Question 184: AC fundamentals

Question

An inductor of 10 mH , capacitor of $0.1 \mu \mathrm{~F}$ and a resistor of $100 \Omega$ are connected in series across an $a . c$ power supply $220 \mathrm{~V}, 70 \mathrm{~Hz}$. 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 $\_\_\_\_$ .

Accepted Answer

For a series $RLC$ circuit, the power factor is
$$ \cos \phi = \frac{R}{Z} $$
Given:
$R = 100\,\Omega$
power factor $= 0.5$
So,
$$ 0.5 = \frac{100}{Z} $$
Hence,
$$ Z = \frac{100}{0.5} = 200\,\Omega $$
Now for a series $RLC$ circuit,
$$ Z = \sqrt{R^2 + (X_L - X_C)^2} $$
Substitute the values:
$$ 200 = \sqrt{100^2 + (X_L - X_C)^2} $$
Squaring both sides,
$$ 200^2 = 100^2 + (X_L - X_C)^2 $$
$$ 40000 = 10000 + (X_L - X_C)^2 $$
$$ (X_L - X_C)^2 = 30000 $$
$$ X_L - X_C = 100\sqrt{3}\,\Omega $$
Given that the difference is $\sqrt{3}\alpha \,\Omega$, we compare:
$$ \sqrt{3}\alpha = 100\sqrt{3} $$
Therefore,
$$ \alpha = 100 $$
So, the required value is
$$ \boxed{100} $$

AC fundamentals

Integer / Numeric Answer Type