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I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The capacitor is charged up to 10 V according to the following formula (when the switch is closed):

$$\text{V}_\text{C}\left(t\right)=\hat{\text{u}}\cdot\left(1-\exp\left(-\frac{t}{\text{CR}_1}\right)\right)\tag1$$

Now, when I open the switch the capacitor will discharge its energy in the resistors that are standing in series with the capacitor.

To find the current that runs in the circuit when I open the switch is:

$$\text{V}_\text{C}\left(t\right)+\text{V}_{\text{R}_1}\left(t\right)+\text{V}_{\text{R}_2}\left(t\right)=0\tag2$$

Writing that in terms of the the current (using Ohm's law and \$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\text{V}'_\text{C}\left(t\right)\$) in the circuit gives:

$$\text{I}_\text{C}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_{\text{R}_1}'\left(t\right)\cdot\text{R}_2+\text{I}_{\text{R}_2}'\left(t\right)\cdot\text{R}_2=0\tag3$$

Now, if I define a new time \$t=0\$ when I open the switch, the initial current in the capacitor is equal to \$0\$ amps. So \$\text{I}_\text{C}\left(0\right)=0\$. Now it is a series circuit (when the switch is open) so I can rewrite:

$$\text{I}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}'\left(t\right)\cdot\text{R}_2+\text{I}'\left(t\right)\cdot\text{R}_2=0\tag4$$

Now, when I use the initial condition (to solve the DE given in equation \$(4)\$), I get the current in the circuit (after the switch is opened) equals \$0\$, but that is not possible. What is mistake here?

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  • \$\begingroup\$ Why wouldn't the instantaneous current simply be V / (R1 + R2) ? \$\endgroup\$ – evildemonic May 17 '19 at 15:06
  • \$\begingroup\$ @evildemonic Well, when I press the switch the instantaneous current is given by: $$\text{I}_{\text{in}}\left(0\right)=10\cdot\left(\frac{1}{\text{R}_1}\cdot\exp\left(-\frac{0}{\text{CR}_1}\right)+\frac{1}{\text{R}_2}\right)=10\cdot\left(\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}\right)$$ But when I redefine time (to \$t=0\$) when I open the switch the initial current in the capacitor is zero amps. And my DE will give that the current in the circuit is equal to zero for all time \$t\ge0\$ \$\endgroup\$ – yuiop May 17 '19 at 15:09
  • \$\begingroup\$ I'm sorry, I thought you were asking about when you open the switch after the capacitor is charged to 10 V. \$\endgroup\$ – evildemonic May 17 '19 at 15:10
  • \$\begingroup\$ @evildemonic, it is okay :) thank for helping anyway. \$\endgroup\$ – yuiop May 17 '19 at 15:11
  • \$\begingroup\$ The time constant, when the switch is opened, is: \$\small \tau =C(R1+R2)\$. The resistors are in series. \$\endgroup\$ – Chu May 17 '19 at 15:46
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Capacitor is charged up to 10 V when the switch is closed.

When the switch opens, the circuit consists of one capacitor (the source, with initial voltage U = 10V) and two resistors (R1 and R2) connected in series.

So

\$V_C + V_{R1} + V_{R2} = 0 \$

Since the current is the same

\$I_C = C\frac{dV_C}{dt} = I_{R1} = I_{R2} = I \$

With this info, it's easy to continue the analysis.


EDIT

You can prove that after the switch opens (t=0) capacitor's voltage is

\$V_C(t) = V_C(0) \cdot e^{-\frac{t}{(R_1+R_2)C}} \$

and from this find current' s equation

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  • \$\begingroup\$ I do not see how this solves the problem, can you help me a bit further \$\endgroup\$ – yuiop May 17 '19 at 16:30
  • \$\begingroup\$ @yuiop see the edit \$\endgroup\$ – thece May 17 '19 at 17:10
  • \$\begingroup\$ also, keep in mind that current's flow will change direction after the switch opens \$\endgroup\$ – thece May 17 '19 at 17:29
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Now, if I define a new time t=0 when I open the switch, the initial current in the capacitor is equal to 0 amps. So IC (0)= 0.

This is not true, capacitor voltage cannot change instantaneously, but current can. The initial condition is actually Ic(0) = Vc(0) / (R1 + R2).

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