The resistance of a galvanometer is \( 30\,\Omega \). It gives full scale deflection for 2 mA. It is to be converted into an ammeter to measure current 0.3 A. The value of the shunt required is approximately
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.2\,\Omega \)
A galvanometer becomes an ammeter when a small shunt resistance \( S \) is joined in parallel with it. At full scale only \( I_g \) passes through the galvanometer and the rest, \( I - I_g \), passes through the shunt. Both have the same potential difference across them.
\( \displaystyle I_g\,G = (I - I_g)\,S \;\Rightarrow\; S = \frac{I_g\,G}{I - I_g} \)
\( G = 30\,\Omega \)
\( I_g = 2\,\mathrm{mA} = 0.002\,\mathrm{A} \)
\( I = 0.3\,\mathrm{A} \)
\( \displaystyle S = \frac{0.002 \times 30}{0.3 - 0.002} = \frac{0.06}{0.298} = 0.201\,\Omega \approx 0.2\,\Omega \)