A voltmeter will not read anything if it is connected in series. It needs to be connected to a hot lead on one side and a neutral or ground on the other.
Connecting in series is typically used for an ammeter to measure the load.
2007-09-25 01:00:45
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answer #1
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answered by Gregory M 3
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Consider that electricity is similar to the flow of water where voltage is electrical pressure and electrical current is the flow of electrons. An open circuit (battery terminals) may have high voltage (pressure) but no current. A pressure gage that measures water pressure has very little flow (as the Bourdon tube expands) and a voltmeter also passes very little current. In series with a load having very little resistance (a bare wire?) the voltmeter will measure the voltage at its terminals (nearly equal to supply voltage) but little or no current will pass. If the load has very high resistance (a good insulator?) the voltmeter should measure nearly zero volts.
2007-09-25 09:23:49
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answer #2
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answered by Kes 7
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A modern voltmeter has a very high input impedance, so almost no current would flow through the meter and the circuit, so the circuit would likely not function correctly. The meter will measure a much higher voltage than would normally exist at that point in the circuit because of the high impedance.
It won't harm the meter or the circuit, but hooking it up this way probably isn't going to give you any useful information.
2007-09-25 08:07:51
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answer #3
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answered by I don't think so 5
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The voltmeter may read more than the scale you had it set on due to being effectively "across the load." It will prevent the load from functioning properly because of the high resistance of the meter.
2007-09-25 08:12:46
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answer #4
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answered by peterngoodwin 6
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an ideal voltmeter is basically having infinite internal resistance.
practically its resistance is in the megohm range.
so connecting it in series will cause a very very small current which would deflect the pointer by a very very small angle.
2007-09-25 08:06:20
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answer #5
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answered by Atul I 2
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It would show its own voltage drop - very small, but not nonexistant.
2007-09-25 08:03:06
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answer #6
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answered by Anonymous
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