question_answer

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

A)  100 k \[\Omega \]                          

B)  75 k \[\Omega \]

C)  50 k \[\Omega \]                            

D)  25 k \[\Omega \]

Correct Answer: C

Solution :

  Potential difference between points A and B \[{{V}_{AB}}=100-\frac{100}{3}=\left( \frac{200}{3} \right)\]volt Current in AB                 \[{{V}_{AB}}=I\times 50\] \[\because \]     \[\frac{200}{3}=I\times 50\] or            \[I=\frac{4}{3}\] Let resultant resistance of voltmeter and BC is R. \[\therefore \]  \[\frac{1}{R}=\frac{1}{50}+\frac{1}{R}=\frac{R+50}{50R}\] \[\Rightarrow \]               \[R=\frac{50R}{50+R}\] By Ohms law      \[V=IR\] \[\therefore \]  \[\frac{100}{3}=\frac{4}{3}\times \frac{50R}{50+R}\] or            \[100+2R=4R\] or            \[R=50k\Omega \]