Q. A galvanometer of resistance 100 $\Omega$ gives a full scale deflection for a current of $10^{- 5}\ A.$ To convert it into a ammeter capable of measuring upto 1A, we should connect a resistance of
A) 1 $\Omega$ in parallel
B) $10^{- 3}\ \Omega$ in parallel
C) $10^{5}\ \Omega$ in series
D) 100 $\Omega$ in series
Explanation
(b) $G = 100\ \Omega\ $
$I_{g} = 10^{- 5}\ $ A
$I\ \ = 1\ $A
$S\ = \ ?$
$I_{g} \times G = (I - I_{g}) \times S$
$S\ \ = \left( \frac{I_{g}}{I - I_{g}} \right) \times G = \frac{10^{- 5}}{1 - 10^{- 5}} \times 100$
Or $= \ \frac{10^{- 3}}{1 - 0.00001} = 10^{- 3}\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.