Q. A galvanometer of resistance $100\Omega$ is converted to a voltmeter of range $10\ V$ by connecting a resistance of $10k\Omega.$ The resistance required to convert the same galvanometer to an ammeter of range $1\ A$ is
A) $0.4\Omega$
B) $0.3\Omega$
C) $1.2\Omega$
D) $0.1\Omega$
Explanation
(d) [image:3_image134.emf]
Here, $V = 10\ V,\ R = 10k\Omega,\ G = 100\ \Omega = 0.1\ k\Omega$
$V = I_{g}(G + R)$
$I_{g} = \frac{V}{G + R} = \frac{10V}{(10 + 0.1)k\Omega} = \frac{10V}{10.1k\Omega} = 0.99mA \approx 1\ mA$
[image:3_image135.emf]
Here, $I = 1A,\ \left( I - I_{g} \right)S = I_{g}G$
$S = \frac{I_{g}G}{I - I_{g}} = \frac{1 \times 10^{- 3} \times 100}{1 - 1 \times 10^{- 3}}$
$\approx \frac{100 \times 10^{- 3}}{1} = \frac{100}{1000}\Omega = \frac{1}{10}\Omega = 0.1\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.