Determine what's in a container

[displayname]

Homework Statement

You are given a container with either a battery and a resistor or a capacitor inside it. There is also a positive and negative terminal sticking out of the container.

Without looking inside, how will you determine what is inside the container? Materials include a voltmeter, a voltage vs. time probe, wires and alligator clips, resistors, and batteries.

If it is a batter and resistor, how will you determine the voltage of the battery and the resistance of the resistor? If it is a capacitor how will you determine the capacitance (hint: graph voltage vs. time)?

Homework Equations

V=IR
V = Vo(e^(-t/RC))

The Attempt at a Solution

Find if battery of capacitor:
- attach a voltmeter in parallel to figure out initial voltage
- If there is 0 voltage, you have a capacitor.
- If voltage is not 0, attach a resistor for a second
- Remove resistor then measure the voltage again. If the voltage dropped significantly to 0 or close to 0, you have discharged a capacitor. Otherwise, if the voltage stayed about the same, you have a battery.

In order to confirm you have a capacitor, attach a battery to the circuit. Disconnect battery. See if voltage is not 0.

If you have a battery:
- find voltage by adding voltmeter in parallel
- find resistance by?
If you have a capacitor:
- add a resistor in series (note resistance)
- use the voltage over time probe and graph voltage over time
- get two voltage points and plug into the voltage of capacitor equation to solve for capacitance

Is this method right? How do i find resistance of a resistor? maybe use V = IR, but we don't know I

 

[SammyS]

How will you charge the capacitor if that's what you have?

 

[displayname]

How will you charge the capacitor if that's what you have?

couldnt you just attach the two ends to the terminals of a battery and then check the voltage to see if the capacitor charged

 

[SammyS]

couldn't you just attach the two ends to the terminals of a battery and then check the voltage to see if the capacitor charged

Yes.

Now, if it's a battery and a resistor, I assume the resistor will be in series with the battery. Connect a known resistance across the terminals of the container. Measure the voltage drop across the resistor. Use Ohm's law to find the current. Do this for a few values of resistance.

 

[displayname]

Connect a known resistance across the terminals of the container. Measure the voltage drop across the resistor. Use Ohm's law to find the current. Do this for a few values of resistance.

Ok. This is my understanding:

So by attaching the voltmeter to the terminals of the container, we find the voltage of the battery (V). Now, we attach a resistor (with resistance r) to the terminals (so the resistor is in series). Let's say the resistance of the resistor attached to the batter is R.

V = I(R+r) -- two unknowns, I and R

Measuring the voltage drop between the known resistance results in solving for I.

Plug in I into the V=I(R+r) and solve for R.

So really we only need to try one resistor but can try multiple resistors to verify answer.

 

[SammyS]

Ok. This is my understanding:

So by attaching the voltmeter to the terminals of the container, we find the voltage of the battery (V). Now, we attach a resistor (with resistance r) to the terminals (so the resistor is in series). Let's say the resistance of the resistor attached to the batter is R.

V = I(R+r) -- two unknowns, I and R

Measuring the voltage drop between the known resistance results in solving for I.

Plug in I into the V=I(R+r) and solve for R.

So really we only need to try one resistor but can try multiple resistors to verify answer.

Yes. Also, a particular resistor, r, may not give a very good result for R. If r has a much larger resistance than R, then voltage drop, V_r that you measure across r will be nearly as large as V, the voltage of the battery. Using too small a value for r can also cause problems.

For a capacitor you may also need to use various resistors. Definitely discharge more than once & compare results.