Question

A voltmeter and an ammeter are connected in series across an ideal cell. When certain resistance is connected in parallel to voltmeter, reading of ammeter increases three times and reading of voltmeter reduces to one third. The ratio of resistance of voltmeter to that of ammeter is

• A. 3
• B. 1
• C. 2
• D. 5

Answer 1

The correct option is A 3

When voltmeter and an ammeter are connected in series across an ideal cell we get,
$V_A+V_V=E$ ...............(1)
where, $V_A$ and $V_V$ are the voltages across ammeter and voltmeter respectively. $E$ is an emf of an ideal cell.
When certain resistance is connected in parallel to voltmeter,
$3V_A+\frac{V_V}{3}=E$ ..............(2)
$9V_A+V_V=3E$
Put,$V_V=E-V_A$ ..........from (1)
$\therefore 9V_A+(E-V_A)=3E$
$\therefore V_A=\left(\frac{1}{4}\right)E$................(3)

Now equation (2) will be,
$\left(\frac{3}{4}\right)E+\frac{V_V}{3}=E$

$\therefore V_V=\left(\frac{3}{4}\right)E$................(4)

From equations (3) and (4) we can write
$\frac{V_V}{V_A}=\frac{\left(\frac{3}{4}\right)E}{\left(\frac{1}{4}\right)E}$
$\Rightarrow \frac{I{R}_V}{I{R}_A}=\frac{3}{1}$
$\therefore \frac{R_V}{R_A}=3$