The current through the solenoid is changing in such way that flux through it is given by $\varphi =\epsilon \text{t}$. The solenoid is surrounded by a loop having resistance ${\text{R}}_{\text{1}}$ and ${\text{R}}_{\text{2}}$ as shown. Then the reading of the two voltmeters ${\mathrm{V}}_1$ and ${\mathrm{V}}_2$ differ by :

• A

  Zero
• B

  $\epsilon$
• Correct answer
  C

  $\left|\frac{\epsilon \left({\text{R}}_1-{\text{R}}_2\right)}{{\text{R}}_1+{\text{R}}_2}\right|$
• D

  $\left|\frac{\epsilon {\text{R}}_1{\text{R}}_2}{{\text{R}}_1+{\text{R}}_2}\right|$

Solution

$\text{i}=\frac{2{\epsilon }_1+2{\epsilon }_2}{{\text{R}}_1+{\text{R}}_2}=\frac{\epsilon }{{\text{R}}_1+{\text{R}}_2}$

Where $\epsilon =\frac{\text{d}\varphi }{\text{dt}}$ is the net emf in the circuit

$\therefore {\text{V}}_1-{\text{V}}_2=\left({\epsilon }_1-{\text{iR}}_1\right)-\left({\epsilon }_1-{\text{iR}}_2\right)=\frac{\epsilon \left({\text{R}}_2-{\text{R}}_1\right)}{{\text{R}}_1+{\text{R}}_2}$

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