Can you help me prepare me for my E&M lab practical?

[rickyman]

I have to perform a "lab practical" tomorrow for my E&M class and it involves using a multimeter to determine what is inside a "black box" - either a battery/resistor combo or a capacitor.

Homework Statement

Taken from the lab practical write-up:

"You’ll be given a box with two electrical connections to the outside
world. One lead is marked plus and the other minus. Do not hook the positive
side of a power supply to the negative terminal on the box. Your task is to
briefly describe and apply a method for finding out what is inside without opening the
boxes.

Tasks:
1. Your box will be one of the above two types (they give two pictures - one for battery/resistor combo and one for capacitor). Briefly describe the strategy you
will use to find out which type you have. If you pick the right measurements,
only one simple one is required.
2. Make measurements and identify your box type.
3. Once you’ve identified your box type, briefly describe the strategy you will use to
do the following:
o If you have the battery and resistor box, determine the battery voltage
and value of the resistance. You will have only one multimeter, and it
must be used to measure voltage only.
o If you have the capacitor box, determine the value of the capacitance.
This will require measuring voltage versus time with the computer. A
separate sheet containing additional information that you may need is on
the front table.
4. Determine values for the data set pertaining to your box type.
Equipment for each setup:
• One multimeter.
• Computer with voltage-versus-time probe.
• Miscellaneous resistors.
• Wires and alligator clips.
• Power supply.

Homework Equations

I know that if I determine that the box contains a capacitor I can find capacitance using RC=(t2-t1)/ln(V1/V2)

The Attempt at a Solution

I am confused about how to approach this but I know that I will need to use the voltage measurement I take to determine what is inside. I just don't know how I will interpret the voltage measurement. I believe that the voltage measurement of the box if it contains a capacitor will steadily increase?

Any help is greatly appreciated.

 

[lswtech]

I believe that if you connect the Battery-Resistor box with some dc supply, you will get a constant pd. only. On the other hand, an exponentially increasing potential differences, which attains a constant in the long run, will result if your box is the capacitor one.