# How to calculate the values of $R$ and $C$?

Greetings, it is asked given the next circuit 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!

• You have to use the information of the current through $R1$ at $t=0$, where is the current going and can you find a relationship between time, current, resistance and capacitance? – Arsenal Oct 4 '16 at 16:00
• +1 for all the MathJAX. I've tidied up some of the markup, particularly the data you were given. I have never heard of "unit step volts". What does that mean? – Transistor Oct 4 '16 at 17:11
• @Transistor I believe he meant the Heaviside step function (also known as "unit step function", from Wikipedia) – Vicente Cunha Oct 4 '16 at 17:13
• @VicenteCunha: Thanks. So what does $V(t)=94u_1(t)V$ look like "on the scope"? A step from 0 to 94 V? – Transistor Oct 4 '16 at 17:17
• @Transistor, thanks for the +1, your are very kind, indeed. Ive learned the formating a bit more to wrote it properly. Thats right! it seems like a step from 0 to 94 v in the second 0 or any other time , based if the function has some delay.\\@VicenteCunha, Thanks for the support on Heaviside function, Im more used to call it "step" – riccs_0x Oct 4 '16 at 17:20

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:

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.

• Ive checked again the response and its different from the proposed, I did take the Laplace transform of $V(t)=CR_1\displaystyle\frac{dV_C}{dt}+\displaystyle\frac{R_1+R_2}{R_2}V_C$ , Im posting the process, and about tau, I have no graph or more data about the response of the system, but what about the $$v_c(t)=\frac{R_2V}{R_1+R_2}\bigg(1-e^{-\dfrac{t}{\tau}})$$ and taking the exponent of $$e$$ as $1$, what does it mean? $$\frac{t}{\tau}=1$$ It can taken to reach a numeric value of C? – riccs_0x Oct 4 '16 at 21:18
• The solution you found for $v_c$ is the same as in the answer I posted...I edited my answer so that you see what happens when $t=\tau$ (it won't help you find C) – Big6 Oct 4 '16 at 22:13
• I make a mistake on the step response and tried to fix the comment said before but I cant do it..yes your response its fine so I corrected the original post – riccs_0x Oct 4 '16 at 23:09
• Lets see, it is considered when the system has a response of $5τ$ the system has the permanent response or almost 100% of this, $t\rightarrow\infty$. So, how valid will be if its asigned τ equal to one and. As you say, there is no condition on it, so it will be $$C=\bigg(\frac{4700+10000}{4700*10000}\bigg)\1=0.000312766$$ or $$312.7 μF$$ – riccs_0x Oct 5 '16 at 17:23
• @riccs_0x That would be a valid solution, any value of $\tau>0$ works. Knowing that $5\tau$ is considered steady state, doesn't help in finding a specific value for C. If you make $\tau=1$, you are saying $\tau=1$ second, that's the units for the time constant. Now, if you let $\tau=1$ second then steady state will be after 5 seconds. if you let $\tau=2$ seconds then steady state will be after 10 seconds, and so on – Big6 Oct 5 '16 at 18:05

Given $V_C(\infty)=76.73V$ and $PR_2(\infty)=588.8mW$ you could have immediately calculated R2: from $P = \frac {V^2}{R}$ we can calculate

$$R_2 = \frac {V^2}{P} = \frac {76.73^2}{0.5888} = 9999.14 ~ \Omega$$

That's close enough to your 10k.

Assuming that $V(t)=94u_1(t)V$ means the voltage steps from 0 to 94 V at t = 0 when C is completely discharged and behaving for an instant like a short circuit then we can use the other bits of information, $iR_1(0^+)=20mA$ and $V(t)=94u_1(t)V$, to calculate

$$R_1 = \frac {V}{I} = \frac {94}{0.02} = 4700 ~ \Omega$$

If you can clarify what $u_1$ and $\delta$ are I might be able to finish this.

Update: While I'm not familiar with Heaviside or delta dirac functions it seems that we don't have enough information to calculate the value of C. We need another voltage or current measurement at t > 0.