A galvanometer shows full scale deflection for a current of 4 mA. It is converted into an ammeter of range from 0 to 8 A by using a shunt. The ratio of resistance of the galvanometer coil to the resistance of the shunt used is nearly:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2000 : 1
The shunt \( S \) is connected in parallel with the galvanometer \( G \), so both have the same potential difference. At full scale, \( I_g \) passes through the galvanometer and the rest \( I - I_g \) through the shunt.
\( \displaystyle I_g\,G = (I - I_g)\,S \;\Rightarrow\; \frac{G}{S} = \frac{I - I_g}{I_g} \)
\( I_g = 4\,\mathrm{mA} = 4\times10^{-3}\,\mathrm{A} \)
\( I = 8\,\mathrm{A} \)
\( \displaystyle \frac{G}{S} = \frac{8 - 0.004}{0.004} = \frac{7.996}{0.004} = 1999 \approx 2000 \)
So \( G : S \approx 2000 : 1 \). The shunt has a very small resistance, as it must carry almost the whole current.