Current Electricity Charging Discharging Of Capacitors Question 289
Question: 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]
Options:
A) $ \frac{10}{9} $
B) 0.1
C) 1.0
D) 10.0
Show Answer
Answer:
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. $
BETA
