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 = \[\frac{{{V}_{i}}-{{V}_{f}}}{{{V}_{i}}}\times 100\] \[=\frac{\frac{10}{11}V-\frac{9}{10}V}{\frac{10}{11}V}\times 100=1.0.\]You need to login to perform this action.
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