Question

In an L-C-R circuit, the inductive reactance of a coil is 400 Ω and the capacitive reactance of a condenser is 100 Ω and \( R = 400\ \Omega \). Find the power factor of the circuit.

Hint:

Compute impedance \( Z=\sqrt{R^2+(X_L-X_C)^2} \); power factor \( \cos\phi = R/Z \).

Answer 1

Step 1: Recall what power factor means.
The power factor \(\cos\phi\) is the cosine of the phase angle between the applied voltage and the current, and equals the ratio of resistance to impedance.

Step 2: Find the phase angle directly.
The phase angle in a series L-C-R circuit satisfies
\(\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{400 - 100}{400} = \dfrac{300}{400} = 0.75.\)

Step 3: Build the right triangle.
With opposite side \(= 300\) and adjacent side \(= 400\), the hypotenuse (impedance) is \(\sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = 500\ \Omega\).

Step 4: Compute the cosine.
\(\cos\phi = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{400}{500}.\)

Step 5: Result.
\[\boxed{\cos\phi = 0.8}\]
Equivalently \(\phi \approx 36.9^{\circ}\), with the voltage leading the current because the circuit is net inductive.