Recent content by gneill

I think you'll want to pay attention to the directions of the currents in the two loop circuit. Right now you've got them labelled without directions given. From your diagram I've put labels on them, such that I1 = I2 + I3: See if you can't write expressions for the two loops, keeping track...

Correct!

When the switch is closed for a long time, the current through the inductor will be ##I## ( given by ##E_0/R_1##). As soon as the switch is opened, that current through the inductor will persist; the inductor will not change its current rapidly. The inductor will create whatever voltage it...

I think it would be simpler to work from the magnitudes: $$\left| \frac{V_R}{V_S} \right| = \frac{\left| R \right|}{\left| R - \omega^2 R L C + j \omega L \right| }$$ Take the magnitudes and clear the square root on the right hand side by squaring; it doesn't matter if you maximize the left...

Take a look at the image supplied in the first post. You have the height and width of the triangle. Solving for θ should be easy. I included θ below:

True. I admit I took a shortcut where ##\left| R \right| = R##, since ##R## is positive real.

Start with: $$ \frac{V_R}{V_S} = \frac{R}{R + \left( j \omega L \right) \left(j \omega R C + 1 \right)}$$ $$\left| \frac{V_R}{V_S} \right| = \left| \frac{R}{R - \omega^2 LRC + j \omega L} \right|$$ But ##\left| \frac{V_R}{V_S} \right| = 1##, so: $$1 = \frac{R}{\left| R - \omega^2 LRC + j \omega...

I think you want to ponder what ##\left| \frac{V_{R}}{V_{S}}\right| ## actually is. It's magnitude is one, but that means you need to take the magnitude of the expression on the right hand side. You should get: $$1=\left| \frac{R}{R+(j\omega L)(j\omega RC+1)}\right|$$ Then solve for C. Edit...

Yes!

It doesn't matter. The core is stated to be spherically symmetric, so that it behave as a point mass to anything beyond it's boundary. It doesn't matter whether it's homogeneous (radially) or not, if it is given as spherically symmetric. It may have alternating layers of dark and normal...

Yes. That would be my answer.

I find it doubtful that dark matter is not fully mixed with the core, and only surrounds it. But anyway, that's that's irrelevant; mass is mass. If you have a mass of dark matter mixed in with normal matter, you cannot distinguish the two. The mass of the core ##M## would not distinguish...

If the core of mass ##M## has a radius of ##r_1##, and the speed of the closest stars is ##v_0##, wouldn't that give you the mass of the core? I presume that the mass of the core includes a mixture of regular as well as dark matter, but that the dark matter extends beyond the core and...

For a Bode plot (sweeping through multiple frequencies) you'll want to specify the voltage and frequencies in a different way. There should be some entries equivalent to LTSPICE's 'Small signal AC analysis' in the setup for the voltage source, probably shown as magnitude and phase? For the...

In LTSPICE the voltage source gives you the option of SINE and you can specify the AC voltage and Frequency, as well as a value for Phi, the angular offset. I don't know what Multisim gives (I don't have it installed, or signed up for the online version). But surely there must be some option...