Current Increase When Connecting Capacitor to Resistor

In summary, the conversation discusses an experiment involving a resistor and a capacitor connected to a direct current source. The initial current reading is 0.9 mA, but it increases to more than 1 mA when the capacitor is added. The experiment procedure suggests reducing the resistance, but the current only reaches 1 mA when the resistance is increased. The participants discuss possible reasons for this, including the capacitance of the capacitor and the accuracy of the instruments being used. The conversation also includes a request for a diagram of the circuit and clarification on the values and
  • #1
ngkamsengpeter
195
0
I have conducted an experiment in school I have some problem with it . I connect a resistor 6600 ohm to a 6V direct current and I read the miliammeter and it show about 0.9 mA . But after I connect a capacitor 3300 F parallel to the resistor , the miliammeter increase to more than 1 mA.Why the current increase when I connect the capacitor parallel to resistor . Furthermore, the experiment procedure told me to reduce the resistance to get 1 mA but I have to increse the resistance in order to get 1 mA. What wrong with my experiment or the experiment procedure is wrong .
 
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  • #2
Initially the capacitor will be a very low resistance parallel path, so you would expect see a higher total current. Within a few seconds, depending on the cap, the current should drop to that passing through the resistor. You will need to reduce the resistance to get a steady state current of 1mA, a 6k Ohm resistor should give a steady state of 1 mA. Your transient current will be higher.
 
  • #3
Integral said:
Initially the capacitor will be a very low resistance parallel path, so you would expect see a higher total current. Within a few seconds, depending on the cap, the current should drop to that passing through the resistor. You will need to reduce the resistance to get a steady state current of 1mA, a 6k Ohm resistor should give a steady state of 1 mA. Your transient current will be higher.
I have wait more than 5 minutes but the miliammeter still show more than 1 mA .
 
  • #4
Measure every few seconds.. (maybe 30sec) see if you can see the change, it should be dropping. This is assuming that you indeed have it wired as you say. And as I said it also depends on the capacitance.
 
  • #5
Integral said:
Measure every few seconds.. (maybe 30sec) see if you can see the change, it should be dropping. This is assuming that you indeed have it wired as you say. And as I said it also depends on the capacitance.
I measure every few seconds for more than 5 minutes but the reading doesn't change at all . I connected another miliammeter series to the capacitor wait until this miliammeter reach zero but the miliammeter connected series with resistor still doesn't change at all , still more than 1 mA.What is the problem?
 
  • #6
Can you post a diagram of your circiut?
 
  • #7
Integral said:
Can you post a diagram of your circiut?
I have attach the diagram of the circuit please help me .
 

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  • #8
Where in the circuit are the ammeters?
 
  • #9
Integral said:
Where in the circuit are the ammeters?
I am sorry I forgot t draw the miliammeter . I am now attaching the newer image of the circuit . As I said the miliammeter is always reading bigger than 1 mA eventhough the other miliammeter(mA2 in the circuit )is already 0 mA.Please help me.
 

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  • #10
Integral are you there ? Why you don't answer my question .Somebody please answer my question
 
  • #11
I don't understand what this experiment is trying to do. The moment you close the switch on the uncharged capacitor, all the current will initially go through the capacitor, it would be like connecting wire between two ends of your power source. What will this achieve?
What is this experiment's aim?
Sam
 
  • #12
ngkamsengpeter said:
I have conducted an experiment in school I have some problem with it . I connect a resistor 6600 ohm to a 6V direct current and I read the miliammeter and it show about 0.9 mA . But after I connect a capacitor 3300 F parallel to the resistor , the miliammeter increase to more than 1 mA.Why the current increase when I connect the capacitor parallel to resistor . Furthermore, the experiment procedure told me to reduce the resistance to get 1 mA but I have to increse the resistance in order to get 1 mA. What wrong with my experiment or the experiment procedure is wrong .
Hopefully, you meant 3300 pF instead of 3300 F, otherwise it will take about 3 and a half years for your capacitor to discharge (plus, I couldn't even begin to imagine how big, physically, a 3300 F capacitor would be).

If you're getting more than 1 mA through the resistor, your voltage at the node has to be higher than 6 V (try measuring it). One thing to check is that your capacitor is fully discharged before you start. You should have zero volts at the node before you close the switch.
 
  • #13
http://home.comcast.net/~integral50/Math/circiut.jpg

To do the experiment you have described you will need an Ammeter as shown in my attachment. Please confirm the values of your circuit components including correct units. Actual measurements of the voltage and resistance would be nice.
 
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  • #14
BobG said:
Hopefully, you meant 3300 pF instead of 3300 F, otherwise it will take about 3 and a half years for your capacitor to discharge (plus, I couldn't even begin to imagine how big, physically, a 3300 F capacitor would be).

If you're getting more than 1 mA through the resistor, your voltage at the node has to be higher than 6 V (try measuring it). One thing to check is that your capacitor is fully discharged before you start. You should have zero volts at the node before you close the switch.
the capacitor is we(i and ngkamsengpeter) use is not 3300pF and also not 3300F but is 3300microfarad.
 
  • #15
Integral said:
http://home.comcast.net/~integral50/Math/circiut.jpg

To do the experiment you have described you will need an Ammeter as shown in my attachment. Please confirm the values of your circuit components including correct units. Actual measurements of the voltage and resistance would be nice.
But my book doesn't show the diagram as your attachment . Actually my experiment is used to calculate the time constant and only has one miliammeter(mA in the diagram I have attached ) and don't have the mA2.This was totally different from your diagram . Is it my book wrong .
 
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  • #16
If your milliameter is reading more than 0.9 mA, then either
1) your voltage source has more that 6V potential difference
2) your resistor has less than 6600 Ohm resistance (even with the meter)
3) your mA meter is slightly out-of-calibration,
. . . maybe even all 3.

Check the source's Voltage ... especially if it a battery!
Check your meter's reading with nothing connected.
Check your resistance with an Ohm-meter.

This is a relative measurement anyway ...
don't get hung up on 10% difference from "lab instructions" values !
 
  • #17
lightgrav said:
If your milliameter is reading more than 0.9 mA, then either
1) your voltage source has more that 6V potential difference
2) your resistor has less than 6600 Ohm resistance (even with the meter)
3) your mA meter is slightly out-of-calibration,
. . . maybe even all 3.

Check the source's Voltage ... especially if it a battery!
Check your meter's reading with nothing connected.
Check your resistance with an Ohm-meter.

This is a relative measurement anyway ...
don't get hung up on 10% difference from "lab instructions" values !
it is really 6V,
and the resistor(if not mistake)is about 8800ohm,
the miliammeter is no problem(new)~
 
  • #18
lightgrav said:
If your milliameter is reading more than 0.9 mA, then either
1) your voltage source has more that 6V potential difference
2) your resistor has less than 6600 Ohm resistance (even with the meter)
3) your mA meter is slightly out-of-calibration,
. . . maybe even all 3.

Check the source's Voltage ... especially if it a battery!
Check your meter's reading with nothing connected.
Check your resistance with an Ohm-meter.

This is a relative measurement anyway ...
don't get hung up on 10% difference from "lab instructions" values !
lifesfun is correct since before I connect the capacitor it shows 0.9mA with resistance 6600 ohm . So, the resistance and the voltage should be no problem but as I mentioned , I don't know why the current increase to more than 1 mA when I connect a capacitor parallel to the resistance .
 
  • #19
can anyone please answer my question.I know that the current increase is due to the internal resistance of the capacitor but I let the capacitor to fully charge and I think that after fully charge the current will back to 0.9mA but it is not the case , the ampere doesn't change even 0.1mA.Please help me !
 
  • #20
anybody please reply th question above~~
 
  • #21
Maybe (just a thought). If you are using a battery for this emf, it is a chemistry problem. An ammeter's resistance is very low, and so dragging the overall resistance of your circuit to a low value while the capacitor is charging. If this value reaches much less than the internal resistance of your battery; a lot of power will be lost in the battery as heat.

Heat causes chemical reactions to happen faster and so might increase the voltage that your battery outputs.

That is (of course) assuming you are using a battery.

Thoughts?

Sam
 
  • #22
BerryBoy said:
Maybe (just a thought). If you are using a battery for this emf, it is a chemistry problem. An ammeter's resistance is very low, and so dragging the overall resistance of your circuit to a low value while the capacitor is charging. If this value reaches much less than the internal resistance of your battery; a lot of power will be lost in the battery as heat.

Heat causes chemical reactions to happen faster and so might increase the voltage that your battery outputs.

That is (of course) assuming you are using a battery.

Thoughts?

Sam
I am using a power supply , not a battery.
 

Related to Current Increase When Connecting Capacitor to Resistor

1. What is the relationship between a capacitor and a resistor in terms of current increase?

When a capacitor is connected to a resistor in a circuit, the current will initially increase due to the discharge of the capacitor's stored energy. However, as the capacitor continues to discharge, the current will decrease and eventually reach a steady state where the current is solely determined by the resistance of the resistor.

2. How does the capacitance value affect the current increase in this circuit?

The capacitance value of the capacitor will affect the rate of current increase in the circuit. A higher capacitance value means that the capacitor can store more energy and therefore discharge more current, resulting in a larger initial increase in current. On the other hand, a lower capacitance value will result in a slower increase in current.

3. What happens to the current if the resistor value is changed?

The current in the circuit will change proportionally to the change in the resistance value. A higher resistance will result in a lower current, while a lower resistance will result in a higher current. However, the initial increase in current when the capacitor is connected will remain the same regardless of the resistor value.

4. Can the current increase be controlled by the capacitor and resistor values?

Yes, the current increase in the circuit can be controlled by choosing appropriate values for the capacitor and resistor. A larger capacitance or smaller resistance will result in a larger initial increase in current, while a smaller capacitance or larger resistance will result in a smaller initial increase in current.

5. Is there a limit to how much the current can increase in this circuit?

Yes, there is a limit to how much the current can increase when a capacitor is connected to a resistor. This limit is determined by the maximum energy that the capacitor can store and the resistance of the circuit. Once the capacitor is fully discharged, the current will reach a steady state and will not increase any further.

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