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The shunt resistance is required to make the $1.00 m A, 20 Ω$ galvanometer into an ammeter with a range of 0 to 50.0mA is $X \times 10^{- 3} Ω$

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Detailed Solution

$i_{g} = 1.00 m A = 10^{- 3} A$

$R_{G} = 20 Ω$

$i_{R} = 50.0 \times 10^{- 3} A$

$R_{s} = (\frac{i_{g}}{i - i_{g}}) R_{G}$

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