In the circuit shown in the figure, reading of voltmeter is ${\mathrm{V}}_1$ when only ${\mathrm{S}}_1$ is closed, reading of voltmeter is ${\mathrm{V}}_2$ when only ${\mathrm{S}}_2$ is closed and reading of voltmeter is ${\mathrm{V}}_3$ when both ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are closed. Then

• A

  ${\mathrm{V}}_3>{\mathrm{V}}_2>{\mathrm{V}}_1$
• Correct answer
  B

  ${\mathrm{V}}_2>{\mathrm{V}}_1>{\mathrm{V}}_3$
• C

  ${\mathrm{V}}_3>{\mathrm{V}}_1>{\mathrm{V}}_2$
• D

  ${\mathrm{V}}_1>{\mathrm{V}}_2>{\mathrm{V}}_3$

Solution

${\mathrm{V}}_2>{\mathrm{V}}_1>{\mathrm{V}}_3$

When ${\mathrm{S}}_1$ closed



${\mathrm{V}}_1=\frac{3\mathrm{R}}{\mathrm{R}+3\mathrm{R}}\mathrm{E}=\frac{3}{4}\mathrm{E}$

When ${\mathrm{S}}_2$ is closed



${\mathrm{V}}_2=\frac{6\mathrm{R}}{\mathrm{R}+6\mathrm{R}}\mathrm{E}=\frac{6}{7}\mathrm{E}$

When both ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are closed



${\mathrm{V}}_3=\frac{2\mathrm{R}}{\mathrm{R}+2\mathrm{R}}\mathrm{E}=\frac{2\mathrm{E}}{3}$

$\therefore \mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{V}}_2>{\mathrm{V}}_1>{\mathrm{V}}_3$

Hence, the correct answer is option: 'B'.