A, B and C are voltmeters of resistance R, 1.5 R 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. [NEET 2015] |
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 :
[a] 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, \[{{V}_{A}}={{V}_{B}}={{V}_{C}}\] |
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