Explanation:
The current range and voltage range of a galvanometer imply the values of current and voltage which produce full scale deflection in the galvanometer.
Thus,
Current for full scale deflection,
\(\,\,\,\,\,\,\)\({I_g} = 15\;mA = 15 \times {10^{ - 3}}\;A\)
Voltage for full scale deflection,
\(\,\,\,\,\,\,\)\({V_g} = 750\,mV = 750 \times {10^{ - 3}}\;V\)
The resistance of the galvanometer is
\(\,\,\,\,\,\,\)\(G = \frac{{{V_g}}}{{{I_g}}} = \frac{{750 \times {{10}^{ - 3}}\;V}}{{15 \times {{10}^{ - 3}}\;A}} = 50\,\Omega \)
Let \(S\) be required shunt to convert the given galvanometer into an ammeter of range \(25\;A\).
\(S = \frac{{{I_g}G}}{{I - {I_g}}} = \frac{{\left( {15 \times {{10}^{ - 3}}\;A} \right)(50\,\Omega )}}{{25\;A - \left( {15 \times {{10}^{ - 3}}} \right)A}} = 0.03\,\Omega \)
