The potential difference across the 100,Omega | vedclass

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(C) Before connecting the voltmeter,the potential difference across the $100\,\Omega$ resistance is given by the voltage divider rule:$V_i = \frac{100

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(C) Before connecting the voltmeter,the potential difference across the $100\,\Omega$ resistance is given by the voltage divider rule:$V_i = \frac{100}{100 + 10} 	imes V = \frac{10}{11}V$After connecting the voltmeter of $900\,\Omega$ in parallel with the $100\,\Omega$ resistance,the equivalent resistance of the parallel combination is:$R_p = \frac{100 	imes 900}{100 + 900} = \frac{90000}{1000} = 90\,\Omega$The final potential difference measured by the voltmeter is:$V_f = \frac{90}{90 + 10} 	imes V = \frac{90}{100}V = 0.9V = \frac{9}{10}V$The percentage error in the measurement is calculated as:$	ext{Percentage Error} = \frac{V_i - V_f}{V_i} 	imes 100$$= \frac{\frac{10}{11}V - \frac{9}{10}V}{\frac{10}{11}V} 	imes 100$$= \frac{\frac{100 - 99}{110}}{\frac{10}{11}} 	imes 100 = \frac{1/110}{10/11} 	imes 100 = \frac{1}{110} 	imes \frac{11}{10} 	imes 100 = \frac{1}{100} 	imes 100 = 1\%$

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

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