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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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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 ...


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The current doesnt pass through the voltmeter so the the resistance of load R is not affected.


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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 ...


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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 ...


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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 = ...


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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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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 ...


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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 ...


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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 ...


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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.


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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 ...


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More the power, more the heat generated. $$P(\textrm{Power})=V\cdot I$$ $$H(\textrm{in joules})\propto V\cdot I\cdot t\,(\textrm{in s})$$ $$ \therefore H(\textrm{in joules})\propto P\cdot t (\textrm{in s})$$ 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.


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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 ...


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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 $H = I^2Rt$ ...


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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 ...


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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


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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 ...


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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. ...


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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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