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A galvanometer coil of resistance 50 ohm shows full deflection at 10 micro ampere.Resistance to be added,for the galvanometer to work as an ammeter of range 10 mega ampere is
Asked by aiswaryapbindhu3 | 08 Aug, 2018, 11:35: PM
Since full scale deflection is obtained by 10 micro-amps through 50Ω coil,
the potential difference across galvanometer coil is 50×10 = 500 micro volt.

To convert the galvanometer to Ammeter, we need to connect a shunt resistance parallel to coil resistance.

Let Req is the equivalent resistance, then when 10 maga amphere is passing through the equivalent resiatance
of parallel combination of coil and shunt, we should get same potentail difference of 500 micro volt.

10×106×Req = 500×10-6    or  Req = 50×10-12 Ω

if R is the required shunt resistance, then Req=    or R = 50×10-12 Ω

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