A) \[100k\,\Omega \]
B) \[75k\,\Omega \]
C) \[50k\,\Omega \]
D) \[25k\,\Omega \]
Correct Answer: C
Solution :
Internal resistance of voltmeter is R. Therefore effective resistance across B and C, R is given by \[\frac{1}{R}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50R}\] or \[R=\left( \frac{50R}{50+R} \right)\] According to Ohms law \[V=IR\] or \[\frac{100}{3}=I.\left( \frac{50R}{50+R} \right)\] or \[\frac{100}{3}=\left( \frac{50+R}{50R} \right)=I\] Now, total resistance of circuit \[R=50+\frac{50R}{50+R}\] or \[R=\frac{(2500+100R)}{(50+R)}\] Now, \[V=IR\] \[\Rightarrow \] \[100=\frac{100}{3}\left( \frac{50+R}{50R} \right)\frac{2500+100R}{(50+R)}\] or \[150R=2500+100R\] or \[50R=2500\] or \[R=50k\Omega \]You need to login to perform this action.
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