Question

An inductor of 10 mH, capacitor of 0.1 µF and a resistor of 100 Ω 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 \(\sqrt{3} a\) Ω. The value of \(a\) is ______.

Answer 1

Correct Answer: 100

Step 1: Understanding the Concept:
In a series RLC circuit, the power factor is the cosine of the phase angle between the voltage and the current. It depends directly on the resistance and the total impedance of the circuit.
Step 2: Key Formula or Approach:
Power factor \(\cos \phi = \frac{R}{Z}\).
The impedance of the series circuit is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
We are given \(\cos \phi = 0.5\) and \(R = 100 \, \Omega\), allowing us to solve for the reactance difference \(|X_L - X_C|\).
Step 3: Detailed Explanation:
Use the power factor to find the total impedance \(Z\):
\(\cos \phi = 0.5 = \frac{1}{2}\)
\(\frac{R}{Z} = \frac{1}{2} \implies Z = 2R\).
Since \(R = 100 \, \Omega\):
\(Z = 2(100) = 200 \, \Omega\).
Substitute \(Z\) and \(R\) into the impedance formula:
\(Z^2 = R^2 + (X_L - X_C)^2\)
\((200)^2 = (100)^2 + (X_L - X_C)^2\)
\(40000 = 10000 + (X_L - X_C)^2\)
\((X_L - X_C)^2 = 30000\)
Take the square root to find the magnitude of the reactance difference:
\(|X_L - X_C| = \sqrt{30000} = \sqrt{10000 \times 3} = 100\sqrt{3} \, \Omega\).
The problem states that this difference is equal to \(\sqrt{3} a \, \Omega\).
Equating the two expressions:
\(\sqrt{3} a = 100\sqrt{3}\).
Step 4: Final Answer:
Dividing both sides by \(\sqrt{3}\), we find that \(a = 100\).