Recent content by guyvsdcsniper

I have, each pole corresponds a 6dB/oct gain. So i'm guessing I would need to take the magnitude of the transfer function and evaluate in the limit of low mid and high frequencies? I calculated the transfer function to be: $$H = \frac{-\omega^2 L C}{j\omega RC-\omega^2 LC+1}$$ and then...

From 0 to ##10^3## ##\omega## there is a dB gain, from ##10^3## to ##10^5## there is another. Finally from ##10^5## to infinity the slope is constant (0). I know the formula $$dbV= 20log_{10}\frac{V_2}{V_1}$$ can give me the slope but that is in terms of Volts, but I have frequency and the...

I am solving #5 in the attached image. So I am graphing the power dissipated by a series circuit which : (a) Contains a constant voltage source and a variable resistor, R (b) Contains a constant current source and a variable resistor. It makes sense to me to just use the power equation that...

I am finding that I only get the same result as the book when I assign the voltage drop polarity with respect to the loop I am following. So when I follow Loop A clockwise, I get the assigned voltage polarity as seen in my previous post. But when I start at node 4, and follow loop B and treat...

Hopefully this makes my thought process clear. I tried breaking down how I viewed each resistor when traveling loop B.

Well I was assuming that the current traveling through this circuit will follow the path as described in figure 1.7. I used that to assign the voltage drop polarities. So the positive end of the resistors are where the current enters and the negative end is where the current exits. So when...

Attached is the example I am working out of a textbook that involves using the Loop Current Method on a bridge circuit. In the pictures attached I am following section 1.6.2 which produces loop equations (1.24) for figure 1.9. Figure 1.7 provides the direction of current. I am having trouble...

TSny was correct, I completely missed the factor of ##10^{-5}##. I am on the right order of magnitude now. I completely missed the units and the ##10^{-5}## on the lab manual, thank you for pointing that out haruspex. Just a bonehead mistake and reading way too fast. You can close this thread...

Falling was about a minute and rise was around 5 seconds. In hindsight, id use a lower voltage to reduce the rise speed to get a more accurate velocity.

Because in S.I. units, 1 electron corresponds to ##1.6 x 10^{-19}## Couloumbs, and what I am calculating is on the order of ##10^{-10-12}## Couloumbs, making my value larger.

Sorry, please refer to page 9 of the lab manual in the link for the formula used to calculate q.

Sorry I should have been more clear. Its Viscosity of Dry Air as a Function of Temperature. To calculate the radius of the oil drop, stokes law is used. But according to the lab manual "Stokes’ Law, however, becomes incorrect when the velocity of fall of the droplets is less than 0.1 cm/s...

I am doing the Millikan Oil Drop experiment to determine the charge of a single electron. I have been following the lab manual provided by the manufacturer, https://hepweb.ucsd.edu/2dl/pasco/Millikans%20Oil%20Drop%20Manual%20(AP-8210).pdf. The manual defines a simple method to calculate for...

Sorry I meant to write index vs fringes, that is what I ended up doing.

I am not sure, the lab manual does state anything other than plotting the two. I have searched online for other lab reports and I dont see any rely on this graph. Given the last equation from the lab manual screen shot I posted, I can simply cancel the common factor of the change in pressure...