question_answer

A, B and C are voltmeters of resistance R, 1.5R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are \[{{V}_{A}},{{V}_{B}}\] and \[{{V}_{C}}\] respectively.                                                                                                                                                                                                      Then,                   

A)  \[{{V}_{A}}={{V}_{B}}={{V}_{C}}\]          

B)  \[{{V}_{A}}\ne {{V}_{B}}={{V}_{C}}\]

C)         \[{{V}_{A}}={{V}_{B}}\ne {{V}_{C}}\]    

D)         \[{{V}_{A}}\ne {{V}_{B}}\ne {{V}_{C}}\]

Correct Answer: A

Solution :

The equivalent resistance between Q and S is given by \[\frac{1}{R'}=\frac{1}{1.5R}+\frac{1}{3R}=\frac{2+1}{3R}\] \[\Rightarrow \,\,\,\,\,R'=R\] Now,          \[{{V}_{PQ}}={{V}_{A}}+IR\] Also            \[{{V}_{QS}}={{V}_{B}}={{V}_{C}}=IR\] Hence,         \[{{V}_{A}}={{V}_{B}}={{V}_{C}}\]