Question

A maximum current of $5mA$ can be passed through a galvanometer of resistance $40\Omega$ . The resistance to be connected in series to convert it into a volt meter of range $0-50V$ is :

• A. $960\Omega$
• B. $9960\Omega$
• C. $99,960\Omega$
• D. $19,960\Omega$

Answer 1

The correct option is B $9960\Omega$

Let resistance connected be $x=R_{eq}=40+x$
$\frac{V_1}{V_2}=\frac{R_1}{R_2};but{V}_1=5\times {10}^{-3}\times 40=0.2$
$⟹\frac{0.2}{50}=\frac{40}{40+x}$
$⟹40+x=10000$
$⟹x=9960\Omega$