A series LCR circuit has an inductor 30 mH and a resistor 1 Ω across an a.c. source of angular frequency 300 rad/s. The capacitance for which the current leads the voltage by 45° is 1/x×10⁻³ F. Find x.

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3. Leading by 45°: tan 45°=(X_C-X_L)/R=1⇒ X_C-X_L=R=1 Ω. X_L=ω L=300×0.03=9 Ω, so X_C=10 Ω. C=1/(ω X_C)=1/(300×10)=1/3×10⁻³ F, so x=3.

A series LCR circuit has an inductor 30 mH and a resistor 1 Ω across an a.c. source of angular frequency 300 rad/s. The capacitance for which the current leads the voltage by 45° is 1/x×10⁻³ F. Find x.

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