Question

The current through the solenoid is changing in such way that flux through it is given by $\phi = \epsilon \text{t}$ . 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 $V_{1}$ and $V_{2}$ differ by :

• A. Zero
• B. $ε$
• C. $\left|\frac{\epsilon \left(\left(\text{R}\right)_{1} - \left(\text{R}\right)_{2}\right)}{\left(\text{R}\right)_{1} + \left(\text{R}\right)_{2}}\right|$
• D. $\left|\frac{ε \text{R}_{1} \text{R}_{2}}{\text{R}_{1} + \text{R}_{2}}\right|$

Answer 1

Answer: (C)

$\text{i} = \frac{2 \epsilon _{1} + 2 \epsilon _{2}}{\text{R}_{1} + \text{R}_{2}} = \frac{\epsilon }{\text{R}_{1} + \text{R}_{2}}$
Where $\epsilon = \frac{\text{d} \phi}{\text{dt}}$ is the net emf in the circuit
$\therefore \left(\text{V}\right)_{1} - \left(\text{V}\right)_{2} = \left(\left(\epsilon \right)_{1} - \left(\text{iR}\right)_{1}\right) - \left(\left(\epsilon \right)_{1} - \left(\text{iR}\right)_{2}\right) = \frac{\epsilon \left(\left(\text{R}\right)_{2} - \left(\text{R}\right)_{1}\right)}{\left(\text{R}\right)_{1} + \left(\text{R}\right)_{2}}$