The figure shows an LCR series circuit with two switches S₁ and S₂, a capacitor C=100 μF, a resistor R and two inductors L₁, L₂, driven by v=V₀ sin(300t). When S₁ is closed keeping S₂ open the phase difference between the current and the source voltage is 30°; it is 60° when S₂ is closed keeping S₁…

Question

The figure shows an LCR series circuit with two switches S₁ and S₂, a capacitor C=100 μF, a resistor R and two inductors L₁, L₂, driven by v=V₀ sin(300t). When S₁ is closed keeping S₂ open the phase difference between the current and the source voltage is 30°; it is 60° when S₂ is closed keeping S₁ open. The value of (3L₁-L₂) is H.

Accepted Answer

(2) 2/9. Angular frequency ω=300 rad/s, so X_C=1/(ω C)=1/(300×10⁻⁴)=(100)/3 Ω. For each switch position the tangent of the phase angle is tan φ=(X_L-X_C)/R, with X_L=ω L. Case S₁: tan 30°=(ω L₁-X_C)/R; Case S₂: tan 60°=(ω L₂-X_C)/R. These two relations involve three unknowns (R,L₁,L₂); the value of R is not legible in the scan, so (3L₁-L₂) cannot be closed independently here. Printed key: option (2), 2/9. Flagged in DEFECTS.md pending a clean copy of the figure (the R value).

Step-by-step solution

Why the other options are wrong