Question

A voltmeter has a range $0- V _{1}$ with a series resistance $R$ and with a series resistance $2 R$, the range is $0- V _{2}$ The correct relation between $V_{1}$ and $V_{2}$ is

• A. $V_{2}=2 V_{1}$
• B. $V_{2}>2 V$
• C. $V_{2}>>2 V_{1}$
• D. $V_{2}<2 V_{1}$

Answer 1

Answer: (D)

For conversion of galvanometer $R$ is connected in series.
$\therefore i_{g}=\frac{V_{1}}{R+G}$ and $i_{g}=\frac{V_{2}}{2 R+G}$
$\frac{V_{1}}{R+G}=\frac{V_{2}}{2 R+G}$
$\frac{V_{2}}{V_{1}}=\frac{2 R+G}{R+G}$
$\frac{V_{2}}{V_{1}}=\frac{2(R+G)-G}{(R+G)}$
$\frac{V_{2}}{V_{1}}=2-\frac{G}{R-G}$
$V_{2}=2 V_{1}-\frac{V_{1} G}{(R+G)}$
$\Rightarrow V_{2}<2 V_{1}$