Question
$$A,B$$ and $$C$$ are voltmeters of resistance $$R,$$ $$1.5\,R$$ and $$3\,R$$ 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,
$$A,B$$ and $$C$$ are voltmeters of resistance $$R,$$ $$1.5\,R$$ and $$3\,R$$ 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.
$${V_A} \ne {V_B} \ne {V_C}$$
Answer :
$${V_A} = {V_B} = {V_C}$$
Solution :
The equivalent resistance between $$Q$$ and $$S$$ is given by
$$\eqalign{ & \frac{1}{{R'}} = \frac{1}{{1.5R}} + \frac{1}{{3R}} = \frac{{2 + 1}}{{3R}} \cr & \Rightarrow R' = R \cr} $$
Now, $${V_{PQ}} = {V_A} = IR$$
Also, $${V_{QS}} = {V_B} = {V_C} = IR$$
Hence, $${V_A} = {V_B} = {V_C}$$
The equivalent resistance between $$Q$$ and $$S$$ is given by
$$\eqalign{ & \frac{1}{{R'}} = \frac{1}{{1.5R}} + \frac{1}{{3R}} = \frac{{2 + 1}}{{3R}} \cr & \Rightarrow R' = R \cr} $$
Now, $${V_{PQ}} = {V_A} = IR$$
Also, $${V_{QS}} = {V_B} = {V_C} = IR$$
Hence, $${V_A} = {V_B} = {V_C}$$
