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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, Im writing down if this helps, dont pretend to have a a question of questions =)

-How much time it takes 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 it yields

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

and the impulse response

$h(t)=\displaystyle\frac{94}{CR_1}e^{-\displaystyle\frac{R_1R_2}{CR_1R_2}(t)}$

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

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  • $\begingroup$ I don't believe you've listed enough information to determine the value of $C_1$. $\endgroup$ Oct 3 '16 at 21:02
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Since the voltage of the source is constant for $t>0$, so for $\lim\limits_{t \to \infty}$ the current through the capacitor should go to zero. Also note that the voltage of wires which are connected together will have the same value.

Given that $V(\infty) = 94\,V$, $V_C(\infty) = 76.73\, V$ and $P_{R_2}(\infty) = V_{R_2}(\infty) \, i_{R_2}(\infty) = 588.8\,mW$. The wires of the capacitor and the second resistor are connected together, so $V_C(\infty) = V_{R_2}(\infty)$ and therefore,

$$ i_{R_2}(\infty) = \frac{P_{R_2}(\infty)}{V_{R_2}(\infty)} = 7.67366\,mA. \tag{1} $$

Using Ohm's law you can indeed conclude that,

$$ R_2 = \frac{V_{R_2}(\infty)}{i_{R_2}(\infty)} = 9999.14\, \Omega. \tag{2} $$

Using Kirchhoff's laws is can be shown that at $\lim\limits_{t \to \infty}$ the voltage across the first resistor is equal to the voltage across the source minus the voltage across the second resistor/capacitor, and the current through the first resistor is equal to the current through the the second resistor (plus the current through the capacitor, which is zero). Using this the value of the first resistor can be found to be,

$$ R_1 = \frac{V(\infty) - V_C(\infty)}{i_{R_2}(\infty)} = 2250.56\,\Omega. \tag{3} $$

It can be noted that for $\lim\limits_{t \to \infty}$ the circuit acts the same as the circuit from your figure without the capacitor, so at as the time goes to infinity the circuit also does not contain any information about the capacitor.


Given that $i_{R_1}(0^+) = 20\,mA$, $V(0^+) = 94\,V$, and the values found for $R_1$ and $R_2$ it is possible to also calculate the voltages across and current through the first and second resistor, and the capacitor using Kirchhoff's laws. This yields, $V_{R_1}(0^+)=45.0111\,V$, $V_{R_2}(0^+)=V_C(0^+)=48.9889\,V$, $i_{R_2}(0^+)=4.8993\,mA$ and $i_C(0^+)=15.1007\,mA$. The value of the capacitor can be found with,

$$ C\frac{dV_C}{dt} = i_C. \tag{4} $$

Using this relation you can indeed find the following differential equation,

$$ \frac{dV_C}{dt} = \frac{V(t)}{C\,R_1} - \frac{R_1+R_2}{C\,R_1\,R_2}V_C, \tag{5} $$

for a constant value for $V(t)$ this has the solution,

$$ V_C(t) = \left(V_C(0) - \frac{R_2\,V}{R_1+R_2}\right) e^{-\frac{R_1+R_2}{C\,R_1\,R_2}t} + \frac{R_2\,V}{R_1+R_2}. \tag{6} $$

However if take the derivative of $V_C(t)$, multiply it by $C$ and evaluate it at $t=0$ you get,

$$ C\left.\frac{dV_C}{dt}\right|_{t=0} = \frac{V}{R_1} - \frac{R_1+R_2}{R_1\,R_2}V_C(0), \tag{7} $$

which should be equal to $i_C(0)$, but as you can see this is not a function of $C$, so it can not be used to solve for $C$. So not enough information is given to derive the value for the capacitor.

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  • $\begingroup$ Whoa! thanks for clarifying the ideas, what would it be needed to determine the C value? $\endgroup$
    – riccs_0x
    Oct 4 '16 at 16:25
  • $\begingroup$ @riccs_0x Soms additional in formation at some $0<t<\infty$ or the derivative of $V_C$ at $t=0^+$. $\endgroup$
    – fibonatic
    Oct 4 '16 at 16:54
  • $\begingroup$ Based in the concept of $$\tau$$, the time constant, in order to get $$C$$, its considered that $$\tau$$ means the $$%63.2$$ of the total response of the sistem (or 0.632), but what about taking the full response of the system, i.e. $$\tau=1$$ and from the model of the system $$ C=\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)\tau$$ or $$ \tau=\frac{CR_1R_2}{R_1+R_2}$$ And $$ C=\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)\tau$$; $$ C=\bigg(\frac{10000+2250}{10000*2250}\bigg)1=544 μF$$ $\endgroup$
    – riccs_0x
    Oct 5 '16 at 20:19
  • $\begingroup$ @riccs_0x I am not entirely sure if I am following you, you could indeed define a characteristic time constant $\tau$ for this system. But you can't just say that this has to be equal to one, because not enough information was provided. $\endgroup$
    – fibonatic
    Oct 5 '16 at 20:32
  • $\begingroup$ Im using the definition of $$\tau$$ time constant, I mean the state of the system (0 to 100 or 0 to 1) and when $$t\rightarrow\infty$$ its supposed to be steady state so the value of $$\tau$$ its understanded, rather than assigned, to 1. Otherwise I dont think this can be solved, as you say there is no information on $$C$$ or $$V_C$$, well I say so...=) $\endgroup$
    – riccs_0x
    Oct 5 '16 at 20:45

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