AMMETER-VOLTMETER
 
AMMETER
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    Ammeter is an electrical measuring device, which is used to measure electric current through the     circuit. It is the modified form of galvanometer
CONNECTION OF AMMETER IN CIRCUIT
 
    An ammeter is always connected in series to a circuit.
SYMBOL
 
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CONVERSION OF GALVANOMETER
INTO AMMETER
    Since Galvanometer is a very sensitive instrument therefore it can’t measure heavy currents. In order to     convert a Galvanometer into an Ammeter, a very low resistance known as "shunt" resistance is     connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes     through the shunt. In this way a Galvanometer is converted into Ammeter and can measure heavy     currents without fully deflected.
VALUE OF SHUNT RESISTANCE
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    Let resistance of galvanometer = Rg and it gives full-scale deflection when current Ig is passed through     it. Then,
Vg = IgRg -------(i)
    Let a shunt of resistance (Rs) is connected in parallel to galvanometer. If total current through the     circuit is I.
    Then current through shunt:
Is = (I-Ig)
    potential difference across the shunt:
Vs= IsRs
or
              Vs = (I – Ig)Rs -------(ii)
    But
Vs =Vg
(I - Ig)Rs = IgRg

VOLT METER
 
    Voltmeter is an electrical measuring device, which is used to measure potential difference between two     points in a circuit.
CONNECTION OF VOLTMETER IN CIRCUIT
 
    Voltmeter is always connected in parallel to a circuit.
SYMBOL
 
CONVERSION OF GALVANOMETER INTO VOLTMETER
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    Since Galvanometer is a very sensitive instrument, therefore it can not measure high potential     difference. In order to convert a Galvanometer into voltmeter, a very high resistance known as "series     resistance" is connected in series with the galvanometer.
VALUE OF SERIES RESISTANCE
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    Let resistance of galvanometer = Rg and resistance Rx (high) is connected in series to it. Then     combined resistance = (Rg + Rx).
    If potential between the points to be measured = V and if galvanometer gives full-scale deflection,     when current "Ig" passes through it. Then,
V = Ig (Rg + Rx)
V = IgRg + IgRx 
V – IgRg = IgRx 
Rx = (V – IgRg)/Ig

    Thus Rx can be found.
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