Question: I am really unsure about how to go about this question and which circuits to consider.

Here's the question: 

Determine the value of the current labeled \$i\$ and the voltage labeled \$v\$ at \$t=0+\$, \$t=1.5ms\$, and \$t = 3.0ms\$.

The circuit is as follows: 

![schematic](https://i.stack.imgur.com/XrbPa.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fXrbPa.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

Relevant equations:

Ohm's law: \$v = iR\$

Decay in source free RC Circuit \$v(t) = v(0) e^{-t/RC}\$

The attempt at a solution:

\$i(0-) = 0.1 A.\$

Therefore, 

\$v\$ across 200 ohm: \$200 \* 0.1 = 20 V\$

Since no current flows through capacitor with constant dc current,

\$v\$ across capacitor: \$20 V @ t < 0\$

\$v\$ across capacitor cannot change immediately as it would require infinite power so \$v\$ across capacitor @ \$t = 0\$ is \$20V\$

Here's where the confusion is: Should I include the 200 ohm resistor in my RC circuit? Why or why not? Also, will my circuit be a source free RC circuit?

Also, what will be the current i through the 200 ohm resistor when t>0?

My last question is a theoretical one: Can a current through a resistor change immediately at one instant of time?

Answer: 
The voltage across a capacitor discharging into a fixed resistance decays exponentially. The time constant is RC, where C is the capacitance, and R is the resistance between the terminals of the resistor.

$$
V(t)=V(0)e^{\frac{-t}{RC}} \\
$$

You're right that the capacitor starts at 20V, because that's the voltage across the 200 ohm resistor after the capacitor is charged. So you know V(0), and you know C. All you're missing is R. To analyze that properly, think of the switch as a resistor of 0 ohms when closed, and don't worry about the current source. The parallel combination of a 200 ohm resistor and a 0 ohm resistor is 0 ohms. The series combination of a 50 ohm resistor and a 0 ohm resistor is 50 ohms. So the 50 ohm resistor is the only one that matters when determining your discharge constant.

The current through the 200 ohm resistor depends on the voltage across the 200 ohm resistor. The voltage across the 200 ohm resistor is the same as the voltage across the 0 ohm resistor (the closed switch). V=IR, so what's the voltage across that pair of resistors when the switch is closed?

And yes, the voltage across an ideal resistor can change instantaneously. Keep in mind though, there's no such thing as an ideal resistor in the physical world. Everything has capacitance to everything else.
The original post mentions current 'i' etc, but the current 'i' etc is not shown in the original circuit diagram. Also, the question is not clearly defined because for these kinds of problems, it's necessary to state the position of the switch (open or closed) at time t < 0. If assuming the switch is open at t < 0, then the initial voltage of the capacitor Vc(0-) is 20V, which also equals Vc(0), which also equals Vc(0+) because the capacitor voltage is continuous. Assuming the switch closes at t = 0, you end up with a capacitor in parallel with the 50 Ohm resistor, so the time constant RC will be based on C and the 50 Ohm resistor, and you get the capacitor voltage equation as being: Vc(t) = Vc(0)e^(t/50.C), where Vc(0) is 20V. The current ic(0) will be zero because voltage can't change instantaneously when the switch closes at t = 0. So the voltage remains constant at t = 0, in which case the dvc(t)/dt = 0 at t = 0, so zero capacitor current at t = 0. But at t = 0+ (a touch greater than t=0), ic(t) = CdVc(t)/dt = So knowing the Vc(t) equation will yield the capacitor current equation ic(t) for t >= 0+. And the reference direction for the current is in the same direction as the voltage drop across the resistor.....in other words through the capacitor and 'downwards' (starting from the + terminal of the capacitor and ending at the - terminal of the capacitor).

At t < 0, if the switch is open, then the 200 Ohm resistor and the current source simply produce the 20V at the upper node, which drives the initial capacitor voltage to 20V. Once the switch closes at t = 0, the 200 Ohm resistor disappears because it gets short circuited (and taken out of the picture) by the closed switch. It all depends on what the switch is doing at t < 0 and t = 0, which is why you need to properly define the conditions of the problem/question.