Electronic – How to calculate the values of \$R\$ and \$C\$

circuit analysis


Greetings, it is asked given the next circuit

circuit diagram


so that it can be calculated the values of the components \$R_1\$, \$R_2\$ and \$C\$.
Ok, Im given the next data and question (besides the values !):

  • \$iR_1(0^+)=20mA\$ ;
  • \$V_C(\infty)=76.73V\$ ;
  • \$PR_2(\infty)=588.8mW\$;
  • \$V(t)=94u_1(t)V\$ (94*unit step volts)

This says the problem, I'm writing down if this helps, don't pretend to have a question of questions. =)

  • How much time does it take to \$R_1\$ to reach \$6.64mA\$?
  • Make a graph for \$V_C(t)\$ if \$V(t)=10δ(t)\$.

Lets says the most interesting part to me is calculating the values of the elements, so the first thing done its to get the mathematical model of the system, taking the variable of \$V_C%\$, it yields as

\$V(t)=CR_1\displaystyle\frac{dV_C}{dt}+\displaystyle\frac{R_1+R_2}{R_2}V_C\$

and rewriting
\$\displaystyle\frac{V(t)}{CR_1}=\displaystyle\frac{dV_C}{dt}+\displaystyle\frac{R_1+R_2}{CR_1R_2}V_C\$

Neat!

Next to obtain the total response of the system using the Laplace transform yields:

\$V(t)=\displaystyle\frac{94R_2}{R_1+R_2}-\displaystyle\frac{-94R_2}{R_1+R_2}e^{-\displaystyle\frac{R_1R_2}{CR_1R_2}(t)}\$

and the impulse response

\$h(t)=\displaystyle\frac{94R_1R_2^2}{C(R_1^2R_2+R_1R_2^2)}e^{-\displaystyle\frac{R_1R_2}{CR_1R_2}(t)}\$

But after that, I have no right idea what to do, so taking the \$V_C\$ value as steady state then \$V_C=VR_2\$ and \$V_R1=V(t)-V_C=94-76.73=17.27V\$ and from the step response taking the permanent part can it say that

\$\displaystyle\frac{94}{R_1}=76.73\$ then \$R_1=1.225Ω\$
; to know \$IR_2\$ it is used the power form of \$P=IV\$ then \$IR_2=\displaystyle\frac{588.8mW}{76.73V}=0.007673A\$

and \$R_2=\displaystyle\frac{76.73V}{0.007673A}=10000Ω\$

But then I dont get how to get the C value.

What am I missing?

Update: Step response
– \$\displaystyle\frac{V(t)}{CR_1}=\displaystyle\frac{dV_C}{dt}+\displaystyle\frac{R_1+R_2}{CR_1R_2}V_C\$ ;

aplying the Laplace transform to the equation and the input:
– \$\displaystyle\frac{94}{SCR_1}=SV_C(S)-V_C(0)+\displaystyle\frac{R_1+R_2}{CR_1R_2}V_C(S)\$ ;
grouping:
\$\displaystyle\frac{\frac{94}{CR_1}}{S(S+\frac{R_1+R_2}{CR_1R_2})}=V_C(S)\$ ;
Using partial fractions
\$\frac{94}{CR_1}=\frac{A}{S}+\frac{B}{S+\frac{R_1+R_2}{CR_1R2}}\$ ;
\$A=\frac{R_2}{R_1+R_2}\$ and \$B=\frac{-R_2}{R_1+R_2}\$

giving the finally
$$v_c(t)=\frac{94R_2}{R_1+R_2}-\frac{94R_2}{R_1+R_2}e^{-\frac{R_1+R_2}{CR_1R_2}t} $$
OK! im refreshing the values!

Best Answer

You are on the right track but your solution to the differential equation is not quite right. I solved it using Wolfram Alpha and this is what I get:

$$v_c(t)=\frac{R_2V}{R_1+R_2}-\frac{R_2V}{R_1+R_2}e^{-\frac{R_1+R_2}{CR_1R_2}t} $$

Or $$v_c(t)=\frac{R_2V}{R_1+R_2}\bigg(1-e^{-\dfrac{t}{\tau}}\bigg) $$

*Where \$V\$ is the voltage from the source (94V).

Which makes sense because when \$t\rightarrow\infty\$ (capacitor is charged) all you have is a voltage divider formed by the two resistors.

Since when the step voltage is applied, the capacitor is a short, \$R_2\$ gets shorted out and all the the current has to flow through \$R_1\$ at that instant. That's why you are given \$i_{R_1}(0^+)\$. Then

$$ R_1 =\frac{V}{i_{R_1}(0^+)}$$

Where \$V\$ is the voltage from the source (94V). (@Transistor answered this in his answer)

You are also given the power on \$R_2\$ when steady state. You got this one right.

Now, in order to find \$C\$, it would be helpful to use the time constant \$\tau\$, which is the time it takes the capacitor to reach 63.2% of its final value.

$$ \tau=\frac{CR_1R_2}{R_1+R_2}$$

And

$$ C=\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)\tau$$

You already have the values for \$R_1\$ and \$R_2\$, and the only unknown is \$\tau\$. From the original question I can't see anything that would force \$C\$ to be a specific value, that is, there is no given constraint on the time constant, other than \$\tau>0\$ so that the problem is nontrivial. No constraint for \$\tau\$ means no constraint for \$C\$. So any value of \$C\$ does the job.

Edit:

Answering the OP question in from the comments.

If you let \$t=\tau\$ then you should get a value for \$v_c(t)\$ equal to 63.2% of it final value. So at \$t=\tau\$:

$$ v_c(\tau)=\frac{R_2V}{R_1+R_2}(1-e^{-1})$$

Or

$$ v_c(\tau)=\frac{R_2V}{R_1+R_2}(0.632)$$

Whatever that gives you won't help you in finding a value for \$C\$. As you can see when you let \$t=\tau\$, the \$C\$ term isn't there anymore (it's embedded in \$\tau\$) so can't solve for \$C\$ from there.

The only place to solve for \$C\$ is in the \$\tau\$ equation.