# Current divider and capacitor

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.

( the source is a current source)

• use circuit lab to evaluate. – JigarGandhi Feb 3 '15 at 19:11
• How are you exciting the circuit? Do you apply a DC voltage, a DC current or something else to the + and - terminals? – 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: $$\large I_c(t) = \frac{U}{R}e^{\frac{-t}{RC}} + I_a$$

• Ia will remain constant if a constant voltage is attached across +/- terminals. Don't know if that is a valid assumption or not. – mkeith Feb 3 '15 at 19:37
• @mkeith that is a valid point, and I've added the assumption to my post. If an arbitrary voltage is applied, things get complicated. – Arsenal Feb 3 '15 at 19:44
• 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. – mkeith Feb 3 '15 at 19:51
• Oh, so a current source as input. Yeah that's another interesting suggestion. Let's wait for the response from @user28324. – Arsenal Feb 3 '15 at 19:55
• sorry i didnt mention that, but this is indeed a dc current source, read the question wrong. sorry again – 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