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In problems of this type, it's required to determine the initial conditions. Often you might have to work them out from analyzing a modified circuit (for example, with a switch in a different position). But it's also possible the initial conditions are simply given as part of the problem statement, and that seems to be what was done here. If your professor ...


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Intuitively, inductors create back EMFs that "resist" changes in current. Anytime there is an inductor in a circuit, it will resist such changes. Mathematically, inductors force the current in a circuit to be continuous. Let's consider an LR circuit where the inductor and resistor are both in series. Regardless of the initial voltages or current, we can ...


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The voltage drop across an inductor is proportional to the change in current, or V = L*dI/dt. When the switch is closed, the circuit is completed and a current starts to increase dramatically. This then causes a time increasing flux of magnetic field in the inductor. According to Lenz's Law, an opposite current will be induced in the inductor to oppose the ...


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The part that I don't get is mathematically how to get from one to the other. $$\frac{d}{dt}U_L(t) = p_L(t) = v_L(t) \cdot i_L(t) = L \cdot i(t) \cdot \frac{di}{dt}$$ but $$\frac{d}{dt}\left\{\frac{1}{2}Li^2(t)\right\} = L \cdot i(t) \cdot \frac{di}{dt} $$ thus $$U_L(t) = \frac{1}{2}Li^2(t)$$


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The 2nd equation defines the ideal inductor circuit element. It is understood that the voltage $v$ and current $i$ in that equation are the voltage across and current through the inductor. The inductor emf is the opposite sign of the inductor voltage. $$\mathcal E_L = -v_L $$ Clearly, when the current 'stabilizes' (the time rate of change of the inductor ...


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In order to calculate the inductance of either solenoid, you calculate the total flux in each solenoid. In order to calculate the mutual inductance you need to calculate either (a) the flux through the smaller solenoid due to a current through the larger solenoid or (b) the flux through the larger solenoid due to a current in the smaller solenoid. In (a) ...


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DC signals do not induce electro-motive forces, for you would need a change in the magnetic flux through your circuit which can only be achieved with a time-varying current (resulting in a time-varying magnetic field). This, of course, assuming that your circuit is stationary, so basically you are not moving the wire around for this would change the area ...


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In the limit of long times, the currents are steady, so the magnetic fields they create are steady so there is no induced EMF. This situation is usually tagged "steady state". That said, there will be a period of time when you have just switched a circuit on or off during which things have not settled down and then there will in general be effects not seen ...


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It seems entirely symmetrical. But it isn't symmetrical. For the voltmeter on the right, the 'outer' loop that encloses the magnetic field consists of the voltmeter, leads, and a $100\Omega$ resistor For the voltmeter on the left, the outer loop that encloses the magnetic field consists of the voltmeter, leads, and a $900\Omega$ resistor. Moreover, ...


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Interesting setup! I assume that you understand that for a short time there is a (induced) current flowing in the loop with the two resistors. You can get the same current by imagining a battery that is in series with the loop (you can put it at A, for example). With that battery in place, it is easy to see that the voltage across the 900r resistor will be ...


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Reluctance = $\dfrac{l_e}{\mu A_e}$ where..... mu is the absolute permeability of the material, $\mu_0 \mu_r $ $l_e$ is the circumference of a circle at a radius r and $A_e$ is a small cross sectional area. The circle I refer to only relates to the cross section of the torus and r is the radius from the centre (where the wire is). All these reluctances ...



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