A galvanometer with resistance 100 Ohm gives full scale deflection with a current of 2 mA. The resistance required to convert galvanometer into ammeter of range 0 to 20 A is nearly :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{-2} \) Ohm in parallel.
To make an ammeter, a small resistance \( S \) (a shunt) is connected in parallel with the galvanometer, so that most of the current bypasses the coil. The potential difference across the two is the same:
\( \displaystyle I_g G = (I - I_g)\,S \;\Rightarrow\; S = \frac{I_g\,G}{I - I_g} \)
\( G = 100\,\Omega \)
\( I_g = 2\,\mathrm{mA} = 2\times10^{-3}\,\mathrm{A} \)
\( I = 20\,\mathrm{A} \)
\( \displaystyle S = \frac{2\times10^{-3}\times100}{20 - 0.002} = \frac{0.2}{19.998} \approx 0.01\,\Omega = 10^{-2}\,\Omega \)
So a resistance of about \( 10^{-2}\,\Omega \) is connected in parallel.