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: Write the given data.
Inductive reactance \( X_L = 400\ \Omega \)
Capacitive reactance \( X_C = 100\ \Omega \)
Resistance \( R = 400\ \Omega \)

Step 2: Net reactance.
In a series L-C-R circuit the net reactance is
\[ X = X_L - X_C = 400 - 100 = 300\ \Omega \]

Step 3: Impedance of the circuit.
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(400)^2 + (300)^2} \]
\[ Z = \sqrt{160000 + 90000} = \sqrt{250000} = 500\ \Omega \]

Step 4: Power factor formula.
The power factor is
\[ \cos\phi = \frac{R}{Z} \]

Step 5: Substitute and evaluate.
\[ \cos\phi = \frac{400}{500} = 0.8 \]
\[\boxed{\cos\phi = 0.8}\]
Since \( X_L > X_C \), the circuit is inductive and the current lags the voltage.