A milli voltmeter of $25$ $milli$ $volt$ range is to be converted into an ammeter of $25$ $ampere$ range. The value (in $ohm$) of necessary shunt will be
AIPMT 2012, Medium
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$S=\frac{V_{g}}{\left(I-I_{g}\right)}$
Neglecting $I_{z}$
$\therefore \quad S=\frac{V_{g}}{I}=\frac{25 \times 10^{-3} \,\mathrm{V}}{25 \,\mathrm{A}}=0.001\, \Omega$
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