Question

The potential difference across the $100\Omega$ resistance in the following circuit is measured by a voltmeter of $900\Omega$ resistance. the percentage error made in reading the potential difference is

• A. $\frac{10}{9}$
• B. $0.1$
• C. $1.0$
• D. $10.0$

Answer 1

The correct option is C $1.0$

Before connecting volt meter, across $100\Omega$ resistance

$V_1=\frac{100}{(100+10)}V=\frac{10}{11}V$

Finally,

Finally after connecting voltmeter across $100\Omega$ equivalent resistance

$\frac{900\times 100}{900+100}=\frac{900\times 100}{1000}=90\Omega$

$V_f=\frac{90}{90+10}\times V=\frac{9}{10}V$ error

percentage $=\frac{V_1-V_f}{V_1}\times 100=\frac{\frac{10}{11}V-\frac{9}{10}V}{\frac{10}{11}V}\times 100=1.0$