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Would you ask why two unequal voltage sources cannot be places in parallel? In that case the answer would that infinte current would flow between them. In your case, with two unequal current sources in series, infinite voltage would be generated.


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The idea is that since the steel beam has conduction electrons that are free to move, the movement of the charges in a magnetic field causes a magnetic force to act. The magnetic force causes the electrons to accumulate at one part of the curved surface of the rod, thereby creating a potential difference. The charges keep accumulating till the potential ...


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Expanding on Jan Dvorak's comment: When you change the magnetic field inside a loop, an emf (electromotive force) will be generated. Now if you have two loops, each of these will experience the same e.m.f. When you put them in series, you have a coil with two loops, or two coils with one loop. No matter which way you look at it the voltage across them ...


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Thanks laying out your work so neatly in the question. I think the solution is the following $$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$ where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge). so, for example, if the point charge goes from $r$ to $2r$ we have two positive ...


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The first equation is when you want to solve for either the voltage, current, or power already knowing the other two, similar for $P=I^2R$, when you want to solve for an unknown already having knowledge of the other two. The last equation you get by noting that, $P=IV=I^2R$, hence $V=IR$ or $I=\frac{V}{R}$ plugging this into, the first equation you get ...


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The voltage of a Galvanic cell can both increases with temperature or decreases with temperature. You can prove using thermodynamics or using your intuition that: for Galvanic cells that get hot when working, the voltage decreases as the temperature increases; for Galvanic cells that get cold when working, the voltage increases as the temperature increases. ...


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But what about the 180° gap in the middle? This implies a negative real part of the impedance, i.e., a negative resistance. Express the impedance $Z$ in polar form $$Z = R + jX = |Z|e^{j\phi}$$ where $$|Z| = \sqrt{R^2 + X^2}$$ and $$\tan \phi = \frac{X}{R}$$ For $R > 0$ $$-90^{\circ} \lt \phi \lt 90^{\circ}$$ but for $R < 0$ $$90^{\circ} ...


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I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its $V$ (notice that $V$ is really just measuring the voltage drop across the ...



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