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In the given circuit, a galvanometer wit...

In the given circuit, a galvanometer with a resistance of 80 `Omega` is converted to an ammeter by a shunt resistance of 0.05 `Omega`, what is the total current measured by this device in ampere?

Text Solution

Verified by Experts

The correct Answer is:
`01.23`

`R_(s)=(R_(G)r_(s))/(R_(G)+r_(s))`
`=(80 xx 0.05)/(80+0.05)`
`=0.05 Omega`
Total resistance in the circuit,
`R=R_(s)+4Omega`
`=(0.05 +4)Omega =4.05Omega`
The current measured by the device,
`I=(V)/(R ) =(5)/(4.05) =1.23A`
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