Practical Physics: Discharging a capacitor

In summary: 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?
  • #1
Taniaz
364
1

Homework Statement


"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."

Homework Equations


C= Q/V and tau = RC

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.
 
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  • #2
Taniaz said:
One more difficulty we faced was the potential difference didn't fall to exactly half its p.d for all the resistances.
I don't understand. At what point did you stop the timer then?
 
  • #3
Taniaz said:
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.
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?

Taniaz said:
One more difficulty we faced was the potential difference didn't fall to exactly half its p.d for all the resistances.
Can you elaborate? What value did it fall to instead? Over how much time? For what resistance values?
 
  • #4
gneill said:
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?

Resistance of the wires or the capacitor itself? The circuit only consisted of the capacitor, wires, resistors and a capacitor.
0.38 s - 2.58 s (the time kept increasing as the resistance went up).

Can you elaborate? What value did it fall to instead? Over how much time? For what resistance values?
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)
 
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  • #5
Taniaz said:
The starting voltage was 5.5 V, some of them fell to 2.5, 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)
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
cnh1995 said:
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?

We were using a digital multimeter to measure the voltage and yes we did stop it when it reached half the voltage.
 
  • #7
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?

Taniaz said:
We were using a digital multimeter to measure the voltage and yes we did stop it when it reached half the voltage.
I'm not sure how to reconcile this statement with your claim that the voltages reached different values.
 
  • #8
Taniaz said:
We were using a digital multimeter to measure the voltage and yes we did stop it when it reached half the voltage.
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
gneill said:
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.

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
Taniaz said:
We stopped the stopwatch when the reading on the multimeter reached close to half of 5.5
How close? Your error will depend on that difference.
Taniaz said:
and we stopped the stopwatch when it reached a voltage of 2.0 V.
You mean 2.75V? That's the half of 5.5V.
 
  • #11
Taniaz said:
Multimeter but isn't the resistance high so a very small current would go through?
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
Taniaz said:
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.
Was the idea to use the video footage to estimate the actual time that the voltage crossed the halfway point?
 
  • #13
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
Taniaz said:
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
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.
Taniaz said:
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.
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
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?
 

Related to Practical Physics: Discharging a capacitor

1. How does a capacitor discharge?

A capacitor discharges by releasing the stored electrical energy in the form of an electric current. This occurs when the capacitor is connected to a circuit and the charges on the two plates of the capacitor flow towards each other, neutralizing each other.

2. What factors affect the rate of discharge of a capacitor?

The rate of discharge of a capacitor is affected by the capacitance of the capacitor, the resistance of the circuit it is connected to, and the voltage across the capacitor. A higher capacitance, lower resistance, and higher voltage will result in a faster discharge.

3. How can I measure the discharge rate of a capacitor?

The discharge rate of a capacitor can be measured by using a multimeter to measure the voltage across the capacitor over time. The rate of change of the voltage will indicate the rate of discharge.

4. What is the purpose of discharging a capacitor?

Discharging a capacitor is important for safety reasons, as a charged capacitor can release a dangerous amount of current. It is also necessary to fully discharge a capacitor in order to use it in a new circuit or to test its capacitance.

5. Can a capacitor be discharged without a circuit?

Yes, a capacitor can be discharged without a circuit by connecting the two plates of the capacitor together with a wire. This will allow the charges to flow towards each other and neutralize, resulting in a discharge.

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