question_answer

The potential difference across the 100W resistance in the following circuit is measured by a voltmeter of 900 W resistance. The percentage error made in reading the potential difference is                                      [AMU (Med.) 2002]

A)         \[\frac{10}{9}\]

B)         0.1

C)         1.0

D)         10.0

Correct Answer: C

Solution :

                                         Before connecting the voltmeter, potential difference across \[100\Omega \] resistance \[{{V}_{i}}=\frac{100}{(100+10)}\times V=\frac{10}{11}V\] Finally after connecting voltmeter across \[100\Omega \] Equivalent resistance  \[\frac{100\times 900}{(100+900)}=90\Omega \] Final potential difference  \[{{V}_{f}}=\frac{90}{(90+10)}\times V=\frac{9}{10}V\] % error = \[\frac{{{V}_{i}}-{{V}_{f}}}{{{V}_{i}}}\times 100\] \[=\frac{\frac{10}{11}V-\frac{9}{10}V}{\frac{10}{11}V}\times 100=1.0.\]