Question

A full scale deflection current of a meter is 2 mA and its internal resistance is $500\Omega$. The instrument is required to be used as voltmeter with full scale deflection of 200 V. The value of series resistor required should be

• A. $90.4k\Omega$
• B. $99.5k\Omega$
• C. $99.9k\Omega$
• D. $90.4k\Omega$

Answer 1

The correct option is B $99.5k\Omega$

For the meter,
$\text{Full scale deflection current = 2 mA}$

$\text{Internal resistance = 500}\Omega$

The voltage drop across voltmeter at full scale current
= 2 mA $\times$ 500 = 1 V

The full scale deflection on required voltmeter = 200 V

Maximum possible voltage around meter = 1 V

Extra potential drop = 200 - 1 V = 199 V

Comparing voltage drop and resistances
$\frac{199}{1}=\frac{x}{500}$

$\text{x = 500}\times \text{199 = 99.5 k}\Omega$

Alternate solution:

Multiplier resistance,

$R_{sc}=R_m(m-1)$

Where,
$m=\frac{V}{v}=multiplier$

$R_m=meterresistance=500\Omega$

$v=i_m{R}_m=2\times {10}^{-3}\times 500=1V$

$R_m=500(200-1)=99.5k\Omega$