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# A galvanometer with a coil resistance $12.0 \Omega$ shows a full scale deflection for a current $2.5\;mA$ How it can be converted into an ammeter of range $7.5 \;A$ What will be resistance of ammeter formed ?

$\begin{array}{1 1} 4 \times 10^3 \;\Omega \\ 4 \times 10^2\;\Omega \\ 4 \times 10^{-3}\;\Omega \\ 4 \times 10^{-2}\;\Omega \end{array}$

$S= \large\frac{I_g G}{I-I_g} =\frac{(2.5 \times 10^{-3}) \times 12}{7.5 -2.5 \times 10^{-3}}$
$\qquad= 4 \times 10^{-3} \Omega$
$R= \large\frac{GS}{G+S} =\frac{12 \times 4 \times 10^{-3}}{12+ 4 \times 10^{-3}}$
$\qquad= 4 \times 10^{-3}\;\Omega$
Answer : $4 \times 10^{-3}\;\Omega$