A galvanometer of \( 60\,\Omega \) resistance shows full scale deflection for a current of 8 mA. The shunt required to convert the galvanometer into an ammeter of range 0 to 6 A will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 8 \times 10^{-2}\,\Omega \)
To make an ammeter, a small resistance \( S \) (the shunt) is joined in parallel with the galvanometer. At full scale the galvanometer carries \( I_g \) and the shunt carries the rest, \( I - I_g \). Both have the same potential difference across them.
\( \displaystyle I_g\,G = (I - I_g)\,S \;\Rightarrow\; S = \frac{I_g\,G}{I - I_g} \)
\( I_g = 8\,\mathrm{mA} = 8\times10^{-3}\,\mathrm{A} \)
\( G = 60\,\Omega \)
\( I = 6\,\mathrm{A} \)
\( \displaystyle S = \frac{8\times10^{-3} \times 60}{6 - 0.008} = \frac{0.48}{5.992} = 0.0801\,\Omega \approx 8\times10^{-2}\,\Omega \)