Question

a galvanometer has resistance of 60 ohm. it is converted in to an ammeter by connecting a shunt resistance of 1.2 ohm it's range becomes​

Answer 1

Answer:

The voltage across the meter is I x R = 2/1000 x 50 = 0.1 volt.

To obtain full scale with 10 amps implies a total resistance of

R = E/I = 0.1 /10 = 0.01 ohms

The shunt resistor in parallel with the meter movement ought to provide 0.01 ohms. In practice, the shunt itself would be 0.01

Answer 2

Explanation:

The voltage across the meter is l × R = 2/1000 x 50 = 0.1 volt.

To obtain full scale with 10 amps implies a total resistance of

R =E/l=0.1/10=0.01 ohms

The shunt resistor in parallel with the meter movement ought to provide 0.01 ohms. In practice,the shunt itself would be 0.01

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