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In order to make the problem symmetric, consider this: What happens, when you take the dodecahedron and drive current $I$ into it from vertex $A$ and drive $I/20$ out from all every vertex (including $A$)? By Kirchhoff's laws and symmetry, there is a current $I_{A-out} = \frac{(I-I/20)}{3} = \frac{19}{60}I$ going from $A$ to neighbouring vertexes. Now ...


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I had read that resistances only lose energy in the form of heat. Correct. Is that what the authors are talking about? Yes, the author is just using confusing language. Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat? The author means that electrical ...


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An electromagnet has a lot of turns of wire, which gives it a property called inductance : the property of resisting a change in the current. You may be familiar with Lenz's Law $$V=-L\frac{d\Phi}{dt}$$ which says that an inductor with inductance $L$ will generate a back e.m.f. $V$ when the flux $\Phi$ through it changes. Now if you drive a current through ...


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You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ...


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It depends on the shape of the medium, but basically you could have an infinitesimal resistance, $\mathrm{d}R$, and integrate that. For example, for a uniform area wire of area $A$ and length $\ell$ in which the resistivity varied sinusoidally about some nominal value, $$\rho(x)=\rho_0(1+0.05\sin\left(\frac{5}{\ell}x\right)),$$ then $$R = ...


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Short answer: "neither". A resistance, in a circuit, causes a voltage drop for a given current. Alternatively, if the voltage is given, it causes a particular current to flow. As soon as a circuit gets a little bit more complex (more than one battery-resistor-light bulb-switch), it becomes completely impractical to use lots of batteries to control ...


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So the spectral approach as I remember is you make an anszats $i=\sum_k a_k f_k(t)$ or something like that, you use some theorem to say that each component of the sum is independent so you get $a_k f_k''(t) + 5 a_k f_k'(t) + 1/4 a_k f_k(t)=0$, to get the factors (characteristic polynomial is what you said I believe) My guesses where the error could be: I ...


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The currents will be the same. Using Kirchoff's current law, the total current going out of any node or any closed contour is 0. Suppose you have two resistances connected in parallel between nodes A and B. Draw a circle around the parallel resistances. The total current going out of the circle is 0. So outgoing current at A + outgoing current at B = 0. In ...


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The electron moving from negative to positive terminal has to suffer lot of collisions with other electrons and has to also interact with the kernels present.Hence greater the length,, greater is resistance.


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Assuming the bulb emits in the visible wavelength regime, then you can use a silicon photodiode. Depending upon the particulars of your "bulb" you could put the photodiode in series with a resistor and put that combination in parallel with the bulb. So as the photodiode receives more light, it's resistance will decrease thereby allowing current to flow in ...



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