Capacitor and circuit simulation

Discussion in 'Homework Help' started by howartthou, Apr 29, 2010.

  1. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi All

    I am using crocodile clips to do a simple DC circuit that has a 9v battery and a capacitor.

    The circuit is initially open so when you connect a multimeter to the open ends of the circuit you should read 9v.

    Many peope think you should read zero because a capacitor blocks DC, but because the capacitor is a new capacitor it is not charged initially, so it initially charges through the multimeter at 9v, then once charged will drop to zero volts.

    This can be tested by replacing the 9v battery with a light globe, removing the multimeter, and closing the circuit. The globe should light briefly while the capacitor discharges and then the globe should go out.

    I am tring to simulate this with crocodile clips software but when I remove the battery and the multimeter, add a globe then close the circuit, the software thinks the capacitor is not charged. But in real life the capacitor would have been charged by the multimeter.

    Is this a limitation in the software? Can I get CC to remember that the capacitor was charged throught the multimeter?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Maybe you can post the schematics because in my crocodile clips every think is alright
     
  3. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Jony

    The circuit attached shows the capacitor charging (I think) because the voltage on the voltmeter works its way to zero ... after a long time.

    But then I detach the battery and the voltmeter, then attach a 10w globe to the capacitor and nothing happens??
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's not really clear to me whether you are conducting a real experiment or trying to simulate (using software?) circuit behaviour in some way.

    In any case you could charge a 100nF capacitor to 9V and then put a 10W (12Volt presumably?) globe across the isolated charged capacitor. You wouldn't expect to see anything. Why?

    The cold resistance of the globe will be relatively low - say a couple of ohms at most if it's a 12V 10W globe.

    The 100nF capacitor will fully discharge through 2ohms in 1 microsecond. No way that could be seen - in fact the globe would barely even start to warm up let alone glow perceptibly.
     
  5. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Thanks TNK.

    Yes I am trying to simulate this with software. Why? Because its faster to experiment with software than to build the circuit. Admittedly this is a simple circuit but I need to be confident that I can indeed do any circuit with the software.

    I am not convinved the capacitor is remembering its charged when I hook it up to a light globe after charging it.

    You explaination make perfect sense. Thank you.

    So any hints on how many mF the capacitor needs to be for a 10w globe?

    Should I use a 1w globe with a bigger capacitor?

    Another problem is the software takes ages tio fully charge the capacitor, probably because thats how long it would really take.

    Can you suggest some sizes for the battery, capacitor and globe so I can charge it quickly and test this circuit (with software)?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The long charging time you see is due to the voltmeter having a very high internal resistance.

    Say the voltmeter in your simulator is modelled with an equivalent resistance of 10MΩ. It would take approximately 5 seconds to charge. If it is modelled with a 100MΩ resistor it will take ~50 seconds.

    Why not just charge the capacitor through a small series resistor rather than the voltmeter? The time constant of the series RC combination is the critical factor. Place the voltmeter across the capacitor if you want to see the charging process.

    If you chose C=1Farad you might see a fleeting flash on a globe - not sure of a suitable wattage.

    Does your simulator have an LED rather than a globe? Discharging C through an LED in series with a resistor would give you a much better chance of "seeing" something.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    CC should have a way to place an initial condition (voltage) on the capacitor. It then starts out already charged.
    Every simulator I have ever used has this capability.
    Your CC multimeter may be ideal, and have infinite input resistance.
     
  8. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    tnk
    Thanks for the tips. I did use a LED (see attached) and a 1W resistor with a 100F capacitor and it lit up. First I charged it with a 9v battery, then took out the battery and put in the resistor and LED.

    What I wanted to do was have the voltmeter charge the capacitor to prove that connecting a voltmeter to an open circuit with a battery and capacitor, will charge the capacitor by passing the full the battery voltage.

    So, my experiment is still not working.

    Ron H
    CC (my software) will not allow me to preset the charge (voltage) but I can charge it with a battery.

    All
    The attached shows 2 circuits. The top circuit does not have a voltmeter and actually charges the capacitor.

    The bottom circuit has a voltmeter and refuses to charge the capacitor. Is that because its an "ideal" voltmeter with infinate resistance? Or is it me not using the CC software properly?

    The experiment is meant to show that the capacitor charges through the voltmeter.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you can insert an ammeter in series with the circuit (with the voltmeter in as well) you would expect to read the capacitor charging current. Depends on the ammeter sensitivity of course.

    You can also easily test if the voltmeter is modelled as an infinite resistance - or open circuit. Include the aforementioned ammeter in the circuit & replace the capacitor with a short and measure the current drawn from a DC source of any suitable magnitude. If you can set the DC to as large a value as possible you would eventually expect to read some current in the circuit if the voltmeter resistance is finite. If you had a 1000V DC source and voltmeter resistance of 1000 MΩ you would read 1μA. If you can't read any current at all you can safely assume the voltmeter is modelled as an open circuit.

    Another option is to forget the ammeter idea and include a high resistance in parallel with your voltmeter. Say you make the parallel resistance 100 Meg ohm. The charging time would be about 50 seconds.

    The reality is if your experiment isn't working as you planned then it's highly likely that the voltmeter resistance is modelled as an infinite value.
     
    Last edited: May 3, 2010
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your capacitor is to small to see anything .
    Circuit may look like this
    [​IMG]
     
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