In the circuit shown below, the voltmeter is of large resistance. The E.M.F. of the cell is $\mathrm{\epsilon }$. The reading of the voltmeter is

• A

  zero
• B

  $\frac{\mathrm{\epsilon }}{10}$
• Correct answer
  C

  $\frac{\mathrm{\epsilon }}{5}$
• D

  $\frac{\mathrm{\epsilon }}{2}$

Solution

From the figure, it is clear that

${\mathrm{V}}_{\mathrm{A}}\mathrm{ }–\mathrm{ }{\mathrm{V}}_{\mathrm{B}}\mathrm{ }=\frac{\mathrm{\epsilon }}{4+6}\mathrm{ }\times \mathrm{ }4=\frac{4\mathrm{\epsilon }}{10}$

${\mathrm{V}}_{\mathrm{A}}\mathrm{ }–\mathrm{ }{\mathrm{V}}_{\mathrm{D}}\mathrm{ }=\frac{\mathrm{\epsilon }}{6+4}\mathrm{ }\times \mathrm{ }6\mathrm{ }=\frac{6\mathrm{\epsilon }}{10}$

${\mathrm{V}}_{\mathrm{B}}\mathrm{ }–\mathrm{ }{\mathrm{V}}_{\mathrm{D}}=\frac{\mathrm{\epsilon }}{5}$

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