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: 
When solving problems like this, it is helpful to draw the two *different* circuits that exist before and after \$t=0\$.

Before, with the switch open, since the circuit is in DC steady state, replace the capacitor with an open circuit and then it is easy to solve for \$v\_C(0)\$, the initial condition.

After, with the switch closed, the current source and 200 \$\Omega \$ resistor are paralleled with a short circuit and so, from the perspective of the capacitor, can be *ignored* (a short circuit in parallel with any other circuit elements is equivalent to a short circuit).

So, the circuit after \$t=0\$ is just the capacitor and the 50 \$\Omega \$ resistor, a simple RC circuit with initial condition \$v\_C(0)\$.

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

With a short circuit in parallel, the voltage across the resistor is ?V and, using Ohm's law, the current is ?A.

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

Theoretically, within ideal circuit theory, the answer is yes. In reality, the answer is no. for example, physical resistors have associated "parasitic" capacitance and inductance.
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.