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)

  • \$\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

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: $$ \large I_c(t) = \frac{U}{R}e^{\frac{-t}{RC}} + I_a $$

  • 1
    \$\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
  • 1
    \$\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

Use the below schematic to do some attempts. I have created in case if you are not aware of tool!


simulate this circuit – Schematic created using CircuitLab


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.