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AMMETER-VOLTMETER
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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 | |||
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CONNECTION
OF AMMETER IN CIRCUIT
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| An ammeter is always connected in series to a circuit. | |||
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SYMBOL
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| For latest information , free computer courses and high impact notes visit www.citycollegiate.com | |||
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CONVERSION
OF GALVANOMETER
INTO AMMETER |
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| 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. | |||
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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, | |||
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Vg
= IgRg -------(i)
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| Let a shunt of resistance (Rs) is connected in parallel to galvanometer. If total current through the circuit is I. | |||
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| Then current through shunt: | |||
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Is
= (I-Ig)
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| potential difference across the shunt: | |||
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Vs=
IsRs
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or
Vs = (I – Ig)Rs -------(ii) |
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| But | |||
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Vs
=Vg
(I - Ig)Rs = IgRg ![]() |
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VOLT
METER
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| Voltmeter is an electrical measuring device, which is used to measure potential difference between two points in a circuit. | |||
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CONNECTION
OF VOLTMETER IN CIRCUIT
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| Voltmeter is always connected in parallel to a circuit. | |||
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SYMBOL
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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. | |||
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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). | |||
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| If potential between the points to be measured = V and if galvanometer gives full-scale deflection, when current "Ig" passes through it. Then, | |||
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V
= Ig (Rg + Rx)
V = IgRg + IgRx V – IgRg = IgRx Rx = (V – IgRg)/Ig |
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| Thus Rx can be found. | |||
| For latest information , free computer courses and high impact notes visit www.citycollegiate.com | |||