Hi,

Regular old circuit analysis solves problems like this and many others and is not "that" hard to learn.

If you are into doing it this way, the voltage across any element in a series circuit is that element impedance divided by the total impedance times the source voltage. So for a given circuit with 3 elements and one voltage source, the 3 elements impedance is:
Z=Z1+Z2+Z3

and the voltage across the three elements is:
v1=Vs*Z1/Z
v2=Vs*Z2/Z
v3=Vs*Z3/Z

and we would go from there, but you see how systematic this is where every one is done using the same basic method.

We could go into this in more detail later.

 

Keep in mind there's no indication that the TS has been exposed to complex impedance yet.

 

Keep in mind there's no indication that the TS has been exposed to complex impedance yet.

Hi,

Yes i was hoping to introduce this idea. Not sure how well it will be accepted yet though.
I like to state general methods of doing things a lot because it helps to solve so many problems that come up. I see so many problems that come up that could be solved using the same methods for each problem but the poster does not seem to have any idea that there is something like general circuit analysis. I feel that is such a shame because these questions can be approached in almost the same way by learning general circuit analysis one step at a time. If i had my way, i would teach everyone at least Nodal analysis

 

Hi TimoG,

I would consider the system as a whole and solve T from there. Add the capacitors (C_tot=C1*C2/(C1+C2)) and resistors (R_tot=R1+R2) and use the formula you provided (T=RC). However, if your question is what the stationary voltage across each of the two capacitors are, its not really needed. It's been previously written, that the stored charge Q is equal in each capacitor. You also know that the voltages across both have to be equal to the input source(5V) at stationary operation. Hence, you can use a set of equation to solve it.
Q=C*V; Q1=Q2=Q; V=5V; C1=56nF; C2=100nF; V1/2=voltage across the respective capacitor.
Q=C1*(V-V2);
Q=C2*(V-V1);
leads to:
C1*(V-V2)=C2(V-V1);
solve one voltage in respect to the other and voilà.

Thank you all for the help! I think I am able to solve the question now.

 

Thank you all for the help! I think I am able to solve the question now.

But PLEASE be sure to not just use the 95% solution that was spoon fed to you and then do a bit of math (no EE) to get the answer and turn it in and call it good. You will get hammered on the exam if you do that.

Be sure that you UNDERSTAND where EVERY part of EVERY step came from.

 

Try to combine (R1 + R2) and (C1 + C2) first.

yep, that works...as unintuitive as it may seem, seeing as the resistors have a capacitor between them.