Engineering DC circuit analysis: conceptual mistake?

[Granger]

Homework Statement

Homework Equations

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I have a circuit that consists on 3 components in parallel: a current source Ia, a resistor R and a capacitor C. Ia is directed upwards, while the currents through the capacitor and the resistor are both directed downwards.
Select the correct option in the following three:
(a) The equation $$\frac{i_a(t)}{C}=\frac{v_2(t)}{RC} + \frac{dv_2(t)}{dt}$$ is valid to this circuit.
(b) The equation $$\frac{Ri_a(t)}{C}= \frac{dv_2(t)}{dt}$$ is valid to this circuit
(c) If the circuit is working for a long time, after dome time we will have $$i_a(t)=i_1(t)$$ and $$i_2(t)=0$$ whatever is the value of $$i_a(t)$$.

The Attempt at a Solution

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So I had no problem concluding that (a) is correct and (b) is not correct.

First, I applied KCL to the upper node

$$i_a=i1+i2$$

We have that:
$$i1=\frac{u1}{R}$$
$$i2=C\frac{du2}{dt}$$

Substituting:

$$i_a=\frac{u1}{R}+C\frac{du2}{dt}$$

Because the components are in parallel: u1=u2. Dividing everything by C we get to the expected expression.

My doubt in the c statement. I don't understand why is it wrong. I thought well if the circuit is working for a long time then we can admit that it starts working in steady state (or is this an incorrect statement only valid if $$i_a(t)$$ remains constant?). Because of this, the voltage across the capacitor will remain constant and because of the expression $$i2=C\frac{du2}{dt}$$ we will have i2=0 (the derivative of a constant is zero).
Because of that we can exchange the capacitor to an open circuit. Therefore all the current Ia(t) will go through the resistor. Or does this depend on Ia(t)?
I'm probably making a conceptual mistake somewhere in here.

Thanks!

 

[cnh1995]

or is this an incorrect statement only valid if

ia(t)ia(t)​

i_a(t) remains constant?

Yes. Since they didn't mention whether it's a constant current source or not in the question, you can't go for option c. It would be correct only if Ia is constant.