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Q. $A$ galvanometer having a resistance of $50 \, \Omega$, gives a full scale deflection for a current of $0.05 \,A$. The length (in metres) of a resistance wire of area of cross section $3\times 10^{-2}\, cm^{2}$ that can be used to convert the galvanometer into an ammeter which can read a maximum of $5 \,A$ current is
(Specific resistance of the wire $\rho=5\times10^{-7}\,\Omega\,m)$

Moving Charges and Magnetism

Solution:

$S=\frac{I_{g}\,G}{I-I_{g}}$
$=\frac{0.05\times50}{5-0.05}$
$=\frac{2.5}{4.95}$
$=\frac{250}{495}$
$=\frac{50}{99}\, \Omega $
$\because S=\frac{\rho l}{A}$ or $ l=\frac{SA}{\rho}$
$\therefore l=\frac{50}{99}\times\frac{3\times10^{-6}}{5\times10^{-7}}$
$=3.0\,m$