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I have this circuit which looks like a current divider but with an extra capacitor. I want to know how Ia , Ib, Ic change as the capacitor begin to charge, i do know one thing after a time of 5 RC the capacitor will be about 99.5% and the first branch ( the one with the capacitor ) will be considered open, and we can consider the whole branch not there. enter image description here

( the source is a current source)

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  • \$\begingroup\$ use circuit lab to evaluate. \$\endgroup\$ – JigarGandhi Feb 3 '15 at 19:11
  • \$\begingroup\$ How are you exciting the circuit? Do you apply a DC voltage, a DC current or something else to the + and - terminals? \$\endgroup\$ – mkeith Feb 3 '15 at 19:53
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Well \$I_c\$ will always be the sum of \$I_a\$ and \$I_b\$.

\$I_c = I_a + I_b \$

Assuming that the current supplied to the circuit is constant:

\$I_c\$ will be constant.

At the first instant, the capacitor is completely empty and acts like a short circuit. So the current is distributed like in a normal current divider. After the capacitor is fully charged, it acts as an open circuit and no current will flow through path B.

Now what happens between those two points: The voltage across both paths must be the same (parallel circuit):

\$U(t) = I_a(t) * R_a\$

\$U(t) = I_b(t) * R_b + U_c(t)\$

\$U_c\$ is the voltage across the capacitor.

with that we end up with:

$$I_a(t) * R_a = I_b(t) * R_b + U_c(t)$$

Substituting the first statement this changes into:

$$(I_c(t)-I_b(t)) * R_a = I_b(t) * R_b + U_c(t)$$

Now we keep two things in mind: \$U_c(t) = \frac{Q_b(t)}{C}\$ and \$I(t)=\dot Q(t)\$ using this we end up with this lovely differential equation:

$$\frac{Q_b(t)}{C*(R_a+R_b)} + \dot Q_b(t) = I_c * R_a$$

Solved to (hopefully):

$$\large I_b(t) = I_c*\frac{R_a}{R_a+R_b}*e^{-\frac{t}{C*(R_a+R_b)}}$$

As for \$I_a(t)\$ well it has to add to \$I_b(t)\$ so that the result is constant.

$$\large I_a(t) = I_c (1-\frac{R_a}{R_a+R_b}*e^{-\frac{t}{C*(R_a+R_b)}})$$

Credits go mainly to this German Wikipedia article as I'm quite rusty with this kind of stuff.


Assuming that the voltage supplied to the circuit is constant:

\$I_a\$ will remain constant as it's just a simple resistor.

\$I_b\$ will behave like the charging current of a RC circuit. That is: $$ \large I_b(t) = \frac{U}{R} e^{\frac{-t}{RC}} $$ So together: $$ \large I_c(t) = \frac{U}{R}e^{\frac{-t}{RC}} + I_a $$

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    \$\begingroup\$ Ia will remain constant if a constant voltage is attached across +/- terminals. Don't know if that is a valid assumption or not. \$\endgroup\$ – mkeith Feb 3 '15 at 19:37
  • \$\begingroup\$ @mkeith that is a valid point, and I've added the assumption to my post. If an arbitrary voltage is applied, things get complicated. \$\endgroup\$ – Arsenal Feb 3 '15 at 19:44
  • \$\begingroup\$ I was thinking that a DC current might be applied since the OP used the term "current divider." In that case, as Ib decayed exponentially, Ib would have to increase by the same amount. \$\endgroup\$ – mkeith Feb 3 '15 at 19:51
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    \$\begingroup\$ Oh, so a current source as input. Yeah that's another interesting suggestion. Let's wait for the response from @user28324. \$\endgroup\$ – Arsenal Feb 3 '15 at 19:55
  • \$\begingroup\$ sorry i didnt mention that, but this is indeed a dc current source, read the question wrong. sorry again \$\endgroup\$ – user28324 Feb 3 '15 at 20:06
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Use the below schematic to do some attempts. I have created in case if you are not aware of tool!

schematic

simulate this circuit – Schematic created using CircuitLab

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