Question

A series RLC circuit has $R = 10 \ \Omega$, $L = 1\text{ H}$, $C = 100 \ \mu\text{F}$. At resonance, the impedance is:

• A. $0 \ \Omega$
• B. $10 \ \Omega$
• C. $100 \ \Omega$
• D. Depends on frequency

Hint:

In a series RLC circuit at resonance:
1. Impedance is minimum and purely resistive (\(Z = R\)).
2. Current is maximum (\(I = V/R\)).
3. The power factor is unity (\(\cos\phi = 1\)).
You do not need to calculate the resonant frequency to find the impedance; it is always simply equal to \(R\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the impedance of a series RLC circuit under the condition of resonance.

Step 2: Key Formula or Approach:
The total impedance (\(Z\)) of a series RLC circuit is given by:
\[ Z = R + j\left(X_L - X_C\right) = R + j\left(\omega L - \frac{1}{\omega C}\right) \] At resonance, the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)) are equal in magnitude but opposite in phase:
\[ X_L = X_C \implies \omega_0 L = \frac{1}{\omega_0 C} \]

Step 3: Detailed Explanation:

• Write down the expression for the magnitude of the impedance of the series RLC circuit:
\[ |Z| = \sqrt{R^2 + (X_L - X_C)^2} \]
• Apply the resonance condition:
\[ X_L - X_C = 0 \]
• Substitute this condition back into the impedance formula:
\[ |Z| = \sqrt{R^2 + 0^2} = R \]
• This shows that at resonance, the impedance is purely resistive and reaches its minimum value.

• The value of resistance \(R\) is given as \(10\ \Omega\).

• Therefore, the impedance at resonance is exactly \(10\ \Omega\).

Step 4: Final Answer:
The impedance of the circuit at resonance is \(10\ \Omega\).