Question

A resistance-meter has five measurement range-settings between $200\Omega \text{and}2M\Omega$ in multiples of 10. The meter measures resistance of a device by measuring the full-range voltage of 2 V across the device by passing an appropriate constant current for each range-setting. If a device having a resistance value in the range $8k\Omega to12k\Omega$ and a maximum power rating of $100\mu W$ is to be measured safely with this meter, the choice for range-setting on the meter for best resolution in measurement, in $k\Omega ,$ is

• A. 2
• B. 20
• C. 200
• D. 2000

Answer 1

The correct option is C 200


Resistance to be measured
$\Rightarrow 8k\Omega to12k\Omega$

$So,minimumrangeshouldbe20k\Omega .$

$With20k\Omega range:$

$I=\frac{2V}{20k\Omega }=\frac{1}{10}mA$

$P_{max}=I^2{R}_{max}$

$={(\frac{1}{10}\times {10}^{-3})}^2\times 12\times {10}^{3}W$

$=120\mu W>100\mu W$

$With200k\Omega range:$

$I=\frac{2V}{200k\Omega }=\frac{1}{100}mA$

$P_{max}=I^2{R}_{max}$

$={(\frac{1}{100}\times {10}^{-3})}^2\times 12\times {10}^{3}W$

$=1.2\mu W$