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The resistance of a galvanometer is $50 \Omega$. When 0.01 A current flows in it, full scale deflection is obtained in galvanometer, the resistance of shunt connected to convert galvanometer into an ammeter of range 5 A , will be
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$0.1 \Omega$
Here: $I_g=0.01 \mathrm{~A}, G=50 \Omega$
To convert the galvanometer in to an ammeter of range 5 A , a low resistance is to be connected in parallel with coil of galvanometer.
$S=\frac{i g \times G}{i-i_g}$
$=\frac{0.01 \times 50}{5-0.01}=\frac{0.5}{4.99}=0.1 \Omega$
To convert the galvanometer in to an ammeter of range 5 A , a low resistance is to be connected in parallel with coil of galvanometer.
$S=\frac{i g \times G}{i-i_g}$
$=\frac{0.01 \times 50}{5-0.01}=\frac{0.5}{4.99}=0.1 \Omega$
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