Question

A voltmeter of variable ranges $3 \, V, \, 15 \, V, \, 150 \, V$ is to be designed by connecting resistances $R_{1}, \, R_{2}, \, R_{3}$ in series with a galvanometer of resistance $G \, = \, 20 \, \Omega $ , as shown in the figure. The galvanometer gives full pass through its coil. Then, the resistances $R_{1}, \, R_{2}$ and $R_{3}$ (in kilo ohms) should be, respectively

• A. $3, \, 12, \, 135$
• B. $2.98, \, 12, \, 135$
• C. $2.98, \, 14.98, \, 149.98$
• D. $3.86, \, 12.55, \, 122.97$

Answer 1

Answer: (B)

$R \, =\frac{V}{I_{g}} \, - \, R_{g}$
$R_{1} \, =\frac{3}{I_{g}}- \, R_{g} \, = \, 2980 \, \Omega \, = \, 2.98 \, k\Omega $
$\left(R_{1} \, + \, R_{2}\right)=\frac{15}{I_{g}}- \, Rg \, = \, 14980\Omega \, = \, 14.98 \, k\Omega $
$\left(R_{1} \, + \, R_{2} \, + \, R_{3}\right)= \, \frac{150}{I_{g}} \, - \, R_{g} \, = \, 149980 \, \Omega $
$= \, 149.98 \, k\Omega $
$\Rightarrow R_{1} \, = \, 2.98 \, k\Omega $
$R_{2} \, = \, 14.98 \, - \, 2.98 \, = \, 12 \, k\Omega $
$R_{3} \, = \, 149.98 \, - \, 14.98 \, = \, 135 \, k\Omega $