# Practical Physics: Discharging a capacitor

1. Jan 22, 2017

### Taniaz

1. The problem statement, all variables and given/known data
"We will concentrate on the time taken for the capacitor to discharge to exactly one-half of its p.d, and how this varies with the value of the resistor. So repeat the experiment with different recorded values of R, and find the corresponding T, the time taken for V to fall from 9 V to 4.5 V (for example). Plot T versus R.

Theory suggests that T versus R should be a straight line through the origin with slope C ln 2. Confirm this."

2. Relevant equations
C= Q/V and tau = RC

3. The attempt at a solution
We were using resistors of 4.7 kilo ohms and we we kept adding one more resistor in series every time. After charging the 100 microFarads capacitor we connected it to the resistors and started the stopwatch in order to record the time for the voltage across it to fall to half it's value.

After we did this with about 5-6 different resistances and plotted our data, T vs. R, it wasn't a straight line through the origin. Why is that the case? We did the same experiment 3 times to confirm but each time our results didn't give us a straight line through the origin.

One more difficulty we faced was the potential difference didn't fall to exactly half its p.d for all the resistances.

2. Jan 22, 2017

### cnh1995

I don't understand. At what point did you stop the timer then?

3. Jan 22, 2017

### Staff: Mentor

Can you think of any other resistances that might have been involved in the circuit for each test? Hint: think of all the characteristics of everything that's connected. What sort of time values were you timing by hand? What was the smallest one?

Can you elaborate? What value did it fall to instead? Over how much time? For what resistance values?

4. Jan 22, 2017

### Taniaz

The starting voltage was 5.5 V, some of them fell to 2.75, some fell to 2.6, some to 2.7 or 2.8 and others were near 3.0 or 2.3. I think it was only for one of them ( 9.4 kilo ohms, the lowest resistance we used)

Last edited: Jan 22, 2017
5. Jan 22, 2017

### cnh1995

Aren't you supposed to stop the timer only when the voltage falls to 2.75V?

Did you use a digital multimeter to measure voltage during this discharge?

Or did you first calculate the time using formulae and discharged the capacitor for that time to see if the voltage falls to half of the initial value?

6. Jan 22, 2017

### Taniaz

We were using a digital multimeter to measure the voltage and yes we did stop it when it reached half the voltage.

7. Jan 22, 2017

### Staff: Mentor

The capacitor may have some small internal resistance, but probably not significant compared to the other resistances in the circuit. The wires, too, will have insignificant resistance. What else is involved? Hint: how are you determining the voltage?

Small stopwatch timings of less than a second are bound to be inaccurate due to human reaction times, which will bias the values upwards. What might that do to your curve for small resistor values?

I'm not sure how to reconcile this statement with your claim that the voltages reached different values.

8. Jan 22, 2017

### cnh1995

Then how did you get different voltages?
Plus, digital multimeter doesn't show continuous readings. It will show discrete values, so it may miss the the exact 2.75V point. That will depend on the circuit time constant. An analog voltmeter would be better IMO. You can see a gradual change in voltage and stop at exactly 2.75V.

9. Jan 22, 2017

### Taniaz

We stopped the stopwatch when the reading on the multimeter reached close to half of 5.5. We actually made a video as the voltage dropped and we stopped the stopwatch when it reached a voltage of 2.0 V.

Multimeter but isn't the resistance high so a very small current would go through?

But we were using such high resistances Our range of resistances was from 4.7 kilo ohms - 28.2 kilo ohms

10. Jan 22, 2017

### cnh1995

How close? Your error will depend on that difference.
You mean 2.75V? That's the half of 5.5V.

11. Jan 22, 2017

### Staff: Mentor

That's the usual assumption. Still, it's best to make sure. What is the Ohms-per-Volt rating of the multimeter you used and what voltage range setting was used? Perhaps you just have the make and model number of the meter and this can be looked up online.

12. Jan 22, 2017

### Staff: Mentor

Was the idea to use the video footage to estimate the actual time that the voltage crossed the halfway point?

13. Jan 22, 2017

### Taniaz

I think I might have figured out the problem, it mentions that the product of R and C should not be less than 60 and not greater than 600.

The capacitor we have available is 220 microFarads and the only resistors we have are 4.7 kilo ohms, so even if we place 10 resistors in series, our product is really small. That's the only thing I can think of.

Yes we used the video to reduce errors.

14. Jan 22, 2017

### cnh1995

Resistance of the multimeter would make a difference.
Plus, the digital multimeter is not fast enough to display half of the initial voltage in such a small time. For 28k resistance, this time should be 0.25s which is very small.
Yes. For 28k, the time constant is 2.8s, way smaller than 60.
I'd suggest you use a properly calibrated analog voltmeter so that you can see the voltage changing gradually and you can stop the timer exactly at the desired value.

15. Jan 22, 2017

### Staff: Mentor

Um, 220 μF x 28 kΩ = 6.2 s. Still a long way from 60 seconds.

@Taniaz: Did you have multiple capacitors available or just the one?