# Need Urgent help. Please someone (Circuits Capacitance, Inductance)

• Engineering

on this diagram, the expression IL(t) should read I(t).

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tiny-tim
Homework Helper
Welcome to PF!

Hi deyiengs! Welcome to PF! Show us what you've tried, and where you're stuck, and then we'll know how to help! Thanks Tiny-tim for taking your time. I planned to post what i had but i forgot.
On the first one, how do you get R- thevinin? If that's the way to go about it?.
On the second one I need the initial and final conditions. I used -5mA and did a source transformation to get -2.1mA as the initial consition. I used 5mA for final and got 2.1mA as the final conditio for the inductor. Then used the formular:
iL = (i(initial) - i(final))e^(-1/tc)+ i(final)
My tc was 3.3333us
Then used VL = L di/dt

Is that close to the correct way or did I deviate

On the following diagram, I have the following for finding Vo:

node analysis on the top nope of the Vn

but i need to replace Vp by Vs.

Hence for the node on the bottom part I have:
I'm stuck here how can I factor out Vp on the above equation. is it possible if it it how.

If not is there another way of solving this equation.

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The Electrician
Gold Member
In the first two equations of the opamp problem, shouldn't the Vp be Vn instead?

In the first two equations of the opamp problem, shouldn't the Vp be Vn instead?

yes but Vp = Vn

The Electrician
Gold Member
Someday you may have to find a transfer function where the opamp isn't ideal. In that case, Vp#Vn, so it's good practice to use the symbol for the actual node that you're writing an equation for. Then at the end, if the opamp is ideal, substitute Vp=Vn. This way, you avoid mistakes in setting up your equations.

The Electrician, could you help me solve it if you can please

The Electrician
Gold Member
You can either solve your two simultaneous differential equations, or you can solve it in the Laplace domain. I would do the latter. Have you studied the use of the s-variable for solving AC circuits?

Set up two equations:

s*C*Vn + (1/R)*(Vn-Vs) = 0

s*C*(Vp-Vs) + Vp/R = 0

Now, you also know that Vp = R/(R + 1/(s*C)) * Vs by the voltage divider rule, and Vp=Vn.

Substitute that that last equation, Vp = R/(R + 1/(s*C))*Vs, for both Vp and Vn in the first two equations. Then you will have two simultaneous equations you can solve with simple algebra.

The solution is Vo = s*R*C*Vs, which in the time domain is Vo = R C d/dt(Vs). This ignores initial values for charge on C.