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 \[R=50k\Omega \]You need to login to perform this action.
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