An inductor of 10 mH, capacitor of 0.1 $$\mu$$F and a resistor of 100 $$\Omega$$ are connected in series across an a.c power supply 220 V, 70 Hz. The power factor of the given circuit is 0.5. The difference in the inductive reactance and capacitance reactance is $$\sqrt{3}\,\alpha$$ $$\Omega$$. The value of $$\alpha$$ is _____.

For a series $$R$$-$$L$$-$$C$$ circuit driven by an a.c. source of angular frequency $$\omega$$, the impedance is $$Z=\sqrt{R^{2}+\left(X_L-X_C\right)^{2}}$$, where the reactances are
  • Inductive: $$X_L=\omega L$$
  • Capacitive: $$X_C=\dfrac{1}{\omega C}$$.

The power factor is defined as the cosine of the phase angle $$\phi$$ between voltage and current:
$$\cos\phi=\dfrac{R}{Z}\qquad -(1)$$

The tangent of the same angle is therefore
$$\tan\phi=\dfrac{X_L-X_C}{R}\qquad -(2)$$

The problem states that the power factor is $$0.5$$, so
$$\cos\phi=0.5\;\;\Longrightarrow\;\;\phi=60^{\circ}$$ (since $$\cos60^{\circ}=0.5$$).

Hence
$$\tan\phi=\tan60^{\circ}=\sqrt{3}$$.

Substituting $$\tan\phi$$ and the given resistance $$R=100\,\Omega$$ into equation $$(2)$$ gives
$$X_L-X_C = R\,\tan\phi = 100 \times \sqrt{3} = 100\sqrt{3}\;\Omega\qquad -(3)$$

The statement of the question further writes this difference as
$$X_L-X_C = \sqrt{3}\,\alpha\;\Omega\qquad -(4)$$

Comparing $$(3)$$ and $$(4)$$:
$$\sqrt{3}\,\alpha = 100\sqrt{3}\;\Longrightarrow\;\alpha = 100$$.

Therefore, the required value of $$\alpha$$ is 100.