An ammeter reads up to 1A. Internal resistance of  ammeter is 0.81 ohm. To increase the range to 10 A, the value of required shunt is

Answers (1)

I is the required current=10A. Ig =1 A , G=0.81 ohm, find S (shunt required)

\frac{S}{S+G}=\frac{I_{g}}{I}\Rightarrow S=\frac{I_{g}G}{I-I_{g}}

\therefore \; \; \; S=\frac{1\times 0.81}{10-1}=\frac{0.81}{9}=0.09\, \Omega \; in\; parallel.

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