Question

Two same voltage rating PMMC type DC voltmeters having figure of merit of $10k\Omega /V$ and $25k\Omega /V$ respectively are connected in series. This series combination of voltmeters can measure maximum of 280 V. The voltage rating of voltmeters will be ______V.

• A. 200

Answer 1

The correct option is A 200
$\text{Figure of merit of voltmeter '1' = 10 k}\Omega \text{/V}$

$\text{Figure of merit of voltmeter '2' = 25 k}\Omega \text{/V}$

$\text{Full scale curent through the meters are}$

$\text{For meter 1,}$

$I_{FSD1}=\frac{1}{S_{v1}}=\frac{1}{10k\Omega /V}=0.1mA$

$\text{For meter 2,}$

$I_{FSD2}=\frac{1}{S_{v2}}=\frac{1}{25k\Omega /V}=0.04mA$

Let the rating of voltmeter be x,

Maximum possible curent in series combination = 0.04 mA

The internal resistance of $M_1=10k\Omega /V\times xV$

The internal resistance of $M_2=25k\Omega /V\times xV$

For series connection total resistance
= (10x + 25x) k $\Omega$

Voltage drop across voltmeter = 280 V

$280=(10x+25x)\times {10}^3\times 0.04\times {10}^{-3}$

$280=35x\times 0.04$

$x=200V$