Well, I missed something pretty obvious.

When T=0, the cap is completely discharged and current begins flowing through the 120 Ohm resistor, 5v/120 Ohms= 41.666...mA. That's the highest current flow that you will obtain.

As the voltage on the capacitor increases, the current through the resistor decreases.
If V(C1) = 1, then current through R1 is (5v-1v)/120 = 4v/120 = 33.333...mA.
By the time V(C1) = 3.5v, R1 current is (5v-3.5)/120 = 1.5/120 = 12.5mA.

You acknowledged the same thing when you said it takes 72 seconds to charge the cap.

So, what you need is some kind of boost circuit that keeps the voltage across R1 as high as possible for maximum current flow.

If he doesn't have access to the 9V battery or the node between the regulator and the resistor, he can't decrease the charge time. I fiddled around with a simulation where I added a series inductor, and was able to shorten the time to 3.5V by a hundred milliseconds or so, but the inductance was something like 800 Henries. Good luck on that.
I don't understand why he doesn't have access to the 9V battery, unless this is a homework problem. I asked if it was, but got no reply.

 

With 5V and 120ohms, the maximum power available to the load is when the load is 120ohm as well, and that is 52mW. 0.5F capacitor@3.5V will store about 3J. To get 3J with 52mW takes 3/0.052= 57 seconds. You can´t get any faster than that.

Of course to get this result you would need some ideal power converter that has 120ohm input impedance and transfers 100% of the power to charge the capacitor, and that is never gonna happen. I think you would need some complex switch-mode circuit to get close to that time.

I simulated the circuit with just the 120ohm resistor and the cap as posted, and the voltage reaches 3.5V at 72 seconds. I think that is close enough to perfection.

 

You acknowledged the same thing when you said it takes 72 seconds to charge the cap.

If he doesn't have access to the 9V battery or the node between the regulator and the resistor, he can't decrease the charge time. I fiddled around with a simulation where I added a series inductor, and was able to shorten the time to 3.5V by a hundred milliseconds or so, but the inductance was something like 800 Henries. Good luck on that.
I don't understand why he doesn't have access to the 9V battery, unless this is a homework problem. I asked if it was, but got no reply.

It's a lab design problem... The source is given and I can only access the two terminals shown in the diagram...

 

It's a lab design problem... The source is given and I can only access the two terminals shown in the diagram...

What defines the start of capacitor charging? Is it when you turn on a switch that powers up the regulator?

 

What defines the start of capacitor charging? Is it when you turn on a switch that powers up the regulator?

yes. also the capacitor has zero charge at the beginning.

 

If you are told that there is a way to speed up charging, or if you discover a way to do it, please post the answer here. We are all very curious.

 

The only thing I can think of in the way of an 'external' device that could increase the charge rate would be another source of current.

Since the OP said the current MUST come from the specified regulator, then the OP has a problem with no possible solution.

Translation: It can't be done the way you have set up the 'rules'. You are stuck with using the regulator and the regulator is stuck with a fixed resistor value. The cap is stuck with a fixed charging current which constantly decreases.

Not every problem can be solved, Especially when the circuit can't be 'modified'.

 

Wow !!!!!

 

I know! Let's take the approach of the overunity crowd! We'll run a boost switcher off the cap, and use it's output to augment the charging.

 

he he he he !