Recent content by energychaser

Could you clarify what you mean by "Now you would assume solutions for V1(t) and V2(t)"

If i were to quickly restate my objective, I want to know how integrate or understand how time acts with respect to reaching equilibrium when one capacitor discharges into another. So if I want to know how long it takes the current to reach 1% of its strength at switch close how would I do so?

Okay, I know that by the KCL concept what is leaving one capacitor 1 must be equal to what is entering capacitor 2. I know that the current at any point is equal to the potential difference between the two capacitors, divided by the total impedance (resistive impedance of the resistor, and...

in answer...Once upon a time yes, I was familiar, now not so. This is an attempt on my part to break the rust loose as it were with an example I have been contemplating. Point the direction?

Hello, I have been pondering how to plot the instantaneous rate of change of one capacitor into another through a resistive impedance R. So for example you would have C1=1 farad (100volts) C2= 10 Farad (Zero volts) R1 = 100ohms So one would have Capacitor 1 at 100 volts, discharge...

Thanks! I was quite perplexed, I had actually solved the problem in about 15 minutes, and went to check with that online applet only to discover I was wrong...I have a decent knowledge base from which to draw, and knew it should not involve calculus, and there should not be two varying...

wow, quite sure the applet was way wrong...I spent all day trying to figure out something i may have done right. if: C1= 1 C2=10 V1=1000 V2=100 Q1=1000 Q2=1000 Then the combined capacitor will have a C=11 and 2000 coulombs of charge. This equates to 181.81 volts across it, thus...

I think perhaps the simple circuit simulator I am using is wrong! could someone please check my numbers and see? http://webphysics.davidson.edu/applets/circuitbuilder/tutorial/circuitbuilder_intro.htm Thanks again for all the kind help.

(ps, I had tried that approach (but am sure I did something wrong) and came up with this. Knowing both capacitors together are 11c, with a combined charge of 2000, the final voltage on them would then be 181, which is not correct according to my simulator!)

I must admit I am quite confused, and have spent several hours in front of a piece of paper trying to figure this out...should not be this difficult! I have never known that you could take the charge on one plate alone as a consideration, would not total charge always be zero then as top and...

Yes! (why didnt I think of that, I should have been more clear) that is exactly what I meant,

There is nothing else in the circuit, simply capacitors connected leg to leg. Here are the parameters for capacitance, voltage, charge and energy on each. C1= 1 C2=10 V1=1000 V2=100 Q1=1000 Q2=1000 j1=500,000 j2=50,000 Before connected we have a total of...

so... if I had 2 capacitors in series equaling 10 farads, at 100 volts, I would have a net of 1000 coulombs. If I looked at a single capacitor of the above the same voltage, say 20 farads at 100v I would have 2000 coulombs. I don't see how this helps me calculate... I really learn...

what should be a simple capacitor question. Hello, I have a question I have been pondering for a bit, me thinks it should be simple, but I cannot seem to get it straight. Imagine two capacitors leg to leg. each has 10 farads. Each is charged to 100 volts. Each should have a charge of 1000...