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Am I doing something wrong calculating the current and voltages at time t for the RC circuit? I put these equations on google sheets and the voltage across all of the elements passes 10V between 4 and 5 seconds. Then it goes all the way to 35V around 20 seconds!

Super thanks

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  • \$\begingroup\$ It may be worth thinking about how it's possible that a 35V potential could measured in this 10V circuit. \$\endgroup\$ – Shadetheartist Oct 17 '19 at 2:30
  • \$\begingroup\$ Observation tells you the current cannot exceed 10V/1k and the voltage on it ends up as 10/11 *10V with a 100R*10us=10ms time constant locally not 10 seconds. Rip up excel and redo. with 10V across 1k just after invisible switch closes after t=0 \$\endgroup\$ – Tony Stewart EE75 Oct 17 '19 at 4:22
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There is a Thevenin equivalent that might help, somewhat. See here:

schematic

simulate this circuit – Schematic created using CircuitLab

The above, right is a very simple RC pair and it should be pretty easy for you to develop the equation for voltage and current with respect to the capacitor. If so, then at least you have your answer for node \$V\$. (Not the rest, but you are a lot closer to getting them, as well, too, as you can trivially develop anything else you need from there.)


The other approach is to just use nodal analysis:

$$\begin{align*} \frac{V_t}{R_1}+\frac{V_t}{R_2}+C_1\frac{\text{d}V_t}{\text{d} t}&=\frac{+10\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}+C_1\frac{\text{d}\left(+10 \:\text{V}\right)}{\text{d} t}\\\\ \frac{V_t}{R_1}+\frac{V_t}{R_2}+C_1\frac{\text{d}V_t}{\text{d} t}&=\frac{10\:\text{V}}{R_1}\\\\ \frac{V_t}{C_1\,R_1}+\frac{V_t}{C_1\,R_2}+\frac{\text{d}V_t}{\text{d} t}&=\frac{10\:\text{V}}{C_1\,R_1}\\\\ \frac{\text{d}V_t}{\text{d} t}+\left(\frac{1}{C_1\,R_1}+\frac{1}{C_1\,R_2}\right)V_t&=\frac{10\:\text{V}}{C_1\,R_1} \end{align*}$$

That's just in standard form for the usual 1st order linear differential equation, which looks like this:

$$V_t^{'}+P_t\:V_t=Q_t$$

For convenience, set \$\tau_{_1}=R_1\cdot C_1\$, \$\tau_{_2}=R_2\cdot C_1\$, and \$\tau_{_0}=\frac{1}{\frac{1}{\tau_{_1}}+\frac{1}{\tau_{_2}}}=C_1\cdot\left(R_1\mid\mid R_2\right)\$.

If you want to see an exposition of how to use an integrating factor, \$\mu_{_t}\$, to solve the above then just look here, for example. There's plenty of explanation there. Here, set \$P_t=\frac{1}{\tau_{_0}}\$ and \$Q_t=10\:\text{V}\cdot \frac{1}{\tau_{_1}}\$ and find that \$\mu_{_t}=e^{^{\int P_t\:\text{d} t}}=e^{^\frac{t}{\tau_{_0}}}\$. \$Q_t\$ is a constant, so: \$V_t=\frac{1}{\mu_t} \int \mu_{_t}\:Q_t\:\text{d}t=\frac{Q_t}{\mu_t} \int \mu_{_t}\:\text{d}t\$. You should be able to solve from there, I think. You know the initial condition is \$V_0=10\:\text{V}\$ because the assumed voltage across \$C_1\$ is \$0\:\text{V}\$ (unless you know a different assumption to use there.)


Either approach gets you to the same place.

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  • \$\begingroup\$ Thanks a lot -- I love how your math equations look and I appreciate that you composed such a high effort answer. I'm familiar with CircuitLab; Awesome schematic. Thevenin's equivalent circuit worked out perfectly for this exercise. \$\endgroup\$ – Renzo M-Svartz Oct 17 '19 at 23:23
  • \$\begingroup\$ @RenzoM-Svartz Thanks very much for your kind words. I'm glad the Thevenin equivalent helped out. If the answer is what you needed and you are done, you may as well select the answer to close this and save the time of others. (I don't care at all about the points, so don't do it if you also don't care to bother. I'm just mentioning this because others may pop in and look over this question if it isn't "answered" and that, unfortunately, winds up unnecessarily wasting the time of others.) \$\endgroup\$ – jonk Oct 18 '19 at 0:07

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