Recent content by Sciurus

I just wanted to place a guess. From the following (probably faulty) reasoning. Imagine a planet orbiting a large body. I think that if the planet follows a stable orbit, the large body does not undergo a displacement over the period of the orbit. So now, let's image the planet breaks up...

Interesting question. http://www.jstor.org/pss/2397270 Apparently someone else had the same question.

You can convert one voltage AC to another with a transformer; that's probably easiest. It depends a little on what you're powering, but voltage usually matters. If your device draws a constant current, you can make a voltage divider, otherwise, no. Again, it depends on what you're...

Hmm...I must misunderstand something. To be honest, my experience with this stuff is from a technician point of view, not from an engineer's, so I often think in terms of mental short-cuts so I can troubleshoot quickly. In this case they may have gotten me into trouble somewhere.

Well, you can get high current through any transformer by decreasing the resistance of the load. (Sometimes it's called increasing the load since the energy the transformer supplies increases). Of course, at some point, either a fuse blows (if you're lucky), or something starts on fire...

OK, cool. Since T1-secondary, T2-secondary, and RL form a closed loop with no branches, the current through those components is the same. The voltage on RL is the sum of the voltages from the T1-secondary and T2-secondary (the AC has the same phase), and by ohm's law, you can work out the...

I always looked at it as the load determined the current, but I guess in an ideal sense, there is also a relationship between turns and current. Is this your circuit?

Hi, toneboy1. What determines the current through a secondary coil?

None that I know of -- an interesting idea, though. Well, the electric field would get weaker, and I suppose at some point it would be negligibly small, but it won't disappear until the charge does. (Unless there's another conductor to interact with.) Yeah, we're dealing with static...

Are you looking at this schematic? Typical diodes have a 0.6-0.7 V drop when forward biased, so you could use a few of those in series.

Try thinking of the eye as a pin-hole camera -- a box with a hole in one side to let light in.

You are correct that capacitance would decrease, but the plates were charged with the higher capacitance when the plates were close together, so they will no loose charge or energy, assuming they were not permitted to discharge. If you wanted to recharge the plates after they were moved farther...

Hi, davisp2012. I'm no expert, but I may be able to help a little. Your transistor should have a datasheet (you can probably just google "TIP31C datasheet" to find one), as should the LEDs. You can use the sheets to find out if the current draw from 50 LEDs exceeds the maximum collector...

Yup, sounds good. The filament center will always be zero volts since the center of the 100 ohm resistors is grounded. It make it a nice little test circuit.

I'm just going to use a schematic from the web to avoid drawing one. Since each series pair of resistors is an independent path, the voltage at the center depends on the resistance of the resistors above and below. If the left two are 500 ohms each, both will have the same magnitude...