little bit of a more complex fashion. ... 1 Impedance expresses a linear response relation between current and voltage, but is usually considered only for fourier-transformed voltages and currents. I.e., it is a complex function Z(\omega) so that:$$V(\omega) = Z(\omega) I(\omega).$$The proper real-time analogue would not be the instantaneous ratio of voltage to current, but rather a relation that ... 1 Wikipedia states In quantitative terms, it is the complex ratio of the voltage to the current in an alternating current (AC) circuit. https://en.m.wikipedia.org/wiki/Electrical_impedance 1 As you have confirmed the markings on the batteries to be, on one hand, 7.4V 2000mAh and on the other, 11.1V 1400mAh, and also considering that I have no knowledge about heated gloves, you can think of it as follows: The 7.4V 2000mAh battery can deliver, say, 200mA for 10 hours at approximately 7.4V. I say can because what determines the actual current, is ... 1 I read some of your comments. Explaining why H = I^2R  is not working out If same power is supplied for same time, equal heating would take place. H = I^2R  is not working here because R in two cases are different. V = IR_{1} 7.4 = 2 R_{1} R_{1} = 3.7 ~\Omega Similarly , R_{2} = 11.1/1.4 = 7.928 ~\Omega Now, if you use H = I^2Rt ... 1 Are you sure the batteries are labelled "2 A" and "1.4 A" rather than "2 Ah" and "1.4 Ah"? A battery doesn't have a fixed intensity. It has an approximately fixed voltage, and the intensity depends on the resistance of the connected load. So usually batteries bear two numbers: voltage and capacity (usually expressed in ampere-hour, 1 Ah=3600 coulombs). If ... 1 More the power, more the heat generated.$$P(\textrm{Power})=V\cdot IH(\textrm{in joules})\propto V\cdot I\cdot t\,(\textrm{in s}) \therefore H(\textrm{in joules})\propto P\cdot t (\textrm{in s})$$The 11.1 volt gloves does not give greater heat just because it has more Potential difference, but because \textrm{Power} is greater in it. 0 Grounding doesn't set the charge to zero - it sets the potential to zero. That means the residual charge will be whatever it needs to be to achieve that. 0 Ohm's law is a misnomer. It is not actually a true law, in the sense of Coulomb's of Ampère's; rather it is a 'rule of thumb' that applies pretty well in most circumstances. You will certainly not get a nobel price for finding an exception! A more general form of Ohm's law is$$\mathbf{J} = \sigma \mathbf{E},$$where \mathbf{J} is the current density, ... 2 If you have a known load R', you can divide the potential of a source with a single resistor. If your source voltage is V and you want a voltage V' across the load, you can write$$V' = \frac{R'}{R_L+R'}V from which you solve for $R_L$, the load. However, if you don't know $R'$ (or if it can change - for example, as a light bulb heats up, its ...

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Are you aware of Kirchoff's Laws? A good analogy is to think of a battery applying an electric pressure difference across the ends of the circuit. For components in parallel, one side of the arrangement is at one potential supplied by the battery, and the other side of the arrangement is at the other potential so that the potential difference across all the ...

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Hint for your second question : Apply kirchoff's loop law in series circuit. And in parallel circuits P.D. remains same as there is no resistance for voltage drop[in ideal wired circuit]

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Adding another battery in parallel could allow the supply of more current, but if the current you need (for your circuit and battery voltage) is already sufficiently supplied by the first battery then you won't see any difference. Note also, that in generally you shouldn't connect batteries in parallel, and it can even be damaging or unsafe to do so. For ...

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