Physics Help Forum Coulombmeter problem
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 Jun 28th 2010, 01:49 AM #1 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 Coulombmeter problem A coulombmeter made from the components shown in the figure is used to measure the charge on a conducting sphere. a) If the charge on the sphere is 1.6 microC, calculate the reading on the voltmeter when the charged sphere touches the terminal P. Explain from first principles each step you make in your calculation and state any assumptions you make. b)It is sugested that if a meter of resistance 1kohm is used for the voltmeter, the capacitor will rapidly discharge. Explain whether you agree or disagree. c) The sphere was charged from a +5000V d.c. supply. Discuss whether all its charge would be measured by the coulombmeter. (a) I got the right answer from V=Q/C where Q is the charge and C the capacitance. But somehow it doesn't seem correct to me, the amplifier is not taken into account. For (b), I with the higher resistance, won't the capacitor take longer to discharge fully? For (c), I have no idea. Thanks! Attached Thumbnails   __________________ To each his own.
 Jun 28th 2010, 10:00 PM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 a) The circuit provided by you is a Voltage follower Voltage follower is used as a buffer amplifier to eliminate loading effects. In this circuit Vin = Vout. b) Usually the voltmeters have high resistance (more than 100 kohm) If you use voltmeter of 1 kohm what will happen? c) No. In a short interval of time, all the charges stored in the capacitor will not discharge through high resistance voltmeter.
Jun 28th 2010, 10:11 PM   #3
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 Originally Posted by sa-ri-ga-ma b) Usually the voltmeters have high resistance (more than 100 kohm) If you use voltmeter of 1 kohm what will happen?
When the resistance decreases, then the full-scale current also decreases, so the meter can be used more accurately for small voltages? I think that's about it, am I right?
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Jun 28th 2010, 10:32 PM   #4
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 Originally Posted by arze When the resistance decreases, then the full-scale current also decreases, so the meter can be used more accurately for small voltages? I think that's about it, am I right?
Voltmeter is connected across a device where the voltage is to be measured. The reading will be more accurate, if most of the current flows throug the devise. If a voltmeter of a smaller resistance in connected, more current will flow through the resistance and the accuracy will be less.

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