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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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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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You set up a circuit which contains the galvanometer and adjust the circuit so that the deflection on the galvanometer is a maximum (for greatest accuracy). A calibrated resistance box is connected in parallel and adjusted until the deflection on the galvanometer is half of the deflection with no resistance box. In this condition the resistance of the ...

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The quality factor is a ration and tells you something about the sharpness of the resonant curve. If you had a resonance at $\omega^{\prime \prime}= 500$ and $\delta \omega = 5$ and compare it with $\omega^{\prime \prime}= 5000000$ and $\delta \omega = 5$ then which has the sharper resonance? What about If you had a resonance at $\omega^{\prime \prime}= ... 1 It depends on the shape of the medium, but basically you could have an infinitesimal resistance,$\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 = ... 1 In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at$a$. At center all have same potential as that at$b$So, all have same potential difference$V_{a} - V_{b}\$ Same potential difference means that they are connected in parallel.

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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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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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