A moving coil galvanometer has a resistance of $50\,\Omega $ and gives full scale deflection for $10\, mA$. How could it be converted into an ammeter with a full scale deflection for $1\,A$
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$S = \frac{{{i_g} \times G}}{{i - {i_g}}} = \frac{{10 \times {{10}^{ - 3}} \times 50}}{{1 - {{10}^{ - 3}} \times 10}} = \frac{{50}}{{99}}\,\Omega $ in parallel.
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