Question

A voltmeter has a range of 50V and has an internal resistance of ${10}^6$ ohm. To convert it into a voltmeter to read up to 100V, the resistance to be connected is :

• A. $2\times {10}^6\Omega$
• B. ${10}^6\Omega$
• C. $3\times {10}^6\Omega$
• D. $0.5\times {10}^6\Omega$

Answer 1

Answer: (B)

${10}^6\Omega$

Sensitivity of Voltmeter = $\frac{\text{internal resistance of voltmeter}}{\text{full scale voltage reading}}$ = $\frac{{10}^6\Omega }{50V}$= $2\times 10{}^4\Omega V^{-1}$
This means for measuring 1V, the resistance should be $2\times 10{}^4\Omega$
Therefore, for measuring 100V, the resistance should be $2\times 10{}^6\Omega$
The voltmeter already has a resistance of $10{}^6\Omega$
Therefore, the resistance to be connected is $(2\times 10{}^6\Omega )-10{}^6\Omega$ =$10{}^6\Omega$.