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

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$\frac{10}{9}$

$0.1$

$1.0$

$10.0$

answer is C.

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Detailed Solution

$V_{a c t u a l} = (\frac{100}{110}) E V_{1} = \frac{10}{11} E$

Equivalent resistance of $100 Ω$ and $900 Ω$ is $90 Ω$ .

$V_{m e a s u r e d} = (\frac{90}{90 + 10}) E V_{2} = \frac{9}{10} E % e r r o r = \frac{V_{1} - V_{2}}{V_{1}} \times 100 = \frac{(\frac{10}{11} E) - (\frac{9}{10}) E}{(\frac{10}{11}) E} = 1 %$

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