Recent content by klouchis

Well if that's the only information you are given in the problem then the correct theoretical answer is infinity. No circuit will ever be left running until infinity though, it is important to know when the current is say 95% or 99% of the max value, so it would be a good idea to calculate...

That would be the answer assuming that's the question being asked. I have a feeling that a different question is being asked because knowing that I=1.6A when t is infinity doesn't seem like a practical problem. What exactly is the question for this part?

I'm still unsure exactly what you are trying to find in part b part ii. The time it takes to reach maximum current is infinity. Usually you want to know when the current is within a certain percentage of the maximum current. An example being, what time will the current be 99% of the maximum...

The two calculations don't agree because one of them is incorrect. It's the first one, I'm not sure what the equation, Growth=I(1-e(-t/T))=34.3(1-e(-0.05/0.05)) is saying. Here you define I as the initial rate of change which is 34.3A/s (which is correct). If I is in units of Amperes per...

Oh my, yes you are correct, I mistyped. The time constant is L/R so that way the ratio or (R/L)*t is unitless. Look back at the last post I had and reverse the R and L values. Henries are equivalent units to Ohm*Seconds, only way to get a time constant it to ratio henries/ohm, or L/R. My...

In your drawing the node V_c is labeled as a the voltage level at a particular node. With the location of the ground in your circuit one would assume that unreferenced node voltages are referenced to ground. This would make V_c the voltage drop across the resistor and the voltage source, not...

No the logic you have used in finding the equation, V_L+V_C+V_R-V_S=0 is incorrect. The voltage V_c does not represent the voltage drop across the resistor that is needed for the KVL loop. It is the voltage at that node relative to ground. So it is the voltage across the resistor plus the...

Sorry for any confusion, I was referring to the KVL equation in the image you posted. The equation reads: V_S=V_L+V+\frac{V_S-V_C}{R} V/R gives a current, that's all I was saying.

Are you saying that the KVL equation you have drawn in that picture is the solution you should arrive to? The equation looks incorrect as the last term is in units of current, not voltage. The summation of voltages should never equal units of current. The solution aside I would recommend...

That's a good idea, I've used a cheap CCD in my optical lab to make a quick IR microscope, those work very well for IR. I'm curious how large are the instructions and registers for your microcontroller?

Sounds like a good project to work on especially if you have the microcontroller handy. There really isn't much of a difference between a photodiode and a phototransistor. As you'll inevitably learn in a device physics course a phototransistor is really just two photodiodes put together. One...

That's right, remember that the Laplace transform is a linear operation. If f(t) has a Laplace transform of F(s), then a*f(t) has a Laplace transform of a*F(s). Assuming "a" is a scalar quantity. The same linearity is true for inverse Laplace transforms.

berkeman is right, it is important to understand what is happening when you use negative feedback with a high gain op amp. There are many applications for op amp circuits that extend beyond resistive feedback networks. I think a good place for you to start to figure this out is to remember...

Oh your right cepheid, I did switch the 100ohm and 200ohm resistor, that's why I didn't understand why vk6kro was saying that 100ohm voltage drop was 1.66V and that it was significant to the circuit working properly. It's hard to be dyslexic and be a circuit designer I suppose :smile: vk6kro...

The initial rate of change of the current is correct, another way to find it would have been to evaluate the derivative of I(t) at t=0. The time constant is incorrect, the time constant of an RL circuit is R/L. If exponential decay is of the form e^at where a is the time constant then the...