question_answer

In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B and C reads \[\frac{100}{3}\]V. Neglecting the internal resistance   of   the battery, the value of R

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 \[R=50k\Omega \]