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Given the systemcircuit calculate the current in \$R_{1}\$, $$R_{1}=R_{2}=10k\Omega $$ and $$C=1.2\mu F$$. Well, I have two approaches.

  • When the capacitor is fully charged, it will open the circuit and there will be a voltage divisor

So $$V_{R_{1}}=20_V\frac{10k\Omega}{20k\Omega}=10_V$$
and then
$$I_{R{1}}=\frac{10_V}{10k\Omega}=1mA$$

  • Model the system, calculate the step response and you will have the capacitor voltage, so $$V_{C}=V_{R_2}$$ and then $$V_{R1}=V-V_{C}$$ System model: $$\frac{V(t)}{R_{1}C}=\dot{V_{C}}+\frac{1}{C}(\frac{R_1+R_2}{R_1R_2})V_{C}$$ the time constant:
    $$\tau=C(\frac{R_1R_2}{R_1+R_2})=0.006s$$ taking the Laplace transform with the input given and aplying the values of the components $$1666.6\frac{1}{s}=V_{C(s)}-V_{C(0)}+166.6V_{C(s)}$$ Grouping, calculating the partial fractions expansion and taking the inverse Laplace transform: $$\frac{1666.6}{s(s+166.6)}=V_{C(s)}$$ $$V_{C}=\frac{10.0036}{s}+\frac{-10.0036}{s+166.6}=10.0036-10.0036e^{-166.6t}$$ with $$t\geq0$$ and when $$\underset{t\rightarrow\infty}{\lim V_{C}}=10.0036V$$ then (again) $$V_{R1}=V-V_{C}$$ $$V_{R1}=20V-10.0036V=9.9964V$$ $$I_{R{1}}=\frac{9.9964_V}{10k\Omega}=0.00099964A$$ or $$I_{R{1}}\approx1mA$$

    I made a graphic of the response. OK, hope this can be reviewed and point any error or comment if this is done right! Thanks in advance. enter image description here

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    \$\begingroup\$ looks good. now repeat until you can do in your head in 10s. 63%Vin at ReqC towards I= V/Rtot \$\endgroup\$ – Sunnyskyguy EE75 Nov 11 '16 at 16:13
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    \$\begingroup\$ Just by simple inspection you can't get the voltage across the capacitor to more than half the supply voltage so there's a little calculating error in there (probably due to approximating values in the calculation). Initially the current in R1 will be 2mA ( capacitor acting as a short circuit ) falling exponentially to 1mA (almost) ( capacitor acting as an open circuit ) after about 5 time constants - assuming no initial charge on the capacitor at t=0. \$\endgroup\$ – JIm Dearden Nov 11 '16 at 16:22
  • \$\begingroup\$ Thanks to Tony and JIm; Im checking it; whoa! I didnt think this way, =) \$\endgroup\$ – riccs_0x Nov 11 '16 at 16:37
  • \$\begingroup\$ And.... consider if you changed your V,R1,R2 circuit to its Thevenin equivalent (10V, 5k) as it charged the capacitor. \$\endgroup\$ – JIm Dearden Nov 11 '16 at 16:59
  • \$\begingroup\$ It is another way; but this time uncle Thevenin doesn’t was allowed in the question. However Im doing this too. =) \$\endgroup\$ – riccs_0x Nov 11 '16 at 17:26
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My first instinct would be to first apply Thevenin to \$V\$, \$R_1\$, and \$R_2\$ in order to reduce this to a simple series RC problem, whose solution is:

$$\begin{align*} R_{th} &= \frac{R_1\cdot R_2}{R_1+R_2} = 5\:\textrm{k}\Omega\\ \\ V_{th} &= V\cdot\frac{R_2}{R_1+R_2} = \frac{1}{2}\:V \\ \\ \tau&= R_{th} \cdot C_1 = 6\:\textrm{ms}\\ \\ V_x=V_{R_1}&=V_{th}\cdot\left(1-e^{-\cfrac{t}{\tau}}\right)=\frac{1}{2}\:V\cdot\left(1-e^{-\cfrac{t}{\tau}}\right) \end{align*}$$

But that's not a general approach. It just works in this case.


Nodal analysis is way, way more general and would be, setting the bottom node arbitrarily to \$0\:\textrm{V}\$:

$$\begin{align*} \frac{V_x}{R_1} + \frac{V_x}{R_2} + C_1\frac{\textrm{d} V_x}{\textrm{d} t}&= \frac{V}{R_1} + \frac{0\:\textrm{V}}{R_2} \\ \\ V_x\cdot\left(\frac{1}{R_1} + \frac{1}{R_2}\right) + C_1\frac{\textrm{d} V_x}{\textrm{d} t}&= \frac{V}{R_1} \\ \\ \frac{\textrm{d} V_x}{\textrm{d} t}+\frac{1}{C_1\frac{R_1\cdot R_2}{R_1+R_2}}V_x&= \frac{V}{R_1 C_1} \\ \\ \frac{\textrm{d} V_x}{\textrm{d} t}+\frac{1}{C_1 R_{th}}V_x&= \frac{V_{th}}{C_1 R_{th}} \end{align*}$$

Which is in a very familiar first-order ODE form. You can apply Laplace to that.

But just using the usual ODE solution method for first-order, gives:

$$\begin{align*} P_t = \frac{1}{C_1 R_{th}},~~Q_t &= \frac{V_{th}}{C_1 R_{th}},~~\therefore~~ \frac{\textrm{d} V_x}{\textrm{d} t}+P_t V_x= Q_t\\ \\ \mu &= e^{\int P_t\:\textrm{d} t} = e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\\ \\ V_x &= \frac{1}{\mu}\int \mu\: Q_t~ \:\textrm{d} t \\ \\ &= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\int e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\: \frac{V_{th}}{C_1 R_{th}}~ \:\textrm{d} t \\ \\ &= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(\frac{V_{th}}{C_1 R_{th}}\right)\cdot \int e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\:~ \:\textrm{d} t \\ \\ &= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(\frac{V_{th}}{C_1 R_{th}}\right)\cdot \left(C_1 R_{th}\: e^{\left[\cfrac{t}{C_1 R_{th}}\right]}+C_0\right) \\ \\ &= V_{th}\:e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(e^{\left[\cfrac{t}{C_1 R_{th}}\right]}+C_0\right) \\ \\ &= V_{th}\left(1 + C_0\:e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\right) \end{align*}$$

Using the initial condition that \$V_x\left(t=0\right) = 0\:\textrm{V}\$, this results in \$C_0=-1\$, so:

$$\begin{align*} V_x &= V_{th}\left(1 -e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\right) = \frac{1}{2}\:V\cdot\left(1-e^{-\cfrac{t}{\tau}}\right) \end{align*}$$

And now you can see why the Thevenin approach would be chosen first.


Or use Laplace applied to the ODE mentioned earlier. Using the known initial condition, you get \$Y_s = \frac{Q_t}{s^2+P_t s}\$, which solves out as \$y_t=\frac{Q_t}{P_t}\left(1-e^{-P_t t}\right)=V_{th}\left(1-e^{\frac{-t}{\tau}}\right)=\frac{1}{2} V\left(1-e^{\frac{-t}{\tau}}\right)\$.


Of course, knowing \$V_x\$ it is trivial to then answer your title question regarding the current in \$R_1\$ over time.

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  • \$\begingroup\$ Thanks to you all, I ve learned today to think a bit different! I never though this can be so rich. \$\endgroup\$ – riccs_0x Nov 11 '16 at 21:24

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