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I think this simple question is an excellent one. The simple answer is: if the slab has the same magnetic constant as freespace, i.e. $\mu_0$, it has negligible effect on the inductor if the frequency is low enough that the problem can be thought of as magnetostatic. I'm not sure how well wonted Maxwell's equations are to you, but the criterion that you ...


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Why is it so? Well, it isn't actually always so. It will depend on the actual circuit configuration and whether the switch opens or closes when $t=0$. But first, here are a couple of crucial results to always keep in mind when solving these type of switched circuits: the current through an inductor must be continuous the voltage across a capacitor ...


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@Floris' answer being good, i'll give another view on the matter. A capacitor is equivalent to an open circuit (since simply put, a capacitor is an element consisting of two plates which do not actually touch but through another medium, the dielectric, the circuit is not connected at that point where the capacitor is located), whereas an inductor is ...


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Circuits with inductors are sensitive to changes in the signal - think of them as differentiators. Circuits with capacitors are responding to the integral of the signal over time. When you first turn on a circuit, the current wants to make a step change - which the capacitor doesn't care about, but the inductor resists vigorously. Thus the current will flow ...


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This is where the 0.7 comes from: Single pole low pass frequency responses are in written in the form: $$ A_v=\dfrac{1}{\sqrt{1+\left( \dfrac{\omega}{\omega_{co}}\right) ^2}}=\dfrac{1}{\sqrt{1+\left( \dfrac{f}{f_{co}}\right) ^2}} $$ for $\omega=2\pi f$ and $A_v=\left(\dfrac{v_{out}}{v_{in}}\right)$. Plots of frequency responses are call Bode plots (try ...


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In electric linear circuits theory, use is made of phasors to represent sinusoidal inputs (most common basic inputs to linear circuits). For linear circuits the total output of a given input is just the sum of each output component associated with each input component (provided there is a way to decompose any given input to compoents) (this is the ...


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There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy. The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$. See the diagram. The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$. Parallel capacitances just add, so ...



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