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A galvanometer has resistance `100 Omega` and it requires current `100 muA` for full scale deflection. A resistor `0.1 Omega` is connected to make it an ammeter. The smallest current required in the circuit to produce the full scale deflection is

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Correct Answer - d
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`I_(g)=0.1/((100+0.1))xxI`
`I=100.1/0.1xxI_(g)`
`I=1001xx100 muA`
`I=100.1mA`

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