Recent content by Kelsi_Jade

Well, since the resistors are in series you can combine them to equal one resistor with a resistance of 3+1= 4kOhm. The capacitors in parallel would equate to one capacitor with 3+C= (3+C)microFerrads. Current with respect to time would be = Q/Δt ?

Homework Statement Here is what I have so far. Am I going the right direction? If so, how do I go about finding the total current? Homework Equations See photo above. The Attempt at a Solution See above.[/B]

Ok! Thanks guys! As for the charges on each capacitor, did I use the wrong equation since the charge for capacitor ln series should be the same?

Homework Statement I am going over problems for exam study - here is the question with my submitted solution. Anything helps, just trying to correct mistakes so I can study the problems. Two capacitors C=3mF, C=2mF are initially discharged. They are connected in series and then the two ends...

Oh sorry! I must have written the title regarding the next problem. Yes, it is about designing an LRC filter!

Well, it's two parts of the same question I was just curious with the information I have how a circuit diagram would actually look. e.g. is there a specific way the resistor, capacitor, and inductor must be oriented? Like, if it's RLC must they be designed Resistor first, then Inductor, then...

Thank you guys! I really appreciate all of the help, I know there are a few of you that have commented a few times on my posts - It's been a big help!

Homework Statement Design a series RLC filter for 10kHz using an 0.01mF capacitor. Homework Equations / 3. The Attempt at a Solution how would the circuit actually look if drawn out here? [/B]

Whoops! So I did. So Ceq=6/5mF and then plug that into the Q=CV equation?

So then the circuit becomes one with one capacitor Ceq= 5/6mF And we know Q=CV So, Q=(5/6mF)(10V) = 25/3 mC = 8.33x10^-6F?

1/Ceq = 1/C1 + 1/C2 = 5/6mF

Hmm...If you start with 10V and on the opposite side is 5V, is it bc of Kirchhoff's voltage law that the voltage drop would be negative over the 2kOhm to 5V battery? Because your overall voltage of the loop should end up =0?

Can I even reduce to it from the right, then? I keep thinking that will just eliminate node A again?

Aren't the currents going in the same direction, though? - They're both moving in a clock-wise direction?

Homework Statement Homework Equations / 3. The Attempt at a Solution [/B] Here is what I attempted for the HW. I only got 2/10 for this here so I'm wanting to correct my mistakes.